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RADIOACTIVITY


A: INTRODUCTION / CAUSES OF RADIOCTIVITY

Radioactivity is the spontaneous disintegration/decay of an unstable nuclide.

A nuclide is an atom with defined mass number (number of protons and neutrons), atomic number and definite energy.

Radioactivity takes place in the nucleus of an atom unlike chemical reactions that take place in the energy levels involving electrons.

A nuclide is said to be stable if its neutron: proton ratio is equal to one (n/p = 1)

All nuclide therefore try to attain n/p = 1 by undergoing radioactivity.

Examples

(i)Oxygen nuclide with 168 O has 8 neutrons and 8 protons in the nucleus therefore an n/p = 1 thus stable and do not decay/disintegrate.

(ii)Chlorine nuclide with 3517 Cl has 18 neutrons and 17 protons in the nucleus therefore an n/p = 1.0588 thus unstable and decays/disintegrates to try to attain n/p = 1.

(ii)Uranium nuclide with 23792 U has 206 neutrons and 92 protons in the nucleus therefore an n/p = 2.2391 thus more unstable than 23592 U and thus more readily decays / disintegrates to try to attain n/p = 1.

(iii) Chlorine nuclide with 3717 Cl has 20 neutrons and 17 protons in the nucleus therefore an n/p = 1.1765 thus more unstable than 3517 Cl and thus more readily decays / disintegrates to try to attain n/p = 1.

(iv)Uranium nuclide with 23592 U has 143 neutrons and 92 protons in the nucleus therefore an n/p = 1.5543 thus more stable than 237 92U but also readily decays / disintegrates to try to attain n/p = 1.

All unstable nuclides naturally try to attain nuclear stability with the production of:

(i)alpha(α) particle decay

The alpha (α) particle has the following main characteristic:

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i)is positively charged(like protons)

ii) has mass number 4 and atomic number 2 therefore equal to a charged Helium atom ( 42He2+)

iii) have very low penetrating power and thus can be stopped /blocked/shielded by a thin sheet of paper.

iv) have high ionizing power thus cause a lot of damage to living cells.

v) a nuclide undergoing α-decay has its mass number reduced by 4 and its atomic number reduced by 2

Examples of alpha decay

210
84 Pb ->  x
82 Pb  +   42He 2+

210
84 Pb ->  206
82 Pb  +   42He 2+

 

226
88 Ra ->  222
y Rn  +   42He 2+

226
88 Ra ->  222
86 Rn  +   42He 2+


x
y U    ->  23490 Th  +   42He 2+

238
92 U    ->  23490 Th  +   42He 2+

x
y U   ->  23088 Ra  +  2 42He 2+

238
92 U ->  23088 Ra  +  2 42He 2+

  210
84 U    ->  xy W + 10 α

210
84 U    ->  17064 W  + 10 α

210
92U    ->  xy W + 6 α

210
92U    ->  18680W  + 6 α

(ii)Beta (β) particle decay

The Beta (β) particle has the following main characteristic:

i)is negatively charged(like electrons)

ii)has no mass number and atomic number negative one(-1) therefore equal to a fast moving electron (0
-1e
)

iii) have medium penetrating power and thus can be stopped /blocked/shielded by a thin sheet of aluminium foil.

iv) have medium ionizing power thus cause less damage to living cells than the α particle.

v) a nuclide undergoing β -decay has its mass number remain the same and its atomic number increase by 1

Examples of beta (β) decay

1.23 x Na ->   2312Mg +  0
-1e

23 11 Na ->   2312Mg +  0
-1e


2. 234 x Th ->   y91 Pa +  0
-1e


234
90 Th ->   y91 Pa +  0
-1e

3. 20770Y ->   x y Pb +  30
-1e


207
70Y ->   207
73Pb +  30
-1e

4.
x y
C ->   147N +  0
-1e



14 6
C ->   147N +  0
-1e

5. 1 x
n ->   y1H +  0
-1e

1 0
n ->   11H +  0
-1e

6. 42He ->   411H +   x
0
-1e



42He    ->  411H +   2 0
-1e

7. 22888Ra    ->  22890Th +   x β


228
88Ra    ->   22892Th +   4 β

8. 23290Th     ->  21282Pb +  2 β  +   x α

  23290Th   ->   21282Pb +  2 β  +   5 α

9. 23892U ->  22688 Ra +  x β   +  3 α


23892U    ->   22688 Ra +  2 β   +  3 α

10. 21884Po   ->  20682Pb +  x β   +  3 α



218
84Po  ->  20682Pb +  4β +  3 α

(iii)Gamma (y) particle decay

The gamma (y) particle has the following main characteristic:

i)is neither negatively charged(like electrons/beta) nor positively charged(like protons/alpha) therefore neutral.

ii)has no mass number and atomic number therefore equal to electromagnetic waves.

iii) have very high penetrating power and thus can be stopped /blocked/shielded by a thick block of lead..

iv) have very low ionizing power thus cause less damage to living cells unless on prolonged exposure..

v) a nuclide undergoing y -decay has its mass number
and its atomic number remain the same.

Examples of gamma (y) decay

  • 3717Cl   ->   3717Cl + y
  • 146C   ->   146C + y

The sketch diagram below shows the penetrating power of the radiations from a radioactive nuclide.

radioactive nuclide sheet of paper aluminium foil thick block of lead

(radiation source) (block α-rays) (block β-rays)   block y-rays)

 

 


 α-rays   β-rays   y-rays

 

The sketch diagram below illustrates the effect of electric /magnetic field on the three radiations from a radioactive nuclide

 

 

 

Image From EcoleBooks.com

Radioactive disintegration/decay naturally produces the stable 20682Pb nuclide /isotope of lead.Below is the 238
92 U natural decay series. Identify the particle emitted in each case

Image From EcoleBooks.com

Write the nuclear equation for the disintegration from :

(i)238
92 U to 23490 T


238

92 U  ->   23490 T  +  
4 2 He 2+


238

92 U  ->   23490 T  +   α

(ii)238
92 U to 222 84 Rn


238

92 U  ->   22284 Rn  +   4 4 2 He 2+


238

92 U  ->   22284 Rn  +  

230
90 Th undergoes alpha decay to
222
86 Rn. Find the number of
α particles emitted. Write the nuclear equation for the disintegration.

Working


230

90 Th  ->   222
86 Rn  +   x 4 2 He

Method 1

Using mass numbers

 230  = 222 + 4 x => 4 x = 230 – 222 = 8

x = 8 / 4 = 2 α

Using atomic numbers

 90   = 86 + 2 x => 2 x = 90 – 86 = 4

  x = 4 / 2 = 2 α

Nuclear equation

230
90 Th  ->   222
86 Rn  +   2 4 2 He

214
82 Pb undergoes beta decay to
214
84 Rn. Find the number of
β particles emitted. Write the nuclear equation for the disintegration.

Working

214
82 Pb  ->   214
84 Rn  +   x 0 -1 e

Using atomic numbers only

  82   = 84 – x => -x = 82 – 84 = -2

  x = 2 β

Nuclear equation

214
82 Pb  ->   214
84 Rn  +   2 0 -1 e

238
92 U undergoes beta and alpha decay to
206
82 Pb. Find the number of
β and
α
particles emitted. Write the nuclear equation for the disintegration.

Working

238
92 U  ->   206
82 Pb  +
x 0 -1 e +   y 4 2 He

Using Mass numbers only

  238   = 206 + 4y => 4y = 238 – 206 = 32

  y =  32  = 8 α

4

Using atomic numbers only and substituting the 8 α(above)

238
92 U  ->   206
82 Pb   +   8 4 2 He +
x 0 -1 e

  92   =   82 +   16 +
– x

  => 92 – (82 + 16) = – x

x = 6 β

Nuclear equation

238
92 U  ->   206
82 Pb  +
6 0 -1 e +   8 4 2 He

298
92 U undergoes alpha and beta decay to
214
83 Bi. Find the number of
α and β particles emitted. Write the nuclear equation for the disintegration.

Working

298
92 U   ->   210
83 Bi +   x 4 2 He +
y 0 -1 e

Using Mass numbers only

298 = 214 + 4x => 4x = 298 – 214 = 84

y =  84  = 21 α

  4

Using atomic numbers only and substituting the 21 α (above)

238
92 U -> 214
83Bi   +
21 4 2 He +
y 0 -1 e

  92 = 83 +   42 +
– y

  => 92 – (83 + 42) = – x

x = 33 β

Nuclear equation

 298
92 U -> 210
83 Bi +
21 4 2 He +
33 0 -1 e

 

B:NUCLEAR FISSION AND NUCLEAR FUSION

Radioactive disintegration/decay can be initiated in an industrial laboratory through two chemical methods:

a) nuclear fission

b) nuclear fusion.

a)Nuclear fission

Nuclear fission is the process which a fast moving neutron bombards /hits /knocks a heavy unstable nuclide releasing lighter nuclide, three daughter neutrons and a large quantity of energy.

Nuclear fission is the basic chemistry behind nuclear bombs made in the nuclear reactors.

The three daughter neutrons becomes again fast moving neutron bombarding / hitting /knocking a heavy unstable nuclide releasing lighter nuclides, three more daughter neutrons each and a larger quantity of energy setting of a chain reaction

Examples of nuclear equations showing nuclear fission

10
n + 235
b U -> 9038
Sr + c
54Xe + 310
n + a

 

10
n + 2713
Al  ->   2813
Al +  y + a

 

10
n + 28a
Al  ->   b11
Na  + 42
He

 

a0 n + 147
N  ->   14b
C +   11
H

 

10 n + 11
H

 ->   21
H  
  + a  

 

10 n + 235
92 U
->
95
42 Mo + 139
57 La  + 210
n
+ 7 a

b) Nuclear fusion

Nuclear fusion is the process which smaller nuclides join together to form larger / heavier nuclides and releasing a large quantity of energy.

Very high temperatures and pressure is required to overcome the repulsion between the atoms.

Nuclear fusion is the basic chemistry behind solar/sun radiation.

Two daughter atoms/nuclides of Hydrogen fuse/join to form Helium atom/nuclide on the surface of the sun releasing large quantity of energy in form of heat and light.

21H + 21H -> abHe + 10
n

 

21H + a -> 32He

 

21H + 21H -> a + 11
H

 

4 11H -> 42He + a

 


147H + a -> 178O + 11
H

C: HALF LIFE PERIOD (t1/2)

The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount.

It is usually denoted t 1/2.

The rate of radioactive nuclide disintegration/decay is constant for each nuclide.

The table below shows the half-life period of some elements.

Element/Nuclide

Half-life period(t 1/2 )

238
92 U

4.5 x 109 years

14
6 C

5600 years

229
88 Ra

1620 years

35
15 P

14 days

210
84 Po

0.0002 seconds

 

The less the half life the more unstable the nuclide /element.

The half-life period is determined by using a Geiger-Muller counter (GM tube)

.A GM tube is connected to ratemeter that records the count-rates per unit time.

This is the rate of decay/ disintegration of the nuclide.

If the count-rates per unit time fall by half, then the time taken for this fall is the half-life period.

Examples

a)A radioactive substance gave a count of 240 counts per minute but after 6 hours the count rate were 30 counts per minute. Calculate the half-life period of the substance.

If t 1/2 = x

then 240 –x–>120 –x–>60 –x—>30

From 240 to 30 =3x =6 hours

=>x = t 1/2 = ( 6 / 3 )

= 2 hours

b) The count rate of a nuclide fell from 200 counts per second to 12.5 counts per second in 120 minutes.

Calculate the half-life period of the nuclide.

If t 1/2 =x

 then

 200 –x–>100 –x–>50 –x—>25 –x—>12.5

 From 200 to 12.5 =4x =120 minutes

  =>x = t 1/2 = ( 120 / 4 )

 = 30 minutes

c) After 6 hours the count rate of a nuclide fell from 240 counts per second to 15 counts per second on the GM tube. Calculate the half-life period of the nuclide.

 If t 1/2 = x

 then 240 –x–>120 –x–>60 –x—>30 –x—>15

 From 240 to 15 =4x =6 hours

  =>x = t 1/2 = ( 6 / 4 )= 1.5 hours

d) Calculate the mass of nitrogen-13 that remain from 2 grams after 6 half-lifes if the half-life period of nitrogen-13 is 10 minutes.

 If t 1/2 = x then:

2 —x–>1 –2x–>0.5 –3x—>0.25 –4x–>0.125–5x—>0.0625–6x—>0.03125

After the 6th half life 0.03125 g of nitrogen-13 remain.

e) What fraction of a gas remains after 1hour if its half-life period is 20 minutes?

 If t 1/2 = x then:

 then 60 /20 = 3x

 1 –x–> 1/2 –2x–> 1/4 –3x—> 1/8

 After the 3rd half-life 1/8 of the gas remain

f) 348 grams of a nuclide A was reduced to 43.5 grams after 270days.Determine the half-life period of the nuclide.

 If t 1/2 = x then:

 348 –x–>174 –2x–>87 –3x—>43.5

  From 348 to 43.5=3x =270days

  =>x = t 1/2 = ( 270 / 3 )

= 90 days

 

g) How old is an Egyptian Pharaoh in a tomb with 2grams of 14C if the normal 14C in a present tomb is 16grams.The half-life period of 14C is 5600years.

If t 1/2 = x = 5600 years then:

   16 –x–>8 –2x–>4 –3x—>2

3x = ( 3 x 5600 )

= 16800years

h) 100 grams of a radioactive isotope was reduced 12.5 grams after 81days.Determine the half-life period of the isotope.

If t 1/2 = x then:

100 –x–>50 –2x–>25 –3x—>12.5

From 100 to 12.5=3x =81days

=>x = t 1/2

= ( 81 / 3 )

= 27 days

A graph of activity against time is called decay curve.

A decay curve can be used to determine the half-life period of an isotope since activity decrease at equal time interval to half the original

Image From EcoleBooks.com

(i)From the graph show and determine the half-life period of the isotope.

From the graph t 1/2 changes in activity from:

( 100 – 50 ) => ( 20 – 0 ) = 20 minutes

( 50 – 25 ) => ( 40 – 20 ) = 20 minutes

Thus t ½
= 20 minutes

(ii)Why does the graph tend to ‘O’?

Smaller particle/s will disintegrate /decay to half its original.

There can never be ‘O’/zero particles

 

D: CHEMICAL vs NUCLEAR REACTIONS

Nuclear and chemical reaction has the following similarities:

(i)-both involve the subatomic particles; electrons, protons and neutrons in an atom

(ii)-both involve the subatomic particles trying to make the atom more stable.

 (iii)-Some for of energy transfer/release/absorb from/to the environment take place.

 

Nuclear and chemical reaction has the following differences:

(i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.

Chemical reactions mainly involve outer electrons in the energy levels an atom.

(ii) Nuclear reactions form a new element.  

Chemical reactions do not form new elements

(iii) Nuclear reactions mainly involve evolution/production of large quantity of heat/energy.

Chemical reactions produce or absorb small quantity of heat/energy.

(iv)Nuclear reactions are accompanied by a loss in mass/mass defect.Do not obey the law of conservation of matter.

Chemical reactions are not accompanied by a loss in mass/ mass defect hence obey the law of conservation of matter.

(v)The rate of decay/ disintegration of the nuclide is independent of physical conditions (temperature/pressure /purityp/article size)

The rate of a chemical reaction is dependent on physical conditions (temperature/pressure/purity/particle size/ surface area)

E: APPLICATION AND USES OF RADIOCTIVITY.

The following are some of the fields that apply and use radioisotopes;

a)Medicine: –Treatment of cancer to kill malignant tumors through radiotherapy.

Sterilizing hospital /surgical instruments /equipments by exposing them to gamma radiation.

b) Agriculture:


If a plant or animal is fed with radioisotope, the metabolic processes of the plant/animal is better understood by tracing the route of the radioisotope.

c) Food preservation:


X-rays are used to kill bacteria in tinned food to last for a long time.

d) Chemistry:


To study mechanisms of a chemical reaction, one reactant is replaced in its structure by a radioisotope e.g.

During esterification the ‘O’ joining the ester was discovered comes from the alkanol and not alkanoic acid.

During photosynthesis the ‘O’ released was discovered comes from water.

e) Dating rocks/fossils:

The quantity of 14C in living things (plants/animals) is constant.

When they die the fixed mass of 14C is trapped in the cells and continues to decay/disintegrate.

The half-life period of 14C is 5600 years .

Comparing the mass of 14C in living and dead cells, the age of the dead can be determined.

F: DANGERS OF RADIOCTIVITY.

All rays emitted by radioactive isotopes have ionizing effect of changing the genetic make up of living cells.

Exposure to theses radiations causes chromosomal and /or genetic mutation in living cells.

Living things should therefore not be exposed for a long time to radioactive substances.

One of the main uses of radioactive isotopes is in generation of large cheap electricity in nuclear reactors.

Those who work in these reactors must wear protective devises made of thick glass or lead sheet.

Accidental leakages of radiations usually occur

In 1986 the Nuclear reactor at Chernobyl in Russia had a major explosion that emitted poisonous nuclear material that caused immediate environmental disaster

In 2011, an earthquake in Japan caused a nuclear reactor to leak and release poisonous radioactive waste into the Indian Ocean.

The immediate and long term effects of exposure to these poisonous radioactive waste on human being is of major concern to all environmentalists.

 

 

 

 

 

 

 

 

 

 

 

 

 

G: SAMPLE REVISION QUESTIONS

The figure below shows the behaviour of emissions by a radioactive isotope x. Use it to answer the question follow


Image From EcoleBooks.com

(a) Explain why isotope X emits radiations. (1mk)

 -is unstable //has n/p ratio greater/less than one

(b) Name the radiation labeled T  (1mk)

alpha particle  

(c) Arrange the radiations labeled P and T in the increasing order of ability to be deflected by an electric filed.  (1mk)

T -> P

a) Calculate the mass and atomic numbers of element B formed after 21280
X has emitted three beta particles, one gamma ray and two alpha particles.

Mass number

  = 212 – (0 beta+ o gamma + (2 x 4 ) alpha = 204

Atomic number

 = 80 – (-1 x3) beta + 0 gamma + (2 x 2 )) alpha =79

b)Write a balanced nuclear equations for the decay of 21280
X to B using the information in (a) above.


212
80
X -> 20479B + 242He + 3 0-1
e + y

Identify the type of radiation emitted from the following nuclear equations.

(i) 146
C    -> 147N   + ………

 β – Beta

  1. 11
    H + 10
    n -> 21H   + ……

y -gamma


(iii) 23592
U   -> 9542Mo + 13957La + 10
n +……

7 β – seven beta particles

  1. 23892
    U   -> 23490Th   + … …

α-alpha

  1. 146
    C   + 11
    H -> 157N   + ……

y-gamma

X grams of a radioactive isotope takes 100 days to disintegrate to 20 grams. If the half-life period isotope is 25 days, calculate the initial mass X of the radio isotope.

Number of half-lifes = ( 100 / 25 ) = 4

20g —–> 40g —-> 80g—–> 160g —–> 320g

Original mass X = 320g

Radium has a half-life of 1620 years.

(i)What is half-life?

The half-life period is the time taken for a radioactive nuclide to spontaneously decay/ disintegrate to half its original mass/ amount

b)If one milligram of radium contains 2.68 x 10 18 atoms ,how many atoms disintegrate during 3240 years.

Number of half-lifes = ( 3240 / 1620 ) = 2

1 mg —1620—> 0.5mg —1620—-> 0.25mg

    If 1mg  ->  2.68 x 1018 atoms

Then 0.25 mg -> ( 0.25 x 2.68 x 1018 ) = 6.7 x 1017

Number of atoms remaining = 6.7 x 1017


Number of atoms disintegrated =

(2.68 x 1018 – 6.7 x 1017 )

= 2.01 x 1018

The graph below shows the mass of a radioactive isotope plotted against time

Image From EcoleBooks.com

Using the graph, determine the half – life of the isotope

From graph 10 g to 5 g takes 8 days

From graph 5 g to 2.5 g takes 16 – 8 = 8 days

Calculate the mass of the isotope dacayed after 32 days

Number of half lifes= 32/8 = 4

Original mass = 10g

10g—1st –>5g—2nd–>2.5—3rd –>1.25—4th –>0.625 g

Mass remaining = 0.625 g

Mass decayed after 32 days = 10g – 0.625 g = 9.375g

A radioactive isotope X2 decays by emitting two alpha (a) particles and one beta (β) to form 214
83Bi  

(a)Write the nuclear equation for the radioactive decay


21286
X -> 214
83Bi + 242He + 0-1
e


(b)What is the atomic number of X2?

86

(c) After 112 days, 1/16 of the mass of X2 remained. Determine the half life of X2


1—x-> 1 /2 –x-> 1 /4 –x-> 1 /8–x-> 1 /16

Number of t 1 /2 in 112 days = 4

 t 1 /2 = 112 = 28 days

    4

1.Study the nuclear reaction given below and answer the questions that follow.

126 C –step 1–>127 N –step 2–> 1211Na

(a)126 C and 146 C are isotopes. What does the term isotope mean?

Atoms of the same element with different mass number /number of neutrons.

(b)Write an equation for the nuclear reaction in step II

 
12
7 N  ->  1211Na   +   0 -1e

(c)Give one use of 146 C

 Dating rocks/fossils:

 Study of metabolic pathways/mechanisms on plants/animals

Study the graph of a radioactive decay series for isotope H below.

Image From EcoleBooks.com

  1. Name the type of radiation emitted when isotope

(i) H changes to isotope J.

AlphaMass number decrease by 4 from 214 to 210(y-axis)

atomic number decrease by 2 from 83 to 81(x-axis)

(ii) J changes to isotope K

 BetaMass number remains 210(y-axis)

atomic number increase by 1 from 81 to 82(x-axis).

(b) Write an equation for the nuclear reaction that occur when isotope

(i)J changes to isotope L

21081 J  ->  21084L   +   3 0 -1e

(i)H changes to isotope M

21483 H  ->  20682M   +   3 0 -1e +   2 4 2He

Identify a pair of isotope of an element in the decay series

K and M

Have same atomic number 82 but different mass number K-210 and M-206

a)A radioactive substance emits three different particles.

Identify the particle:

 (i)with the highest mass.

Alpha/ α

(ii) almost equal to an electron

Beta/ β

1.a)State two differences between chemical and nuclear reactions(2mks)

(i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical reactions mainly involve outer electrons in the energy levels an atom.

(ii) Nuclear reactions form a new element. Chemical reactions do not form new elements

(iii) Nuclear reactions mainly involve evolution/production of large quantity of heat/energy.Chemical reactions produce or absorb smaller quantity of heat/energy.

(iv)Nuclear reactions are accompanied by a loss in mass /mass defect.

Chemical reactions are not accompanied by a loss in mass.

(v)Rate of decay/ disintegration of nuclide is independent of physical conditionsThe rate of a chemical reaction is dependent on physical conditions of temperature/pressure/purity/particle size/ surface area

b)Below is a radioactive decay series starting from 21483 Bi and ending at 20682 Pb. Study it and answer the question that follows.

Image From EcoleBooks.com

Identify the particles emitted in steps I and III (2mks)

 I – α-particle

 III – β-ray

ii)Write the nuclear equation for the reaction which takes place in (a) step I

21483Bi
-> 21081Bi
+ 4 2 He

(b) step 1 to 3

21483Bi
-> 21081Bi
+ 4 2 He +
2 0 -1 e

(c) step 3 to 5

21082Pb
-> 20682Pb
+ 4 2 He + 2 0 -1 e

(c) step 1 to 5

21483Bi -> 20682Pb
+ 2 4 2 He + 3 0 -1 e

The table below give the percentages of a radioactive isotope of Bismuth that remains after decaying at different times.

Time (min)

0

6

12

22

38

62

100

Percentage of Bismuth

100

81

65

46

29

12

3

i)On the grid below , plot a graph of the percentage of Bismuth remaining(Vertical axis) against time.

ii)Using the graph, determine the:

 I. Half – life of the Bismuth isotope

 II. Original mass of the Bismuth isotope given that the mass that remained after 70 minutes was 0.16g  (2mks)

d)  Give one use of radioactive isotopes in medicine (1mk)

14.a)Distinguish between nuclear fission and nuclear fusion. (2mks)

Describe how solid wastes containing radioactive substances should be disposed of.  (1mk)

 b)(i)Find the values of Z1 and Z2 in the nuclear equation below

Z1  1  94   140   1

U +   n ->   Sr + Xe + 2 n

92  0  38   Z2   0  

iii)What type of nuclear reaction is represented in b (i) above?

A radioactive cobalt 6128Co undergoes decay by emitting a beta particle and forming Nickel atom,

Write a balanced decay equation for the above change 1 mark

If a sample of the cobalt has an activity of 1000 counts per minute, determine the time it would take for its activity to decrease to 62.50 if the half-life of the element is 30years 2 marks

Define the term half-life.

 

 

The diagram below shows the rays emitted by a radioactive sample

Image From EcoleBooks.com

  1. Identify the rays S,R and Q

    S- Beta ( β )particle/ray

 R- Alpha (α )particle/ray

 Q- Gamma (y )particle/ray

b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q:

 R-is blocked/stopped/do not pass through

 Q-is not blocked/pass through

  S-is blocked/stopped/do not pass through

(c)The diagram bellow is the radioactive decay series of nuclide A which is 24194Pu.Use it to answer the questions that follow. The letters are not the actual symbols of the elements.

Image From EcoleBooks.com

 

(a)Which letter represent the : Explain.

(i)shortest lived nuclide

L-has the shortest half life

 (ii)longest lived nuclide

 P-Is stable

(iii) nuclide with highest n/p ratio

 L-has the shortest half life thus most unstable thus  easily/quickly decay/disintegrate

(iv) nuclide with lowest n/p ratio

 P-is stable thus do not decay/disintegrate

(b)How long would it take for the following:

(i)Nuclide A to change to B

 10 years (half life of A)

(ii) Nuclide D to change to H

  27days +162000years+70000years+16days

 232000 years and 43 days

(iii) Nuclide A to change to P

  27days +162000years+70000years+16days

 232000 years and 43 days

Study

A.THE RATE OF CHEMICAL REACTION

(CHEMICAL KINETICS)

1.Introduction

The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up.

Some reactions are too slow to be determined. e.g rusting ,decomposition of hydrogen peroxide and weathering.

Some reactions are too fast and instantaneous e.g. neutralization of acid and bases/alkalis in aqueous solution and double decomposition/precipitation.

Other reactions are explosive and very risky to carry out safely e.g. reaction of potassium with water and sodium with dilute acids.

The study of the rate of chemical reaction is useful in knowing the factors that influence the reaction so that efficiency and profitability is maximized in industries.

Theories of rates of reaction.

The rate of a chemical reaction is defined as the rate of change of concentration/amount of reactants in unit time. It is also the rate of formation of given concentration of products in unit time. i.e.

Rate of reaction = Change in concentration/amount of reactants

Time taken for the change to occur

 

Rate of reaction = Change in concentration/amount of products formed

Time taken for the products to form

For the above, therefore the rate of a chemical reaction is rate of decreasing reactants to form an increasing product.

The SI unit of time is second(s) but minutes and hours are also used.

 (a)The collision theory

The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules.

The
collision theory proposes that

(i)for a reaction to occur, reacting particles must collide.

(ii)not all collisions between reacting particles are successful in a reaction. Collisions that initiate a chemical reaction are called successful / fruitful/ effective collisions

(iii)the speed at which particles collide is called collision frequency.

The higher the collision frequency the higher the chances of successful / fruitful/ effective collisions to form products.

(iv)the higher the chances of successful collisions, the faster the reaction.

(v)the average distance between solid particles from one another is too big for them to meet and collide successfully.

(vi)dissolving substances in a solvent ,make the solvent a medium for the reaction to take place.

The solute particle distance is reduced as the particle ions are free to move in the solvent medium.

(vii)successful collisions take place if the particles colliding have the required energy and right orientation which increases their vibration and intensity of successful / fruitful/ effective collisions to form products.

(b)The Activation Energy(Ea) theory

The Enthalpy of activation(Ha) /Activation Energy(Ea) is the minimum amount of energy which the reactants must overcome before they react.
Activation Energy(Ea) is usually required /needed in bond breaking of the reacting particles.

Bond breaking is an endothermic process that require an energy input.

The higher the bond energy the slower the reaction to start of.

Activation energy does not influence whether a reaction is exothermic or endothermic.

The energy level diagrams below shows the activation energy for exothermic and endothermic processes/reactions.

Energy level diagram showing the activation energy for exothermic processes /reactions.

 Activated complex

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Energy level diagram showing the activation energy for endothermic processes /reactions.

  Activated complex

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Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com The activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions ,but can theoretically exist.

Exothermic reaction proceeds without further heating /external energy because it generates its own energy/heat to overcome activation energy.

Endothermic reaction cannot proceed without further heating /external energy because it does not generates its own energy/heat to overcome activation energy. It generally therefore requires continuous supply of more energy/heat to sustain it to completion.

3. Measuring the rate of a chemical reaction.

The rate of a chemical reaction can be measure as:  

(i)Volume of a gas in unit time;

– if reaction is producing a gas as one of the products.

– if reaction is using a gas as one reactants

(ii)Change in mass of reactants/products for solid products/reactants in unit time.

(iii)formation of a given mass of precipitate in unit time

(iv)a certain mass of reactants to completely form products/diminish.

Reactants may be homogenous or heterogenous.

-Homogenous reactions involve reactants in the same
phase/state e.g. solid-solid,gas-gas,liquid-liquid.

 -Heterogenous reactions involve reactants in the different
phase/state e.g. solid-liquid,gas-liquid,solid-gas.

 4. Factors influencing/altering/affecting/determining rate of reaction

The following factors alter/influence/affect/determine the rate of a chemical reaction:

(a)Concentration

(b)Pressure

(c) Temperature

(d)Surface area

(e)Catalyst

  1. Influence of concentration on rate of reaction

The higher the concentration, the higher the rate of a chemical reaction. An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products.

Practically an increase in concentration reduces the time taken for the reaction to take place.

Practical determination of effect of concentration on reaction rate

Method 1(a)

Reaction of sodium thisulphate with dilute hydrochloric acid

Procedure:

Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the acid and adding the volumes of the distilled water to complete table 1. Sample results:Table 1.    

Volume of acid(cm3)

Volume of water(cm3)

Volume of sodium thiosulphate(cm3)

Time taken for mark ‘X’ to be invisible/obscured(seconds)

Reciprocal of time

1

t

20.0

0.0

20.0

20.0

5.0 x 10-2

18.0

2.0

20.0

23.0

4.35 x 10-2

16.0

4.0

20.0

27.0

3.7 x 10-2

14.0

6.0

20.0

32.0

3.13 x 10-2

12.0

8.0

20.0

42.0

2.38 x 10-2

10.0

10.0

20.0

56.0

1.78 x 10-2

For most examining bodies/councils/boards the above results score for:

(a) complete table as evidence for all the practical work done and completed.

(b) (i)Consistent use of a
decimal point on time as evidence of understanding/knowledge of the degree of accuracy of stop watches/clock.

(ii)Consistent use of a minimum of four
decimal points on inverse/reciprocal of time as evidence of understanding/knowledge of the degree of accuracy of scientific calculator.

(c) accuracy against a school value based on candidate’s teachers-results submitted.

(d) correct trend (time increase as more water is added/acid is diluted) in conformity with expected theoretical results.

Sample questions  


1. On separate graph papers plot a graph of:

(i)volume of acid used(x-axis) against time. Label this graph I

(ii) volume of acid used(x-axis) against 1/t. Label this graph II

2. Explain the shape of graph I

Diluting/adding water is causes a decrease in concentration.

Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products.

 

 

 

 

 

 

 

 

 

Sketch sample Graph I

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 Sketch sample Graph II

 

 

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is:

(i) 13cm3

From a correctly plotted graph

1/t at 13cm3 on the graph => 2.75 x 10-2

t = 1 / 2.75 x 10-2 = 36.3636 seconds

(ii) 15cm3

From a correctly plotted graph

1/t at 15cm3 on the graph => 3.35 x 10-2

t = 1 / 3.35 x 10-2 = 29.8507 seconds

(iii) 15cm3

From a correctly plotted graph

1/t at 17cm3 on the graph => 4.0 x 10-2

t = 1 / 4.0 x 10-2 = 25.0 seconds

(iv) 19cm3

From a correctly plotted graph

1/t at 19cm3 on the graph => 4.65 x 10-2

t = 1 / 4.65 x 10-2 = 21.5054 seconds

4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is:

 (i)25 seconds

 1/t => 1/25 = 4.0 x 10-2

Reading
from a correctly plotted graph;

4.0 x 10-2 correspond to 17.0 cm3

 (ii)30 seconds

 1/t => 1/30 = 3.33 x 10-2

Reading
from a correctly plotted graph;

3.33 x 10-2 correspond to 14.7 cm3

 (iii)40 seconds

 1/t => 1/40 = 2.5 x 10-2

Reading
from a correctly plotted graph;

2.5 x 10-2 correspond to 12.3 cm3

4. Write the equation for the reaction taking place

Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l)

Ionically:

S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l)

5.Name the yellow precipitate

 Colloidal sulphur

Method 1(b)

Reaction of sodium thisulphate with dilute hydrochloric acid

You are provided with

 2.0M Hydrochloric acid

 0.4M sodium thiosulphate solution

Procedure:

Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it.

Add 5.0cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate.

Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above.

Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1.

Sample results:Table 1.  

 
 

Volume of acid(cm3)

Volume of water

(cm3)

Volume of sodium thiosulphate

(cm3)

Concentation of sodium thisulphate in molesdm-3

Time(T) taken for mark ‘X’ to be invisible/ obscured(seconds)

T-1

5.0

0.0

25.0

0.4

20.0

5.0 x 10-2

5.0

5.0

20.0

0.32

23.0

4.35 x 10-2

5.0

10.0

15.0

0.24

27.0

3.7 x 10-2

5.0

15.0

10.0

0.16

32.0

3.13 x 10-2

Note concentration of diluted solution is got:

C1V1=C2V2 => 0.4 x 25 = C2x 25 =0.4M

C1V1=C2V2 => 0.4 x 20 = C2x 25 =0.32M

C1V1=C2V2 => 0.4 x 15 = C2x 25 =0.24M

C1V1=C2V2 => 0.4 x 10 = C2x 25 =0.16M

Sample questions  


1. On separate graph papers plot a graph of:

(i)Concentration of sodium thiosulphate against time. Label this graph I

(ii)Concentration of sodium thiosulphate against against T-1.Label this graph II

2. Explain the shape of graph I

Diluting/adding water causes a decrease in concentration.

Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products.

From graph II

Determine the time taken if

(i)12cm3 of sodium thisulphate is diluted with 13cm3 of water.

At 12cm3 concentration of sodium thisulphate

= C1V1=C2V2 => 0.4 x 1 2 = C2x 25 =0.192M

From correct graph at concentration 0.192M => 2.4 x10-2

I/t = 2.4 x10-2 t = 41.6667seconds

(ii)22cm3 of sodium thisulphate is diluted with 3cm3 of water.

At 22cm3 concentration of sodium thisulphate

= C1V1=C2V2 => 0.4 x 22 = C2x 25 =0.352M

From correct graph at concentration 0.352M => 3.6 x10-2

I/t = 3.6 x10-2 t = 27.7778seconds

Determine the volume of water and sodium thiosulphate if T-1 is 3.0 x10-1

From correct graph at T-1 = 3.0 x10-1 => concentration = 0.65 M

= C1V1=C2V2 => 0.4 x 25 = 0.65 M x V2 = 15.3846cm3

Volume of water = 25 – 15.3846cm3 = 9.6154cm3

Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used.

At 12cm3 concentration of sodium thisulphate

= C1V1=C2V2 => 0.4 x 1 2 = C2x 25 =0.192M

Mole ratio Na2S2 O3 :HCl =1:2

 

Moles of Na2S2 O3 =  0.192M x 12  => 2.304 x 10-3 moles

  1000

 Mole ratio HCl =2.304 x 10-1 moles   = 1.152 x 10-3 moles

2

Molarity o f HCl = 1.152 x 10-3 moles x 1000 = 0.2304M

5.0

Method 2

Reaction of Magnesium with dilute hydrochloric acid

Procedure

Scub 10centimeter length of magnesium ribbon with sand paper/steel wool. Measure 40cm3 of 0.5M dilute hydrochloric acid into a flask .Fill a graduated gas jar with water and invert it into a trough. Stopper the flask and set up the apparatus to collect the gas produced as in the set up below:

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Carefully remove the stopper, carefully put the magnesium ribbon into the flask . cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar. Immediately start off the stop watch and determine the volume of the gas produced after every 30 seconds to complete table II below.

Sample results: Table II

Time(seconds)

0

30

60

90

120

150

180

210

240

Volume of gas produced(cm3)

0.0

20.0

40.0

60.0

80.0

90.0

95.0

96.0

96.0

Sample practice questions

1.Plot a graph of volume of gas produced (y-axis) against time

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2.Explain the shape of the graph.

The rate of reaction is faster when the concentration of the acid is high .

As time goes on, the concentration of the acid decreases and therefore less gas is produced.

When all the acid has reacted, no more gas is produced after 210 seconds and the graph flattens.

3.Calculate the rate of reaction at 120 seconds

From a tangent at 120 seconds rate of reaction = Change in volume of gas

Change in time

=> From the tangent at 120seconds V2 – V1 = 96-84 = 12 = 0.2cm3sec-1

T2 – T1 150-90 60

4. Write an ionic equation for the reaction taking place.

 Mg2+(s) + 2H+(aq) -> Mg2+(aq) + H2 (g)

5. On the same axis sketch then explain the curve that would be obtained if:

 (i) 0.1 M hydrochloric acid is used –Label this curve I

 (ii)1.0 M hydrochloric acid is used –Label this curve II

Observation:

Curve I is to the right

Curve II is to the left

Explanation

A decrease in concentration shift the rate of reaction graph to the right as more time is taken for completion of the reaction.

An increase in concentration shift the rate of reaction graph to the left as less time is taken for completion of the reaction.

Both graphs flatten after some time indicating the completion of the reaction.

b)Influence of pressure on rate of reaction

Pressure affects only gaseous reactants.

An increase in pressure reduces the volume(Boyles law) in which the particles are contained.

Decrease in volume of the container bring the reacting particles closer to each other which increases their chances of effective/successful/fruitful collision to form products.

An increase in pressure therefore increases the rate of reaction by reducing the time for reacting particles of gases to react.

At industrial level, the following are some reactions that are affected by pressure:

(a)Haber process for manufacture of ammonia

N2(g) + 3H2(g) -> 2NH3(g)

(b)Contact process for manufacture of sulphuric(VI)acid

2SO2(g) + O2(g) -> 2SO3(g)

(c)Ostwalds process for the manufacture of nitric(V)acid

4NH3(g) + 5O2(g) -> 4NO (g) + 6H2O (l)

The influence of pressure on reaction rate is not felt in solids and liquids.

This is because the solid and liquid particles have fixed positions in their strong bonds and therefore no degree of freedom (Kinetic Theory of matter)

c)Influence of temperature on rate of reaction

An increase in temperature increases the kinetic energy of the reacting particles by increasing their collision frequency.

Increase in temperature increases the particles which can overcome the activation energy (Ea).

A 10oC rise in temperature doubles the rate of reaction by reducing the time taken for the reaction to complete by a half.

Practical determination of effect of Temperature on reaction rate

Method 1

Reaction of sodium thisulphate with dilute hydrochloric acid

Procedure:

Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker.

Place the beaker on a white piece of filter paper with ink mark ‘X’ on it.

Determine and record its temperature as room temperature in table 2 below.

Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder.

Put the acid into the beaker containing sodium thisulphate.

Immediately start off the stop watch/clock.

Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above.

Measure another 20cm3 separate portion of the thisulphate into a beaker, heat the solution to 30oC.

Add the acid into the beaker and repeat the procedure above. Complete table 2 below using different temperatures of the thiosulphate.

Sample results:Table 2.

Temperature of Na2S2O3

Room temperature

30

40

50

60

Time taken for mark X to be obscured /invisible (seconds)

50.0

40.0

20.0

15.0

10.0

Reciprocal of time(1/t)

0.02

0.025

0.05

0.0667

0.1

Sample practice questions

  1. Plot a graph of temperature(x-axis) against 1/t

 

 

 

2(a)From your graph determine the temperature at which:

(i)1/t is

I. 0.03

Reading directly from a correctly plotted graph = 32.25 oC

II. 0.07

Reading directly from a correctly plotted graph = 48.0 oC

(ii) t is;

I. 30 seconds

   30 seconds => 1/t =1/30 =0.033

Reading directly from a correctly plotted graph 0.033 => 33.5 oC

II. 45 seconds

    45 seconds => 1/t =1/45 =0.022

  Reading directly from a correctly plotted graph 0.022 => 29.0 oC

III. 25 seconds

    25 seconds => 1/t =1/25 =0.04

  Reading directly from a correctly plotted graph 0.04 => 36.0 oC

(b) From your graph determine the time taken for the cross to become invisible at:

(i) 57.5 oC

Reading directly from a correctly plotted graph at 57.5 oC= 0.094

=>1/t = 0.094

t= 1/0.094 => 10.6383 seconds

(ii) 45 oC

Reading directly from a correctly plotted graph at 45 oC = 0.062

=>1/t = 0.062

t= 1/0.094 => 16.1290 seconds

(iii) 35 oC

Reading directly from a correctly plotted graph at 35 oC = 0.047

=>1/t = 0.047

t= 1/0.047 => 21.2766 seconds

Method 2

Reaction of Magnesium with dilute hydrochloric acid

Procedure

Scub 5centimeter length of magnesium ribbon with sand paper/steel wool.

Cut the piece into five equal one centimeter smaller pieces.

Measure 20cm3 of 1.0M dilute hydrochloric acid into a glass beaker .

Put one piece of the magnesium ribbon into the acid, swirl.

Immediately start off the stop watch/clock.

Determine the time taken for the effervescence/fizzing/bubbling to stop when viewed from above.

Record the time in table 2 at room temperature.

Measure another 20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker.

Heat separately one portion to 30oC, 40oC , 50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above .

Record each time to complete table 2 below using different temperatures of the acid.

 

 

 

Sample results:Table 1.

Temperature of acid(oC)

Room temperature

30

40

50

60

Time taken effervescence to stop (seconds)

80.0

50.0

21.0

13.5

10.0

Reciprocal of time(1/t)

0.0125

0.02

0.0476

0.0741

0.1

Sample practice questions

  1. Plot a graph of temperature(x-axis) against 1/t

 

 

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2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0)

Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles

  Molar mass of Mg   24

  (b)Calculate the number of moles of hydrochloric acid used

Moles of acid = molarity x volume of acid

1000

=> 1.0 x 20  = 2.0 x 10 -2 moles

1000  

  (c)Calculate the mass of magnesium that remain unreacted

Mole ratio Mg: HCl = 1:2

Moles Mg = ½ moles HCl

=> ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles

Mass of reacted Mg = moles x molar mass

=> 1.0 x 10 -2 moles x 24 = 0.24 g

Mass of unreacted Mg = Original total mass – Mass of reacted Mg

=> 1.0 g – 0.24 = 0.76 g

  (b)Calculate the total volume of hydrogen gas produced during the above reactions.

 Mole ratio Mg : H2 = 1:1

Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles

Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3

=> 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3

Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5

 = 1.2 dm3


3.(a)At what temperature was the time taken for magnesium to react equal to:

 (i)70seconds

 70 seconds => 1/t =1/70 =0.01429

Reading directly from a correctly plotted graph 0.01429 => 28.0 oC

 (ii)40seconds

 40 seconds => 1/t =1/40 =0.025

Reading directly from a correctly plotted graph 0.025 => 32.0 oC

(b)What is the time taken for magnesium to react if the reaction was done at:

 (i) 55.0 oC

Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2

=> t = 1/8.0 x 10-2 = 12.5 seconds


(ii) 47.0 oC

Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2

=> t = 1/6.0 x 10-2 = 16.6667 seconds

 (iii) 33.0 oC

Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2

=> t = 1/2.7 x 10-2 = 37.037 seconds

4. Explain the shape of the graph.

Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products.

d)Influence of surface area on rate of reaction

Surface area is the area of contact. An increase in surface area is a decrease in particle size. Practically an increase in surface area involves chopping /cutting solid lumps into smaller pieces/chips then crushing the chips into powder. Chips thus have a higher surface area than solid lumps but powder has a highest surface area.

An increase in surface area of solids increases the area of contact with a liquid solution increasing the chances of successful/effective/fruitful collision to form products. The influence of surface area on rate of reaction is mainly in heterogeneous reactions.

Reaction of chalk/calcium carbonate on dilute hydrochloric acid


Procedure

Measure 20cm3 of 1.0 M hydrochloric acid into three separate conical flasks labeled C1 C2 and C3 .

Using a watch glass weigh three separate 2.5g a piece of white chalk. Place the conical flask C1 on an electronic balance.

Reset the balance scale to 0.0.

Put one weighed sample of the chalk into the acid in the conical flask. Determine the scale reading and record it at time =0.0.

Simultaneously start of the stop watch.

Determine and record the scale reading after every 30 seconds to complete Table I .Repeat all the above procedure separately with C2 and C3 to complete Table II and Table III by cutting the chalk into small pieces/chips for C2 and crushing the chalk to powder for C3

Sample results:Table 1.

Time(seconds)

0.0

30.0

60.0

90.0

120.0

150.0

180.0

210.0

240.0

Mass of CaCO3

2.5

2.0

1.8

1.4

1.2

1.0

0.8

0.5

0.5

Loss in mass

0.0

0.5

0.7

1.1

1.3

1.5

1.7

2.0

2.0

Sample results:Table 1I.

Time(seconds)

0.0

30.0

60.0

90.0

120.0

150.0

180.0

210.0

240.0

Mass of CaCO3

2.5

1.9

1.5

1.3

1.0

0.8

0.5

0.5

0.5

Loss in mass

0.0

0.6

1.0

1.2

1.5

1.7

2.0

2.0

2.0

Sample results:Table III.

Time(seconds)

0.0

30.0

60.0

90.0

120.0

150.0

180.0

210.0

240.0

Mass of CaCO3

2.5

1.8

1.4

1.0

0.8

0.5

0.5

0.5

0.5

Loss in mass

0.0

0.7

1.1

1.5

1.7

2.0

2.0

2.0

2.0

Sample questions:

1.Calculate the loss in mass made at the end of each time from the original to complete table I,II and III

2.On the same axes plot a graph of total loss in mass against time (x-axes) and label them curve I, II, and III from Table I, II, and III.

3.Explain why there is a loss in mass in all experiments.

Calcium carbonate react with the acid to form carbon(IV)oxide gas that escape to the atmosphere.

4.Write an ionic equation for the reaction that take place

CaCO3(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) + CO2(g)

5.Sulphuric(VI)acid cannot be used in the above reaction. On the same axes sketch the curve which would be obtained if the reaction was attempted by reacting a piece of a lump of chalk with 0.5M sulphuric(VI)acid. Label it curve IV. Explain the shape of curve IV.

Calcium carbonate would react with dilute 0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate stopping the reaction from reaching completion.

 

6.Calculate the volume of carbon(IV)oxide evolved(molar gas volume at room temperature = 24 dm3, C= 12.0, O= 16.O Ca=40.0)

Method I


Mole ratio CaCO3(s) : CO2(g) = 1:1

Moles CaCO3(s) used = Mass CaCO3(s) = 0.025 moles

  Molar mass CaCO3(s)

Moles CO2(g) = 0.025 moles

 Volume of CO2(g)  = moles x molar gas volume

=>0.025 moles x 24 dm3 = 0.600 dm3/600cm3

Method II


Molar mass of CaCO3(s) = 100g produce 24 dm3 of CO2(g)

Mass of CaCO3(s) =2.5 g produce 2.5 x 24 = 0.600dm3

100

7.From curve I ,determine the rate of reaction (loss in mass per second)at time 180 seconds on the curve.

From tangent at 180 seconds on curve I

  Rate = M2-M1 => 2.08 – 1.375 = 0.625 = 0.006944g sec-1

T2– T1 222-132   90

8.What is the effect of particle size on the rate of reaction?

A larger surface area is a reduction in particle size which increases the area of contact between reacting particles increasing their collision frequency.

Theoretical examples

1. Excess marble chips were put in a beaker containing 100cm3 of 0.2M hydrochloric acid. The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below.

Time(minutes)

0.0

2.0

4.0

6.0

8.0

10.0

12.0

Loss in mass(g)

0.0

1.80

2.45

2.95

3.20

3.25

3.25

(a)Why was there a loss in mass?

Carbon (IV) oxide gas was produced that escape to the surrounding

(b)Calculate the average rate of loss in mass between:

 (i) 0 to 2 minutes

 Average rate =M2-M1 => 1.80 – 0.0 = 1.8 = 9.00g min-1

T2– T1 2.0 – 0.0   2

 (i) 6 to 8 minutes

 Average rate =M2-M1 => 3.20 – 2.95 = 0.25 = 0.125g min-1

T2– T1 8.0 – 6.0   2

(iii) Explain the difference between the average rates of reaction in (i) and(ii) above.

Between 0 and 2 minutes , the concentration of marble chips and hydrochloric acid is high therefore there is a higher collision frequency between the reacting particles leading to high successful rate of formation of products.

Between 6 and 8 minutes , the concentration of marble chips and hydrochloric acid is low therefore there is low collision frequency between the reacting particles leading to less successful rate of formation of products.


(c)Write the equation for the reaction that takes place.

 CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g)


(d)State and explain three ways in which the rate of reaction could be increased.

 (i)Heating the acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency.

 (ii)Increasing the concentration of the acid-increasing in concentration reduces the distances between the reacting particles increasing their chances of effective/fruitful/successful collision to form products faster.

 (iii)Crushing the marble chips to powder-this reduces the particle size/increase surface area increasing the area of contact between reacting particles.


(e)If the solution in the beaker was evaporated to dryness then left overnight in the open, explain what would happen.

It becomes wet because calcium (II) chloride absorbs water from the atmosphere and form solution/is deliquescent.

(f)When sodium sulphate (VI) was added to a portion of the contents in the beaker after the reaction , a white precipitate was formed .

 (i)Name the white precipitate.

Calcium(II)sulphate(VI)

 (ii)Write an ionic equation for the formation of the white precipitate

  Ca2+(aq) + SO42-(aq)->CaSO4(s)

 (iii)State one use of the white precipitate

-Making plaster for building

-Manufacture of plaster of Paris

-Making sulphuric(VI)acid

(g)(i) Plot a graph of total loss in mass(y-axes) against time

(ii)From the graph, determine the rate of reaction at time 2 minutes.

From a tangent/slope at 2 minutes;

Rate of reaction = Average rate =M2-M1 => 2.25 – 1.30 = 0.95 = 0.3958g min-1

T2– T1 3.20 – 0.8   2.4

(iii)Sketch on the same axes the graph that would be obtained if 0.02M hydrochloric acid was used. Label it curve II

 

 

e) Influence of catalyst on rate of reaction

Catalyst is a substance that alter the rate /speed of a chemical reaction but remain chemically unchanged at the end of a reaction. Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. from solid to powder to fine powder. Like biological enzymes, a catalyst only catalyse specific type of reactions

Most industrial catalysts are transition metals or their compounds. Catalyst works by lowering the Enthalpy of activation(∆Ha)/activation energy (Ea) of the reactants .The catalyst lowers the Enthalpy of activation(∆Ha)/activation energy (Ea) by:

 (i) forming short lived intermediate compounds called activated complex that break up to form the final product/s

  (ii) being absorbed by the reactants thus providing the surface area on which reaction occurs.

A catalyst has no effect on the enthalpy of reaction ∆Hr but only lowers the Enthalpy of activation(∆Ha)/activation energy (Ea)It thus do not affect/influence whether the reaction is exothermic or endothermic as shown in the energy level diagrams below.

Energy level diagram showing the activation energy for exothermic processes /reactions.

 Activated complex

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Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com   Ea Catalysed

 

 

 

 

Energy level diagram showing the activation energy for endothermic processes /reactions.

 

 

 

 

 

Activated complex

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Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com  

 

 

 

 

 

The following are some catalysed reaction processes.

(a)The contact process

Vanadium(V) Oxide(V2O5) or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process.

 SO2(g)  +   O2(g) —-V2O5–> SO3(g)

To reduce industrial cost of manufacture of sulphuric (VI) acid from contact process Vanadium(V) Oxide(V2O5) is used because it is cheaper though it is easily poisoned by impurities.

 (b)Ostwalds process

Platinum promoted with Rhodium catalyses the oxidation of ammonia to nitrogen(II)oxide and water during the manufacture of nitric(V)acid

4NH3(g)  +  5O2(g) —-Pt/Rh–> 4NO (g) + 6H2O(l)

(c)Haber process

Platinum or iron catalyses the combination of nitrogen and hydrogen to form ammonia gas

 N2(g) + 3H2(g) —Pt or Fe—> 2NH3(g)

(d)Hydrogenation/Hardening of oil to fat

Nickel (Ni) catalyses the hydrogenation of unsaturated compound containing – C=C- or –C=C- to saturated compounds without double or triple bond

This process is used is used in hardening oil to fat.

(e)Decomposition of hydrogen peroxide

Manganese(IV)oxide speeds up the rate of decomposition of hydrogen peroxide to water and oxygen gas.

This process/reaction is used in the school laboratory preparation of Oxygen.

2H2O2 (g) —-MnO2–> O2(g) + 2H2O(l)

 (f)Reaction of metals with dilute sulphuric(VI)acid

Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid.

This process/reaction is used in the school laboratory preparation of Hydrogen.

H2 SO4 (aq) + Zn(s) —-CuSO4–> ZnSO4 (aq) + H2(g)

(g) Substitution reactions

When placed in bright sunlight or U.V /ultraviolet light , a mixture of a halogen and an alkane undergo substitution reactions explosively to form halogenoalkanes. When paced in diffused sunlight the reaction is very slow.

e.g. CH4(g) + Cl2(g) —u.v. light–> CH3Cl(g) + HCl(g)

(h)Photosynthesis

Plants convert carbon(IV)oxide gas from the atmosphere and water from the soil to form glucose and oxygen as a byproduct using sunlight / ultraviolet light.

6CO2(g) + 6H2O(l) —u.v. light–> C6H12O6(g) + O2(g)

(i)Photography

Photographic film contains silver bromide emulsion which decomposes to silver and bromine on exposure to sunlight.

 2AgBr(s) —u.v/sun light–> 2Ag(s) + Br2(l)

When developed, the silver deposits give the picture of the object whose photograph was taken depending on intensity of light. A picture photographed in diffused light is therefore blurred.

Practical determination of effect of catalyst on decomposition of hydrogen peroxide

Measure 5cm3 of 20 volume hydrogen peroxide and then dilute to make 40cm3 in a measuring cylinder by adding distilled water.

Divide it into two equal portions.

(i)Transfer one 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask. Cork and swirl for 2 minutes. Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask.

(ii)Put 2.0g of Manganese (IV) oxide into the clean conical/round bottomed/flat bottomed flask. Stopper the flask.

Transfer the second portion of the 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask through the dropping/thistle funnel. Connect the delivery tube to a calibrated/graduated gas jar as in the set up below.

Start off the stop watch and determine the volume of gas in the calibrated/graduated gas jar after every 30 seconds to complete Table 1.

(iii)Weigh a filter paper .Use the filter paper to filter the contents of the conical conical/round bottomed/flat bottomed flask. Put the residue on a sand bath to dry. Weigh the dry filter paper again .Determine the new mass Manganese (IV) oxide.

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Time(seconds)

0.0

30.0

60.0

90.0

120.0

150.0

180.0

210.0

240.0

270.0

Volume of gas (cm3)

0.0

20.0

40.0

60.0

80.0

90.0

95.0

96.0

96.0

96.0

 

Mass of MnO2
before reaction(g)

Mass of MnO2
after reaction(g)

2.0

2.0

Plot a graph of volume of gas produced against time(x-axes)

 

Image From EcoleBooks.com

Image From EcoleBooks.com

 

 

 

 

 

b) On the same axes, plot a graph of the uncatalysed reaction.

(c) Explain the changes in mass of manganese(IV)oxide before and after the reaction.

The mass of MnO2 before and after the reaction is the same but a more fine powder after the experiment. A catalyst therefore remains unchanged chemically but may physically change.

 

 

 

B.EQUILIBRIA (CHEMICAL CYBERNETICS)

Equilibrium is a state of balance.

Chemical equilibrium is state of balance between the reactants and products.

As reactants form products, some products form back the reactants.

Reactions in which the reactants form products to completion are said to be reversible i.e.

A  +   B  ->   C  +   D

Reactions in which the reactants form products and the products can reform the reactants are said to be reversible.

A  +   B C  +   D

Reversible reactions may be:

(a)Reversible physical changes

(b)Reversible chemical changes

(c)Dynamic equilibrium

(a)Reversible physical changes

Reversible physical change is one which involves:

 (i) change of state/phase from solid, liquid, gas or aqueous solutions. States of matter are interconvertible and a reaction involving a change from one state/phase can be reversed back to the original.

 (ii) colour changes. Some substances/compounds change their colours without change in chemical substance.

Examples of reversible physical changes

(i)
colour change on heating and cooling:

 I. Zinc(II)Oxide changes from white when cool/cold to yellow when hot/heated and back.

ZnO(s)  ZnO(s)

 (white when cold)   (yellow when hot)

 II. Lead(II)Oxide changes from yellow when cold/cool to brown when hot/heated and back.

PbO(s)  PbO(s)

 (brown when hot)   (yellow when cold)

(ii)Sublimation

 I. Iodine sublimes from a grey crystalline solid on heating to purple vapour. Purple vapour undergoes deposition back to the grey crystalline solid.

I2(s)  I2(g)

(grey crystalline solid (purple vapour

undergo sublimation) undergo deposition)

II. Carbon (IV)oxide gas undergoes deposition from a colourless gas to a white solid at very high pressures in a cylinder. It sublimes back to the colourless gas if pressure is reduced

  CO2(s) CO2(g)

(white powdery solid (colourless/odourless gas

undergo sublimation) undergo deposition)

(iii)Melting/ freezing and boiling/condensation

Ice on heating undergo melting to form a liquid/water. Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid.

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com Melting boiling  

Freezing condensing

(iv)Dissolving/ crystallization/distillation

Solid crystals of soluble substances (solutes) dissolve in water /solvents to form a uniform mixture of the solute and solvent/solution. On crystallization /distillation /evaporation the solvent evaporate leaving a solute back. e.g.

 NaCl(s) + aq NaCl(aq)

(b)Reversible chemical changes

These are reactions that involve a chemical change of the reactants which can be reversed back by recombining the new substance formed/products.

Examples of Reversible chemical changes

(i)Heating Hydrated salts/adding water to anhydrous salts.

When hydrated salts are heated they lose some/all their water of crystallization and become anhydrous.Heating an unknown substance /compound that forms a colourless liquid droplets on the cooler parts of a dry test/boiling tube is in fact a confirmation inference that the substance/compound being heated is hydrated.

When anhydrous salts are added (back) some water they form hydrated compound/salts.

Heating Copper(II)sulphate(VI)pentahydrate and cobalt(II)chloride hexahydrate

(i)Heat about 5.0g of Copper(II)sulphate(VI) pentahydrate in a clean dry test tube until there is no further colour change on a small Bunsen flame. Observe any changes on the side of the test/boiling tube. Allow the boiling tube to cool.Add about 10 drops of distilled water. Observe any changes.

(ii)Dip a filter paper in a solution of cobalt(II)chloride hexahydrate. Pass one end the filter paper to a small Bunsen flame repeatedly. Observe any changes on the filter paper. Dip the paper in a beaker containing distilled water. Observe any changes.

Sample observations

Hydrated compound

Observation before heating

Observation after heating

Observation on adding water

Copper(II)sulphate

(VI) pentahydrate

Blue crystalline solid

(i)colour changes from blue to white.

(ii)colourless liquid forms on the cooler parts of boiling / test tube

(i)colour changes from white to blue

(ii)boiling tube becomes warm /hot.

Cobalt(II)chloride hexahydrate

Pink crystalline solid/solution

(i)colour changes from pink to blue.

(ii) colourless liquid forms on the cooler parts of boiling / test tube (if crystal are used)

(i)colour changes from blue to pink

(ii)boiling tube becomes warm/hot.

When blue Copper(II)sulphate (VI) pentahydrate is heated, it loses the five molecules of water of crystallization to form white anhydrous Copper(II)sulphate (VI).Water of crystallization form and condenses as colourless droplets on the cooler parts of a dry boiling/test tube.

This is a chemical change that produces a new substance. On adding drops of water to an anhydrous white copper(II)sulphate(VI) the hydrated compound is formed back. The change from hydrated to anhydrous and back is therefore reversible chemical change.Both anhydrous white copper(II)sulphate(VI) and blue cobalt(II)chloride hexahydrate are therefore used to test for the presence of water when they turn to blue and pink respectively.

 CuSO4(s) + 5H2 O(l) CuSO4.5H2 O(s/aq)


(white/anhydrous)   (blue/hydrated)

CoCl2(s) + 6H2 O(l)   CoCl2.6H2 O(s/aq)


(blue/anhydrous)   (pink/hydrated)

(ii)Chemical sublimation

Some compounds sublime from solid to gas by dissociating into new different compounds. e.g.

Heating ammonium chloride

(i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made.

When a glass rod containing hydrogen chloride gas is placed near ammonia gas, they react to form ammonium chloride solid that appear as white fumes.

This experiment is used interchangeably to test for the presence of hydrogen chloride gas (and hence Cl ions) and ammonia gas (and hence NH4+ ions)

(ii)Put 2.0 g of ammonium chloride in a long dry boiling tube. Place wet / moist /damp blue and red litmus papers separately on the sides of the mouth of the boiling tube. Heat the boiling tube gently then strongly. Explain the observations made.

When ammonium chloride is heated it dissociates into ammonia and hydrogen chloride gases. Since ammonia is less dense, it diffuses faster to turn both litmus papers blue before hydrogen chloride turn red because it is denser. The heating and cooling of ammonium chloride is therefore a reversible chemical change.

NH3(g) + HCl(g)  NH4Cl(s)

(Turns moist (Turns moist   (forms white fumes)

litmus paper blue) litmus paper red)

(c)Dynamic equilibria

For reversible reactions in a closed system:

(i) at the beginning;

 -the reactants are decreasing in concentration with time

 -the products are increasing in concentration with time

(ii) after some time a point is reached when as the reactants are forming products the products are forming reactants. This is called equilibrium.

Image From EcoleBooks.comSketch showing the changes in concentration of reactants and products in a closed system

 

Image From EcoleBooks.com

 

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com For a system in equilibrium:

 (i) a reaction from left to right (reactants to products) is called forward reaction.

 (ii) a reaction from right to left (products to reactants) is called backward reaction.

 (iii)a reaction in which the rate of forward reaction is equal to the rate of backward reaction is called a dynamic equilibrium.

A dynamic equilibrium
is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered.

The influence of different factors on a dynamic equilibrium was first investigated from 1850-1936 by the French Chemist Louis Henry Le Chatellier. His findings were called Le Chatelliers Principle which states that:

if a stress/change is applied to a system in dynamic equilibrium, the system readjust/shift/move/behave so as to remove/ reduce/ counteract/ oppose the stress/change

 

Le Chatelliers Principle is applied in determining the effect/influence of several factors on systems in dynamic equilibrium. The following are the main factors that influence /alter/ affect systems in dynamic equilibrium:

(a)Concentration

(b)Pressure

(c)Temperature

(d)Catalyst

(a)Influence of concentration on dynamic equilibrium

An increase/decrease in concentration of reactants/products at equilibrium is a stress. From Le Chatelliers principle the system redjust so as to remove/add the excessreduced concentration.

Examples of influence of concentration on dynamic equilibrium

(i)Chromate(VI)/CrO42- ions in solution are yellow. Dichromate(VI)/Cr2O72- ions in solution are orange. The two solutions exist in equilibrium as in the equation:

2H+ (aq) + 2CrO42- (aq)
Cr2O72- (aq) + H2O(l)

(Yellow) (Orange)

I. If an acid is/H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already H+ ions. The equilibrium shift forward to the right to remove/reduce the excess H+ ions added. Solution mixture becomes More
Cr2O72- ions formed in the solution mixture make it to be more orange in colour.

II. If a base/OH (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH (aq) to form water.

H+ (aq) +OH (aq) -> H2O(l)

 

The equilibrium shift backward to the left to add/replace the H+ ions that have reacted with the OH (aq) ions . More of the CrO42- ions formed in the solution mixture makes it to be more yellow in colour.

 

2OH (aq) + 2Cr2O72- (aq)  CrO42- (aq) + H2O(l)

(Orange)  (Yellow)

 

 I. If an acid/ H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side on the OH (aq). H+ ions react with OH (aq) to form water.

H+ (aq) +OH (aq) -> H2O(l)

 

The equilibrium shift backward to the left to add/replace the 2OH (aq) that have reacted with the H+ (aq) ions . More Cr2O72- (aq)ions formed in the solution mixture makes it to be more Orange in colour.

 

II. If a base /OH (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already OH (aq) ions. The equilibrium shift forward to the right to remove/reduce the excess OH (aq) ions added. More
of the Cr2O72- ions are formed in the solution mixture making it to be more orange in colour.

(i)Practical determination of the influence of alkali/acid on Cr2O72- / CrO42- equilibrium mixture

 

Measure about 2 cm3 of Potassium dichromate (VI) solution into a test tube.

Note that the solution mixture is orange.

Add three drops of 2M sulphuric(VI) acid. Shake the mixture carefully.

  Note that the solution mixture is remains orange.

Add about six drops of 2M sodium hydroxide solution. Shake carefully.

 Note that the solution mixture is turns yellow.

Explanation

The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium. Dichromate(VI) ions are stable in acidic solutions while chromate(VI)ions are stable in basic solutions. An equilibrium exist thus:

Image From EcoleBooks.comImage From EcoleBooks.com    OH-  

H+

When an acid is added, the equilibrium shift forward to the right and the mixture become more orange as more Cr2O72- ions exist.

When a base is added, the equilibrium shift backward to the left and the mixture become more yellow as more CrO42- ions exist.

(ii)Practical determination of the influence of alkali/acid on bromine water in an equilibrium mixture

Measure 2cm3 of bromine water into a boiling tube. Note its colour.

Bromine water is yellow

Add three drops of 2M sulphuric(VI)acid. Note any colour change

Colour becomes more yellow

Add seven drops of 2M sodium hydroxide solution. Note any colour change.

Solution mixture becomes colourless/Bromine water is decolourized.

Explanation

When added distilled water,an equilibrium exist between bromine liquid (Br2(aq)) and the bromide ion(Br), hydrobromite ion(OBr) and hydrogen ion(H+) as in the equation:

 

H2O(l) + Br2(aq) OBr (aq) + H+ (aq) + Br (aq)

If an acid (H+)ions is added to the equilibrium mixture, it increases the concentration of the ions on the product side which shift backwards to the left to remove the excess H+ ions on the product side making the colour of the solution mixture more yellow.

If a base/alkali OH is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water.

H+ (aq)+ OH(aq) -> H2O(l)

This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized)

(iii)Practical determination of the influence of alkali/acid on common acid-base indicators.

Place 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes.

To each test tube add two drops of water. Record your observations in Table 1 below.

To the same test tubes, add three drops of 2M sulphuric(VI)acid. Record your observations in Table 1 below.

To the same test tubes, add seven drops of 2M sodium hydroxide solution. Record your observations in Table 1 below.

To the same test tubes, repeat adding four drops of 2M sulphuric(VI)acid. Table 1Indicator

Colour of indicator in

Water

Acid(2M sulphuric (VI) acid)

Base(2M sodium hydroxide)

Phenolphthalein

Colourless

Colourless

Pink

Methyl orange

Yellow

Red

Orange

Litmus solution

Colourless

Red

Blue

Explanation

An indicator is a substance which shows whether another substance is an acid , base or neutral.

Most indicators can be regarded as very weak acids that are partially dissociated into ions.An equilibrium exist between the undissociated molecules and the dissociated anions. Both the molecules and anions are coloured. i.e.

 

 HIn(aq) H+ (aq) + In (aq)

(undissociated indicator (dissociated indicator

molecule(coloured))   molecule(coloured))

When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears.

When a base/alkali OH is added to the indicator, the OH reacts with H+ ions from the dissociated indicator to form water.

H+ (aq)  +  OH(aq)  -> H2O(l)

 (from indicator) (from alkali/base)

The equilibrium shift forward to the right to replace the H+ ion and therefore the colour of dissociated (In) molecule shows/appears.

The following examples illustrate the above.

(i)Phenolphthalein indicator exist as:

HPh H+ (aq)  +  Ph(aq)

 (colourless molecule)  (Pink anion)

On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore colourless.

When a base/alkali OH is added to the indicator, the OH reacts with H+ ions from the dissociated indicator to form water.

H+ (aq)  +  OH(aq)  -> H2O(l)

 (from indicator) (from alkali/base)

The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The pink colour of dissociated (Ph) molecule shows/appears.


(ii)Methyl Orange indicator exists as:

HMe H+ (aq)  +  Me(aq)

 (Red molecule)  (Yellow/Orange anion)

On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore red.

When a base/alkali OH is added to the indicator, the OH reacts with H+ ions from the dissociated indicator to form water.

H+ (aq)  +  OH(aq)  -> H2O(l)

 (from indicator) (from alkali/base)

The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The Orange colour of dissociated (Me) molecule shows/appears.

(b)Influence of Pressure on dynamic equilibrium

Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules. Decrease in pressure shift the equilibrium towards the side with more volume/molecules. More yield of products is obtained if high pressures produce less molecules / volume of products are formed.

If the products and reactants have equal volume/molecules then pressure has no
effect on the position of equilibrium

The following examples show the influence of pressure on dynamic equilibrium:

(i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture

Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas.

Chemical equation : 2NO2(g) ===== N2 O4 (g)

Gay Lussacs law 2Volume   1Volume  

Avogadros law 2molecule   1molecule  


2 volumes/molecules of Nitrogen(IV)oxide form 1 volumes/molecules of dinitrogen tetraoxide

Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules.The equilibrium mixture become more yellow.

Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The equilibrium mixture become more brown.

(ii)Iodine vapour-Hydrogen gas/Hydrogen Iodide mixture.

Pure hydrogen gas reacts with Iodine vapour to form Hydrogen Iodide gas.

Chemical equation :   I2(g)  +   H2(g) ===== 2HI (g)

Gay Lussacs law  1Volume 1Volume    2Volume    

Avogadros law 1molecule 1molecule 2molecule  

(1+1) 2 volumes/molecules of Iodine and Hydrogen gasform 2 volumes/molecules of Hydrogen Iodide gas.

Change in pressure thus has no effect on position of equilibrium.

(iii)Haber process.

Increase in pressure
of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process.

The yield of ammonia is thus favoured by high pressures

Chemical equation : N2(g)
+ 3H2 (g) -> 2NH3 (g)

Gay Lussacs law  1Volume   3Volume   2Volume

Avogadros law 1molecule   3molecule   2molecule

(1 + 3) 4 volumes/molecules of Nitrogen and Hydrogen react to form 2 volumes/molecules of ammonia.

Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules.

The yield of ammonia increase.

Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules.

The yield of ammonia decrease.

(iv)Contact process.

Increase in pressure
of the Sulphur(IV)oxide/Oxygen mixture favours the formation of more molecules of Sulphur(VI)oxide gas in Contact process. The yield of Sulphur(VI)oxide gas is thus favoured by high pressures.

 

Chemical equation : 2SO2(g)
+ O2 (g) -> 2SO3 (g)

Gay Lussacs law  2Volume   1Volume   2Volume

Avogadros law 2molecule   1molecule   2molecule

(2 + 1) 3 volumes/molecules of Sulphur(IV)oxide/Oxygen mixture react to form 2 volumes/molecules of Sulphur(VI)oxide gas.

Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of Sulphur(VI)oxide gas increase.

Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of Sulphur(VI)oxide gas decrease.

(v)Ostwalds process.

Increase in pressure
of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures.

Chemical equation : 4NH3(g)
+ 5O2 (g) -> 4NO(g) + 6H2O
(g)

Gay Lussacs law  4Volume   5Volume   4Volume 6Volume

Avogadros law 4molecule   5molecule   4molecule 6Molecule

(4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 10 volumes/molecules of Nitrogen(II)oxide gas and water vapour.

Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour decrease.

Decrease in pressure shift the equilibrium forward to the right where there is more volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour increase.

Note

If the water vapour is condensed on cooling, then:

Chemical equation : 4NH3(g)
+ 5O2 (g) -> 4NO(g) + 6H2O
(l)

Gay Lussacs law  4Volume   5Volume   4Volume 0Volume

Avogadros law 4molecule   5molecule   4molecule 0Molecule

(4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 4 volumes/molecules of Nitrogen(II)oxide gas and no vapour.

Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules. The yield of Nitrogen(II)oxide gas increase.

Decrease in pressure shift the equilibrium backward to the left where there is more volume/molecules. The yield of Nitrogen(II)oxide gas decrease.

(c)Influence of Temperature on dynamic equilibrium

A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΔH).

An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΔH).

Endothermic reaction are thus favoured by high temperature/heating

Exothermic reaction are favoured by low temperature/cooling.

If a reaction/equilibrium mixture is neither exothermic or endothermic, then a change in temperature/cooling/heating has no effect on the equilibrium position.

(i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture

Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas.

Chemical equation : 2NO2(g) ===== N2 O4 (g)

On heating /increasing temperature, the mixture becomes more brown. On cooling the mixture become more yellow.

This show that

(i)the forward reaction to the right is exothermic(-ΔH).

On heating an exothermic process the equilibrium shifts to the side that generate /liberate less heat.

(ii)the backward reaction to the right is endothermic(+ΔH).

On cooling an endothermic process the equilibrium shifts to the side that do not generate /liberate heat.

(c)Influence of Catalyst on dynamic equilibrium

A catalyst has no effect on the position of equilibrium. It only speeds up the rate of attainment. e.g.

Esterification of alkanols and alkanoic acids naturally take place in fruits.In the laboratory concentrated sulphuric(VI)acid catalyse the reaction.The equilibrium mixture forms the ester faster but the yield does not increase.

CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l)

(d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes

Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium.

For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount.

The conditions required to obtain the highest yield of products within the shortest time at minimum cost are called optimum conditions

Optimum condition thus require understanding the effect of various factors on:

 (i)rate of reaction(Chemical kinetics)

 (ii)dynamic equilibrium(Chemical cybernetics)

1.Optimum condition in Haber process

Chemical equation


N2 (g) + 3H2 (g) ===Fe/Pt=== 2NH3 (g) ΔH = -92kJ

Equilibrium/Reaction rate considerations

(i)Removing ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia. More/higher yield of ammonia is attained.

(ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of ammonia is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 500atmospheres is normally used.

(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -92kJ) . Ammonia formed decomposes back to Nitrogen and Hydrogen to remove excess heat therefore a less yield of ammonia is attained. Very low temperature decrease the collision frequency of Nitrogen and Hydrogen and thus the rate of reaction too slow and uneconomical.

An optimum temperature of about 450oC is normally used.

(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation.

2.Optimum condition in Contact process

Chemical equation

2SO2 (g) + O2 (g) ===V2O5/Pt=== 2SO3 (g) ΔH = -197kJ

Equilibrium/Reaction rate considerations

(i)Removing sulphur(VI)oxide gas once formed shift the equilibrium forward to the right to replace the sulphur(VI)oxide. More/higher yield of sulphur(VI) oxide is attained.

(ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of sulphur(VI)oxide is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 1-2 atmospheres is normally used to attain about 96% yield of SO3.

(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -197kJ) . Sulphur(VI)oxide formed decomposes back to Sulphur(IV)oxide and Oxygen to remove excess heat therefore a less yield of Sulphur(VI)oxide is attained. Very low temperature decrease the collision frequency of Sulphur(IV)oxide and Oxygen and thus the rate of reaction too slow and uneconomical.

An optimum temperature of about 450oC is normally used.

(iv)Vanadium(V)Oxide and platinum can be used as catalyst. Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation.

3.Optimum condition in Ostwalds process

 

Chemical equation


4NH3 (g) + 5O2 (g) ===Pt/Rh=== 4NO (g) + 6H2O (g) ΔH = -950kJ

Equilibrium/Reaction rate considerations

(i)Removing Nitrogen(II)oxide gas once formed shift the equilibrium forward to the right to replace the Nitrogen(II)oxide. More/higher yield of Nitrogen(II) oxide is attained.

(ii)Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules . Less/lower yield of Nitrogen(II)oxide is attained. Very low pressures increases the distance between reacting NH3and O2 molecules.

An optimum pressure of about 9 atmospheres is normally used.

(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -950kJ) . Nitrogen(II)oxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of Nitrogen(II)oxide is attained. Very low temperature decrease the collision frequency of Ammonia and Oxygen and thus the rate of reaction too slow and uneconomical.

An optimum temperature of about 900oC is normally used.

(iv)Platinum can be used as catalyst. Platinum is very expensive.It is:

-promoted with Rhodium to increase the surface area/area of contact.

-added/coated on the surface of asbestos to form platinized –asbestos to reduce the amount/quantity used.

The catalyst does not increase the yield of Nitrogen (II)Oxide but it speed up its rate of formation.

 

 

C.SAMPLE REVISION QUESTIONS

 

1.State two distinctive features of a dynamic equilibrium.

(i)the rate of forward reaction is equal to the rate of forward reaction

(ii)at equilibrium the concentrations of reactants and products do not change.

2. Explain the effect of increase in pressure on the following:

(i) N2(g)  +  O2(g) ===== 2NO(g)

Gay Lussacs law 1Volume 1Volume 2 Volume

Avogadros law 1 molecule 1 molecule 2 molecule

2 volume on reactant side produce 2 volume on product side.

Increase in pressure thus have no effect on position of equilibrium.

(ii) 2H2(g)   +  CO(g) ===== CH3OH (g)

Gay Lussacs law 2Volume 1Volume 1 Volume

Avogadros law 2 molecule 1 molecule 1 molecule

3 volume on reactant side produce 1 volume on product side.

Increase in pressure shift the equilibrium forward to the left. More yield of CH3OH is formed.

4. Explain the effect of increasing temperature on the following:


2SO2(g)   +   O2 (g) ===== 2SO3 (g) ΔH = -189kJ Forward reaction is exothermic. Increase in temperature shift the equilibrium backward to reduce the excess heat.

5.120g of brass an alloy of copper and Zinc was put it a flask containing dilute hydrochloric acid. The flask was placed on an electric balance. The readings on the balance were recorded as in the table below

Time(Seconds)

Mass of flask(grams)

Loss in mass(grams)

0

600

 

20

599.50

 

40

599.12

 

60

598.84

 

80

598.66

 

100

598.54

 

120

598.50

 

140

598.50

 

160

598.50

 

(a)Complete the table by calculating the loss in mass

(b)What does the “600” gram reading on the balance represent

The initial mass of brass and the acid before any reaction take place.

(c)Plot a graph of Time (x-axes) against loss in mass.

(d)Explain the shape of your graph

The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the react is complete. No more hydrogen is evolved.The mass of flask remain constant.

(d)At what time was the loss in mass equal to:

 (i)1.20g

Reading from a correctly plotted graph =

(ii)1.30g

Readng from a correctly plotted graph =

 (iii)1.40g

Reading from a correctly plotted graph =

(e)What was the loss in mass at:

 (i)50oC

Reading from a correctly plotted graph =

 (ii) 70oC

Reading from a correctly plotted graph =

 

 (iii) 90oC g

Reading from a correctly plotted graph =


 




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EcoleBooks | Chemistry Form 4 Notes : RADIOACTIVITY

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