hkdse chemistry – book 4a ans

©Aristo Educational Press Ltd. 2010

HKDSE CHEMISTRY – A Modern View

(Chemistry)

Coursebook 4A

Suggested answers

Chapter 38 Rate of chemical reaction Page Number

Class Practice 1

Chapter Exercise 4

Chapter 39 Factors affecting rate of reaction

Class Practice 8

Chapter Exercise 11

Chapter 40 Molar volume of gases at room temperature and pressure (r.t.p.)

Class Practice 14

Chapter Exercise 17

Part Exercise 20

Chapter 41 Dynamic equilibrium

Class Practice 25

Chapter Exercise 27

Chapter 42 Equilibrium constant

Class Practice 30

Chapter Exercise 33

Chapter 43 The effect of changes in concentration and temperature on chemical equilibria

Class Practice 37

Chapter Exercise 40

Part Exercise 43

HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 4A

©Aristo Educational Press Ltd. 2010 1

Chapter 38 Rate of chemical reaction Class Practice A38.1 (a) H2 + O2 (ignition) (b) K + H2O (c) AgNO3 solution + NaCl solution A38.2

(a) (i) Average rate of decrease in mass (g) of Mg =90s

g 0.36= 4.0 103 g s1

(ii) Average rate of decrease in no. of moles of Mg

=1

13

mol g 24.3

s g104.0

= 1.65 104 mol s1

According to the equation, 1 mole of Mg reacts with 2 moles of HCl. average rate of decrease in concentration of HCl

=3

14

dm 1000

50.0s mol101.65

2 = 6.60 103 mol dm3 s1

(Assume the volume of resultant solution to be 50.0 cm3.)

(iii) Average rate of increase in volume of H2 =s90

cm 360 3

= 4.0 cm3 s1

(b) No. Rate varies during the whole course of the reaction. A38.3 (a) The rate at A is the slope of the tangent to the curve at this point.

Instantaneous rate at A (t 0) =min01.6

dm mol 0 6.0 3

= 3.75 mol dm3 min1

The rate at B is the slope of the tangent to the curve at this point.

Instantaneous rate at B =min0.13.0

dm mol 7.2 5.1 3

= 1.2 mol dm3 min1

The rate at C is the slope of the tangent to the curve at this point. Instantaneous rate at C is zero as the line is horizontal. (b) At the start of the reaction, the tangent is the steepest. This indicates that the rate

of reaction is the highest at this time. As the reaction proceeds, the slope of the tangents decreases (point B), until it becomes zero (point C). This indicates that the rate of reaction becomes lower and lower, until it reaches zero when the reaction finishes.

(c) Average rate of the reaction =min7.0

dm mol 45. 3

= 0.77 mol dm3 min1



A38.4 (a) Pipette (10.0 cm3 type) (b) To quench the reaction. Since hydrochloric acid is used as a catalyst to speed up

the reaction, excess sodium hydrogencarbonate is used to remove the acid and stop the reaction. NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)

(c) (i) 2S2O32(aq) + I2(aq) S4O6

2(aq) + 2I(aq) (ii) No. of moles of Na2S2O3(aq) used

= 0.005 1000

6.5mol = 3.25 105 mol

From the equation in (i), mole ratio of S2O32 : I2 = 2: 1.

no. of moles of I2 present =2

mol103.25 5= 1.625 105 mol

Concentration of I2 in the third portion

=3

5

1010.0

101.625

mol dm3 = 1.625 103 mol dm3

(d)

A38.5 It is because SO2 is very soluble in water. The SO2 generated cannot cause an obvious volume or pressure change. A38.6 (a)

Time of reaction t (min) 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Mass of CO2 produced (g) 0 0.80 1.50 2.05 2.50 2.85 3.05 3.25 3.40

Time of reaction t (min) 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5

Mass of CO2 produced (g) 3.51 3.59 3.64 3.67 3.69 3.70 3.70 3.70 3.70

Con

cent

ratio

n of

I2

(mol

dm

3)

0 Time (min)



(b)

(c) Average rate of reaction =(min)reaction of timetotal

(g) producedCO of mass total 2

=min 7.0

g 3.70= 0.53 g min1

A38.7 (a) Since O2 is the only gas formed, the progress of reaction can be followed by

measuring the volume of O2 gas collected in a gas syringe at regular time intervals.

(Use a pipette to withdraw a small but known volume of the reaction mixture and put it in a conical flask. Quench the reaction by adding excess dilute sulphuric acid which removes MnO2 catalyst. Then titrate the small portion of reaction mixture with standard acidified potassium permanganate solution until the end point has been reached. Repeat the withdrawal and titration of small portions of the reaction mixture at regular time intervals. This method measures the change in the concentration of H2O2 with time.)

(b) Br2(aq) is brown in colour. Its colour intensity decreases as Br2(aq) is converted to colourless Br(aq) during the reaction. The change in colour intensity of the reaction mixture can be followed by a colorimeter.

(As H+(aq) ions are formed during the course of reaction, there will be a change in pH of the resultant solution. This change in pH can be followed by a pH meter.)

(c) As there is a decrease in the number of moles of the gases at the product side, the pressure inside the closed container will change. This can be followed by a pressure sensor connected to a data-logger.

Time (min)

End of reaction

Total mass of CO2 produced

Total time of reaction

Mas

s of

CO

2 p

rodu

ced

(g)



Chapter 38 Rate of chemical reaction Chapter Exercise 1. rates 2. quickly 3. verage, nstantaneous, urve 4. Average, product, reactant 5. Instantaneous rate, tangent 6. Rate curve 7. steep, steep, horizontal 8. Initial, fastest 9. experiments 10. (a) titrimetric, quenched (b) olume, pressure, ass, colour, transmittance 11. quench, cooling, diluting, removing 12. data-logger 13. colorimeter 14. C 15. A 16. A 17. B 18. D 19. D 20. D 21. A 22. C 23. B 24. D

25. (a) No. of moles of KMnO4 used = 0.05 1000

25mol = 0.00125 mol

No. of moles of K2C2O4 used = 0.125 1000

25mol = 0.003125 mol

Mole ratio of KMnO4 : K2C2O4 = 0.00125 : 0.003125 = 1 : 2.5 The equation indicates that the mole ratio of KMnO4 : K2C2O4 = 2 : 5 = 1 : 2.5. So, none is in excess. They just react completely. (b) Since all K2C2O4 reacted, its molarity changed from 0.125 M to 0 M in 30 s.

Average rate of decrease in molarity of C2O42 =

s30

M 0.125= 4.17 103 M

s1 (c) MnO4

was completely reduced to Mn2+ in 30 seconds.

Average rate of decrease in molarity of KMnO4 =s30

M 0.05= 1.67 103 M

s1 From the equation, mole ratio of MnO4

: Mn2+ = 1 : 1. average rate of increase in molarity of Mn2+ in the mixture = 1.67 103 M s1



(d) From the equation, mole ratio of MnO4 : CO2 = 2 : 10.

no. of moles of CO2 formed in 30 s = 0.00125 mol 5 = 0.00625 mol Mass of CO2 formed in 30 s = 0.00625 mol (12.0 + 16.0 2) g mol1 = 0.275 g

Average rate of formation of CO2 =s30

g 0.275= 0.00917 g s1

26. (a) (i) Average rate of decrease of CaCO3 = s120

g 2.0 = 0.0167 g s1

(ii) Average rate of decrease of CaCO3 in mol s1

=30.160.121.40

0167.0

mol s1 = 1.67 104 mol s1

From the equation, mole ratio of CaCO3 : HCl = 1 : 2. average rate of decrease of HCl in mol dm3 s1

=

1000

5021067.1 4

mol dm3 s1

= 0.00668 mol dm3 s1

(b) (i) No. of moles of CaCO3 used =30.160.121.40

0.2

mol = 0.0200 mol

No. of moles of HCl used = 2 1000

50mol = 0.100 mol

Mole ratio of CaCO3 : HCl = 1 : 5 From the equation, mole ratio of CaCO3 : HCl = 1 : 2. hydrochloric acid is in excess. (ii) From the equation, mole ratio of CaCO3 : CO2 = 1 : 1. rate of production of CO2 in g s1 = 1.67 104 (12.0 + 16.0 2) g s1 = 0.00735 g s1 27. (a) 150 cm3 (b) At about the 22nd minute

(c) Average rate of O2 produced =min22

cm 150 3

= 6.82 cm3 min1

(d) From the graph, about 112 cm3 of O2 were produced at the 4th minute. When the reaction completed, 150 cm3 of O2 were produced.

percentage of H2O2 decomposed at the 4th minute =150

112 100% = 74.7%



(e)

The tangent at the second minute is steeper (has a greater value) than that at

the sixth minute. This suggests that the instantaneous reaction rate at the second minute is higher than that at the sixth minute.

(f) The graph begins steeply, and slowly levels off until it is horizontal. This is because as the reaction proceeds, the concentration of the reactant is decreasing with time, and hence the reaction rate becomes lower and lower. When all the reactant has turned into products, the rate of reaction becomes zero.

(g)

28. (a)

(b) (i) It is because carbon dioxide gas escaped from the flask.

(ii) CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) (c) It is to prevent the acid spray from spilling out of the conical flask. (Acid

spray will be trapped by the cotton wool.) But it allows the escape of carbon dioxide gas.

(d) As seen from the curve, the initial rate is lower than expected. This is because the solution has to be saturated with carbon dioxide gas before the

collected gas gas syringe

string

hydrogen peroxide solution

Time (min)

Vol

ume

of O

2 (c

m3 )

Time (min)

Lo

ss in

ma

ss (

g)



gas can fully escape from the solution. (e) 0.540 g

(f) No. of moles of CO2 formed =20.160.12

540.0

mol = 0.0123 mol

From the equation in (b)(ii), mole ratio of HCl : CO2 = 2 : 1. no. of moles of HCl needed = 0.0123 2 mol = 0.0246 mol

Molarity of HCl =

1000

250.0246

M = 0.984 M

29. (a) S2O32(aq) + 2H+(aq) SO2(g) + H2O(l) + S(s)

(b) 1. Gas bubbles evolve and a pungent choking smell can be detected. This is due to the production of colourless SO2 gas.

2. The solution becomes cloudy. This is due to the production of creamy yellow precipitate of sulphur.

(c) As the reaction goes on, more and more sulphur precipitate is formed. When the solution turns more and more cloudy, the transmittance of light through the solution also becomes lower. To measure the rate of the reaction, we can measure the time taken for the cloudy solution to ‘blot out’ a reference mark under the beaker of the reaction mixture. The time for ‘blot out’ is inversely proportional to the average rate of reaction.

30. As nitrogen (N2) is the only gaseous product formed in the reaction, the rate

of reaction can be followed by measuring the volume of N2 formed with time.

The volume of N2 can be measured by using a gas syringe. If the reaction is performed in a closed container, the rate of reaction can be

followed by measuring the increase in pressure with time. The pressure inside the closed container can be measured by using a

pressure sensor connected to a data-logger. If the reaction mixture is put on an electronic balance, the rate of reaction

can be followed by measuring the loss in mass (as a result of the escape of N2) of the mixture with time.



Chapter 39 Factors affecting rate of reaction Class Practice A39.1 1. (a) 2.0 M HCl(aq) (b) Yes. This is because same mass of zinc strip is used up each time. (c)

2. (a) The reaction of Mg with 2.0 M hydrochloric acid. Hydrochloric acid is a strong acid while ethanoic acid is a weak acid. The

H+(aq) concentration in 2.0 M HCl(aq) is higher than that in 2.0 M CH3COOH(aq). Since reaction rate increases with an increase in concentration of a reactant, reaction in 2.0 M HCl(aq) takes a shorter time to complete.

(b) No. of moles of HCl(aq) and CH3COOH(aq) used = MV = 100

500.2 mol =

0.10 mol

No. of moles of Mg used = 3.24

1.0mol = 4.12 103 mol

The equation of reaction is Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) Mole ratio of Mg : H+ = 1:2 The amount of HCl(aq) used is in great excess and Mg is the limiting

reagent. As the same mass of Mg is used in the two reactions, the same amount of hydrogen gas is produced.

0.1 M

Time of reaction (min)

Vol

ume

of h

ydro

gen

coll

ecte

d (c

m3 )



A39.2 (a) 2MnO4

(aq) + 5C2O42(aq) + 16H+(aq) 2Mn2+(aq) + 10CO2(g) + 8H2O(l)

(b) From purple to colourless / disappearance of purple colour. (c) Add ice to lower the temperature of the water in the water bath. (d) The average rate of reaction is inversely proportional to the time of the reaction.

That is, average rate time

1.

(e)

A39.3 (a) CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O(l) + CO2(g) (b) 275 cm3 (c) The actual volume of CO2 collected is less than expected. The difference is 85

cm3 (360 – 275 cm3). It is because some of the carbon dioxide has dissolved in water and the powdered calcium carbonate used is not pure.

(d)

Temperature (C)

Time

1(s1)

Time (min)

Vol

ume

of c

arbo

n di

oxid

e ga

s (c

m3 )

A

B



(e) As carbon dioxide is formed as the only gaseous product, we can carry out the reaction in a closed system. The increase in pressure inside the apparatus at different times can be monitored using a pressure sensor connected to a data-logger.

As carbon dioxide is formed, it is allowed to escape from the reaction flask. The decrease in mass of the reaction mixture at different times can be monitored by weighing with an electronic balance.

A39.4 (a) MnO2(s) formed acts as a catalyst in the decomposition of hydrogen peroxide. (b) Sulphuric acid is used to remove the MnO2(s) catalyst in the reaction mixture.

This is to quench the decomposition of H2O2 and allow the analysis of H2O2 concentration in the samples at different time intervals.

Sulphuric acid is also used to acidify the potassium permanganate solution for the redox reaction with H2O2.

(c) 6H+(aq) + 2MnO4(aq) + 5H2O2(aq) 2Mn2+(aq) + 5O2(g) + 8H2O(l)

(d) and (e)

without MnO2(s) catalyst

with MnO2(s) catalyst

Time (min)

Con

cent

rati

on o

f H

2O2

(M)



Chapter 39 Factors affecting rate of reaction Chapter Exercise 1. oncentration, urface area, emperature, atalyst 2. collide, effective collisions 3. effective 4. oncentration, urface area, emperature, atalyst 5. Catalysts 6. Enzymes 7. A 8. A 9. D 10. B 11. C 12. B 13. D 14. C 15. A 16. B

17. (a)

Sample Concentration of

Na2S2O3(aq) (M)

Volume of Na2S2O3(aq) used (cm3)

Volume of distilled

water added (cm3)

1 0.100 20 0 2 0.075 15 5 3 0.050 10 10 4 0.025 5 15

(b) The amount of HCl(aq) added was the same in all the experiments. All the experiments were carried out at room conditions. (c)

Time

1(s1)

Conc. of Na2S2O3 (aq) (M)



(d) From the graph, when the rate of reaction was 0.015 s1, the concentration

of Na2S2O3(aq) was 0.058 M. (e) Based on the graph, the rate of the reaction increased when there was an

increase in concentration of Na2S2O3(aq). (f) When we stopped the stopwatch, there was just enough sulphur precipitate

produced in the solution to mask the cross. The reaction kept going to produce more sulphur.

18. (a) Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

(b) No. of moles of Zn used = 0.65

1mol = 0.0154 mol

No. of moles of HCl(aq) present = MV = 2.0 1000

50mol = 0.100 mol

From the equation in (a), mole ratio of Zn : HCl = 1 : 2 0.0154 mol of Zn requires only 0.0154 2 mol = 0.0308 mol of HCl(aq) for

reaction, so HCl(aq) is in excess. Zn is the limiting reactant. (c) (i) Initial rate of experiment 2 is higher than that of experiment 1 because

of larger surface area of zinc in the form of powder. (ii) Volume of H2 formed is unchanged / the same because the same mass

of Zn is used. (d)

19. (a) S2O32(aq) + 2H+(aq) S(s) + SO2(g) + H2O(l)

(b) This is the reacting temperature of the experiment. It is necessary to note it for comparison.

(c) As the reaction proceeds, more and more sulphur precipitate is produced until it is enough to blot out the cross.

(d) The smell comes from the sulphur dioxide gas produced. (e) Immerse the small beaker containing the thiosulphate solution in an

ice-water bath. (f) The longer the time to blot out the cross, the slower the reaction. (The

shorter the time to blot out the cross, the faster the reaction.)

the reciprocal of time (i.e.time

1) is a measure of reaction rate.

Vol

ume

of H

2(g)

Time

3



(g)

20. In order to investigate the effect of temperature on rate of chemical

reactions, several experiments should be done. In all the experiments, same mass of marble chips is allowed to react with same volume but excess of 2 M hydrochloric acid at different temperatures.

As the reaction involves the production of gaseous product, carbon dioxide, we can collect the gas with a gas syringe and measure the volume of carbon dioxide gas produced with time.

Knowing the volume of carbon dioxide gas produced with time, we can find out the rate of carbon dioxide gas produced. By plotting a graph of reaction rate against temperature, we can understand the effect of temperature on rate of chemical reactions.

Temperature (C)

Time

1




Class Practice A40.1 Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

No. of moles of Mg used = 1mol g 24.3

g 243.0

= 0.01 mol

From the equation, mole ratio of Mg : H2 = 1 : 1,

molar volume of hydrogen at r.t.p. = mol0.01

cm 240 3

= 24 000 cm3 mol1 or 24.0

dm3 mol1 A40.2 1. No. of moles of Cl2 molecules

=13

3

molcm 000 24

cm 120

= 0.0050 mol No. of Cl2 molecules = 0.0050 mol 6.02 1023 mol1 = 3.01 1021 2. (a) Volume of nitrogen = 0.1 mol 24.0 dm3 mol1 = 2.4 dm3

(b) No. of moles of CO2 = 123

21

mol 1002.6

1041.2

= 4.00 103 mol Volume of CO2 = 4.00 103 mol 24.0 dm3 mol1 = 0.096 dm3 (96.0 cm3) A40.3 1. From the equation, mole ratio of CO : CO2 = 2 : 2 = 1 : 1 No. of moles of CO needed = 0.20 mol volume of CO needed = 0.20 mol 24.0 dm3 mol1 = 4.8 dm3 (4800 cm3) From the equation, mole ratio of O2 : CO2 = 1 : 2 No. of moles of O2 needed = 0.10 mol volume of O2 needed = 0.10 24.0 dm3 = 2.4 dm3 (2400 cm3)



2. (a) (i) 2SO2(g) + O2(g) 2SO3(g) (from equation) 2 moles 1 mole 2 moles (By Avogadro’s Law) 2 volumes 1 volume 2 volumes 15.0 cm3 ? cm3 ? cm3

By simple proportion, volume of O2 required = 2

10.15 cm3 = 7.5 cm3

(ii) Volume of SO3 formed = 1

10.15 cm3 = 15.0 cm3

(b) From the equation, volume ratio of SO3 : SO2 = 1 : 1 volume of SO2 needed = 24.0 cm3 From the equation, volume ratio of SO3 : O2 = 2 : 1

volume of O2 needed = 2

10.24 cm3 = 12.0 cm3

3. In 100 dm3 of town gas, volume of H2 = 49.0 dm3, volume of CH4 = 28.5 dm3, volume of CO = 3.0 dm3.

For H2 and by applying Avogadro’s Law, 2H2(g) + O2(g) 2H2O(l) volume ratio 2 1 49.0 dm3 ? dm3

volume of O2 needed to react with H2 = 2

0.49dm3 = 24.5 dm3

For CH4 and by applying Avogadro’s Law, CH4(g) + 2O2(g) CO2(g) + 2H2O(l) volume ratio 1 2 28.5 dm3 ? dm3 volume of O2 needed to react with CH4 = 28.5 2 dm3 = 57.0 dm3 For CO and by applying Avogadro’s Law, 2CO(g) + O2(g) 2CO2(g) volume ratio 2 1 3.0 dm3 ? dm3

volume of O2 needed to react with CO = 2

0.3dm3 = 1.5 dm3

Total volume of O2 needed = 24.5 + 57.0 + 1.5 dm3 = 83.0 dm3 A40.4

1. No. of moles of nitrogen gas = 0.24

464.0mol = 0.0193 mol

Mass of nitrogen gas = 0.0193 (14.0 2) g = 0.540 g

2. (a) No. of moles of oxygen = 23

21

1002.6

10204.1

mol = 2.00 103 mol

Volume of oxygen (at r.t.p.) = 2.00 103 24.0 dm3 = 0.0480 dm3 (48.0 cm3)

(b) Mass of oxygen = 2.00 103 (16.0 2) g mol1 = 0.0640 g



A40.5

1. No. of moles of CO2 formed at r.t.p. = 00024

360mol = 0.0150 mol

From the equation, mole ratio of NaHCO3 : CO2 = 2 : 1, no. of moles of NaHCO3 needed = 0.0150 2 mol = 0.0300 mol mass of NaHCO3 needed = 0.0300 (23.0 + 1.0 + 12.0 + 16.0 3) g = 2.52 g

2. No. of moles of Mg used = 3.24

43.2mol = 0.100 mol

No. of moles of HCl(aq) used = 2.0 (1000

30) mol = 0.060 mol

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) From the equation, mole ratio of Mg : HCl = 1 : 2,

0.060 mol of HCl(aq) needs 2

060.0= 0.030 mol of Mg for complete reaction.

Since Mg is in excess, HCl is the limiting reactant. From the equation, mole ratio of HCl : H2 = 2 : 1,

no. of moles of H2 produced = 2

060.0mol = 0.030 mol

Volume of H2 produced at r.t.p. = 0.030 24.0 dm3 = 0.72 dm3




Chapter exercise 1. molar 2. dm3 mol1 3. room temperature and pressure, 25, 1, dm3 mol1, cm3 mol1 4. molar 5. Avogadro’s, same 6. volume, temperature, pressure 7. B 8. A 9. A 10. B 11. A 12. B 13. C 14. C 15. C 16. D 17. D

18. (a) No. of moles of propane present = 8)1.03(12.0

88.0

mol = 0.0200 mol

Molar volume of propane = 0200.0

465cm3 mol1 = 23 250 cm3 mol1 (23.25

dm3 mol1) (b) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) From the equation, mole ratio of C3H8 : CO2 = 1 : 3, no. of moles of CO2 formed = 0.0200 3 mol = 0.0600 mol Volume of CO2 obtained = 0.0600 23 250 cm3 = 1395 cm3

19. (a) No. of moles of SO2 = 00024

250mol = 0.0104 mol

(b) No. of moles of C2H4 = 0.24

85.0mol = 0.0354 mol

20. No. of moles of N2 = 000 24

750mol = 0.0313 mol

No. of moles of O2 = 000 24

500mol = 0.0208 mol

total no. of moles of gases = 0.0313 + 0.0208 mol = 0.0521 mol (Total volume of gases = (750 + 500) cm3 = 1250 cm3

Total no. of moles of gases = 00024

1250mol = 0.0521 mol)



21. (a) Volume of H2S = 0.5 24.0 dm3 = 12.0 dm3 (b) Volume of Cl2 = 4.2 24.0 dm3 = 100.8 dm3 (c) I2 is a solid at r.t.p. Density of solid I2 = 4.93 g cm3 Mass of 10 mol of I2 = 10 127 g = 1270 g

Volume of 10 mol of I2 = 93.4

1270cm3 = 257.6 cm3

22. (a) No. of moles of H2S = 32.1)2(1.0

41.3

mol = 0.100 mol

No. of moles of Cl2 = 2)(35.5

10.7

mol = 0.100 mol

From the given equation, mole ratio of H2S : Cl2 = 1 : 1, both reactants are all used up. As the number of moles of H2S : Cl2 : HCl = 1 : 1 : 2, no. of moles of HCl formed = 0.100 2 mol = 0.200 mol mass of HCl = 0.200 (1.0 + 35.5) g = 7.30 g Volume of HCl (at room conditions) = 0.200 24.0 dm3 = 4.8 dm3 (4800

cm3) (b) From the given equation, mole ratio of H2S : Cl2 : HCl = 1 : 1 : 2 By Avogadro’s Law, volume ratio of H2S : Cl2 : HCl = 1 : 1: 2 100 cm3 of H2S react with 100 cm3 of Cl2 to produce 200 cm3 of HCl at

25C and 1 atm.

(c) No. of moles of S formed = 1.32

21.3mol = 0.100 mol

From the given equation, mole ratio of S : Cl2 = 1 : 1, no. of moles of Cl2 needed = 0.100 mol Volume of 0.100 mol of Cl2 = 0.100 24.0 dm3 = 2.4 dm3 (2400 cm3) 23. For methane, CH4(g) + 2O2(g) CO2(g) + 2H2O(l) 100 cm3 ? cm3 From the equation, volume ratio of CH4 : O2 = 1 : 2, volume of O2 needed = 100 2 cm3 = 200 cm3 For hydrogen, 2H2(g) + O2(g) 2H2O(l) 200 cm3 ? cm3 From the equation, volume ratio of H2 : O2 = 2 : 1,


200cm3 = 100 cm3

For carbon monoxide, 2CO(g) + O2(g) 2CO2(g) 250 cm3 ? cm3 From the equation, volume ratio of CO : O2 = 2 : 1,


250cm3 = 125 cm3

Total volume of O2 needed = (200 + 100 + 125) cm3 = 425 cm3



24. From the given equation (1), volume ratio of NH3 : O2 : NO = 4 : 5 : 4,

volume of O2 needed to oxidize NH3 to NO = 40 4

5cm3 = 50 cm3

Volume of NO produced = 40 cm3 From the given equation (2), volume ratio of NO : O2 = 2 : 1,

volume of O2 needed to react with NO = 2

40cm3 = 20 cm3

Minimum volume of O2 needed = (50 + 20) cm3 = 70 cm3

25. (a) (i) No. of moles of HCl used = 1.0 1000

50mol = 0.050 mol

(ii) No. of moles of H2 formed = 00024

600mol = 0.025 mol

(b) Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) H2(g) formed came from the zinc reacted.

From the equation, the mole ratio of Zn : H2 = 1 : 1, no. of moles of Zn reacted = 0.025 mol Mass of Zn reacted = 0.025 65.4 g = 1.635 g (c) No. of Zn atoms reacted = 0.025 6.02 1023 = 1.505 1022 (d) From the equation, mole ratio of ZnCl2 : H2 = 1 : 1, no. of moles of ZnCl2 formed = 0.025 mol

Molarity of ZnCl2(aq) = 31050

025.0

mol dm3 = 0.50 mol dm3 (M)



Part IX Rate of reaction Part Exercises 1. C 2. B 3. B 4. A 5. B 6. D 7. B 8. C 9. D 10. A 11. A 12. D 13. C 14. B 15. D 16. C 17. A 18. C 19. B 20. D 21. A

22. (a) The progress of the reaction can be followed by using a colorimeter. The

reactant iodine, I2, is brown in colour. The other reactant and all the products are colourless. The colour intensity of the reaction mixture decreases as the reaction proceeds.

(b) The progress of the reaction can be followed by determining the volume of nitrogen gas formed. Only nitrogen is a gas and it is not quite soluble in water. The volume of gas released from the reaction mixture increases as the reaction proceeds.

(c) The progress of the reaction can be followed by a chemical method. Known volumes of the reaction mixture are withdrawn at regular intervals and the reaction is quenched by lowering the temperature. The concentration of iodine can be determined by titration with standard thiosulphate (S2O3

2) solution.

(d) The reaction can be followed by measuring the time taken when the same turbidity due to sulphur precipitate appears in the reaction mixture. The precipitate can mask a reference mark placed underneath the reaction flask.



23. (a) The CO2 formed is quite soluble in water especially when the pressure of the closed system is increasing.

(b) 1. By using a colorimeter, we can measure the change in colour intensity of the reaction mixture with time. As the reaction proceeds, the concentration of the purple coloured MnO4

(aq) decreases. Hence, the colour intensity of the reaction mixture decreases too.

2. By using a gas syringe, we can collect and measure the volume of CO2 formed with time.

3. By using an electronic balance, we can measure the loss in mass (as a result of the escape of the CO2 formed) of the reaction mixture with time. (Note: CO2 is quite soluble in water, so measurements at the start of the experiment will have some errors.)

24. (a) HCO3

(aq) + H+(aq) H2O(l) + CO2(g) (b) Initial rate is the rate of a reaction at the start.

This is the highest rate for the whole reaction because the concentration(s) of reactant(s) is/are the highest at the beginning of a reaction.

(c) Reaction 3 will have a higher initial rate because for the same concentration, hydrochloric acid is a stronger acid/ionizes more completely than ethanoic acid to give a higher concentration of H+(aq). Thus, the initial rate is higher.

(d) Reaction 2 will have a higher initial rate because for the same concentration, sulphuric acid is a dibasic acid/has two ionizable hydrogen atoms per molecule. As a result, sulphuric acid has a higher concentration of hydrogen ions. Thus, the initial rate is higher.

(e) No. of moles of NaHCO3 present = 1 50 103 mol = 5 102 mol

No. of moles of HCl used = 1 1000

10mol = 1 102 mol

HCl is the limiting reactant, so it is used to calculate the volume of CO2 formed.

From the equation, mole ratio of HCl : CO2 = 1 : 1, no. of moles of CO2 formed = 1 102 mol Volume of CO2 formed = 1 102 24.0 dm3 = 0.24 dm3 (240 cm3) (f)

2

1 Pre

ssur

e (k

Pa)

reaction 3

Time (min)



(g) If the reaction is quite rapid, the use of data-logger can collect enough data for analysis within a short period of time. Besides, data is stored in digital format which allows for more convenient analysis and presentation.

25. Na2CO3nH2O(s) + 2HNO3(aq) 2NaNO3(aq) + CO2(g) + (n+1)H2O(l)

No. of moles of CO2 produced = 00024

600mol = 0.0250 mol

From the equation, mole ratio of Na2CO3nH2O : CO2 = 1 : 1, no. of moles of Na2CO3nH2O = 0.0250 mol

Molar mass of Na2CO3nH2O = 0250.0

15.7g mol1 = 286 g mol1

286 = (23.0 2 + 12.0 + 16.0 3) + n(1.0 2 + 16.0) 286 = 106 + 18.0n 180 = 18.0n n = 10 26. (a) Na2CO3(s) + 2HCl(aq) 2NaCl(aq) + CO2(g) + H2O(l)

No. of moles of HCl used = 1000

402 mol = 0.080 mol

No. of moles of Na2CO3 used = 3)16.012.02(23.0

3.5

mol = 0.050 mol

From the equation, mole ratio of Na2CO3 : HCl = 1 : 2, 0.050 mol of Na2CO3 needs 0.050 2 = 0.10 mol of HCl to react. HCl is the limiting reactant, so some Na2CO3 is left unreacted. (b) From the equation in (a), mole ratio of HCl : CO2 = 2 : 1,

no. of moles of CO2 evolved = 2

080.0mol = 0.040 mol

Volume of CO2 evolved at room conditions = 0.040 24.0 dm3 = 0.96 dm3

(960 cm3) 27. ZnCO3(s) + H2SO4(aq) ZnSO4(aq) + H2O(l) + CO2(g)

No. of moles of ZnCO3 used = 3)16.012.0(65.4

5.8

mol = 0.0678 mol

No. of moles of H2SO4 used = 1.20 1000

50mol = 0.0600 mol

Mole ratio of ZnCO3 : H2SO4 = 1 : 1, H2SO4 is the limiting reactant.

From the equation, mole ratio of H2SO4 : CO2 = 1 : 1, no. of moles of CO2 = 0.0600 mol Volume of CO2 formed = 0.0600 24.0 dm3 = 1.44 dm3 (1440 cm3)



28. (a) Volume of oxygen in 500 cm3 air = 500 20% cm3 = 100 cm3

From the equation, volume ratio of CH3NO2 : O2 = 4 : 3,

volume of CH3NO2 needed = 3

4100 cm3 = 133.33 cm3

(b) From the equation, volume ratio of O2 : CO2 : N2 : H2O = 3 : 4 : 2 : 6

volumes of CO2, N2 and H2O formed respectively are 133.3 cm3, 66.67 cm3 and 200 cm3.

Total volume of gaseous products = 133.33 + 66.67 + 200 cm3 = 400 cm3

(c) Total volume of gases consumed = 100 + 133.33 cm3 = 233.33 cm3

Overall volume change

= total volume of gaseous products formed – total volume of gases consumed

= 400 – 233.33 cm3

= 166.67 cm3

29. (a) No. of moles of FeS2 used = 2)32.1(56.0

100

mol = 0.832 mol

From the given equation, mole ratio of FeS2 : SO2 = 4 : 8 = 1 : 2, no. of moles of SO2 produced = 0.832 2 mol = 1.664 mol Volume of SO2 produced = 1.664 24.0 dm3 = 39.94 dm3 (b) From the given equation, mole ratio of FeS2 : O2 = 4 : 11,

no. of moles of O2 used = 4

11832.0 mol = 2.288 mol

Volume of O2 needed = 2.288 24.0 dm3 = 54.91 dm3

Volume of air needed = 20

10091.54 dm3 = 274.55 dm3

30. (a) MgCO3(s) MgO(s) + CO2(g) CaCO3(s) CaO(s) + CO2(g) (b) Let x g and (7.0 – x) g be the mass of MgCO3 and CaCO3 respectively in the

mixture. For MgCO3,

no. of moles of MgCO3 decomposed = 3)16.012.0(24.3

xmol =

3.84

xmol

From the equation in (a), mole ratio of MgCO3 : CO2 = 1 : 1,

no. of moles of CO2 produced = 3.84

xmol



For CaCO3,

no. of moles of CaCO3 decomposed = 3)16.012.0(40.1

)(7.0

x mol =

100.1

)(7.0 x mol

From the equation in (a), mole ratio of CaCO3 : CO2 = 1 : 1,

no. of moles of CO2 produced = 100.1

)(7.0 x mol

Total volume of CO2 produced

= {(3.84

x) + [

100.1

)(7.0 x ]} 24 000 cm3 = 1880 cm3

Solving for x, x = 4.49 mass of MgCO3 in the mixture = 4.49 g

31.

The rate of a reaction tells how quickly a chemical reaction occurs. We can only determine reaction rates by doing experiments. The rate of a chemical reaction can be followed by measuring the characteristic change in some physical properties of the reaction with time.

In the decomposition of ammonium nitrite, gaseous product, N2, is formed which can be conveniently collected by using a gas syringe. We can measure the volume of N2 gas produced with time.

If the reaction is performed in a closed system, as more and more N2 is produced, the higher the pressure inside the reaction vessel. We can measure the increase in pressure with time by using pressure sensor which is connected to a data-logger.

As N2 gas is produced, it will escape into the atmosphere resulting in a loss in mass of the reaction mixture. If the reaction vessel containing the reaction mixture is put on an electronic balance, the change in mass with time can be measured.



Chapter 41 Dynamic equilibrium Class Practice A41.1 (a) Yes, it is a reversible reaction. The sign ‘ ’ shows that the reaction can

proceed in either direction. (b) Backward reaction. (c) Forward reaction: reactants are SO2 and O2; product is SO3 Backward reaction: reactant is SO3; products are SO2 and O2 A41.2 (a) (i) or

Time

H2, I2

HI

Conc.

Time

HI

H2, I2

Conc.



(ii) (b) As time passes, the colour of the mixture changes from colourless to violet.

The colour intensity keeps on increasing with time. When the equilibrium state is reached, the colour intensity remains constant.

H2 + I2 → 2HI

2HI → H2 + I2

Reaction rate

Time



Chapter 41 Dynamic equilibrium Chapter Exercise 1. rreversible, ompletion 2. eversible, ompletion 3. (a) closed

(b) orward, ackward (c) equal (d) eactants, roducts, oncentrations, nchanged, bservable

4. B 5. D 6. A 7. D 8. B 9. C 10. A 11. C 12. (a) (i) 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g) Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g) (ii) The rates of these two reactions are equal. (b) 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g) (c) Yes. (d) Equilibrium can only be established in a closed system. Equilibrium can be reached from either the forward or the backward

direction of the reversible reaction. At equilibrium, rate of forward reaction is equal to rate of backward

reaction. At equilibrium, the concentrations of both reactants and products remain

unchanged.



13.

14. (a) Measure the pH of solution with a pH meter. The pH of the solution should

remain constant. (b) Measure the colour intensity (brown colour of NO2(g)) of the reaction

mixture using a colorimeter. The colour intensity of the equilibrium mixture should remain constant.

OR Measure the pressure of the reaction mixture using a pressure sensor connected to a data-logger. The pressure of the equilibrium mixture should remain constant.

(c) Measure the colour intensity (purple colour of I2(g)) of the reaction mixture using a colorimeter. The colour intensity of the equilibrium mixture should remain constant.

15.

forward reaction

backward reaction

Rea

ctio

n ra

te

Time Time to reach equilibrium

Y

X

dynamic equilibrium is established

100%

50%

Con

cent

rati

on

Time



16. Chemical equilibrium only occurs in reversible reactions. e.g. esterification,

thermal decomposition of calcium carbonate, Haber process, etc. The general equation of chemical equilibrium can be expressed as:

A + B C + D where the process from A + B C + D is called the

forward reaction, from C + D A + B is called the backward reaction.

Chemical equilibrium is dynamic in nature during which the rates of forward and backward reactions are not zero.

Characteristics of dynamic equilibrium: (a) Only established in closed system with no exchange of materials

between the system and the surroundings. (b) Can be established by starting from either the forward or the

backward reaction. (c) The rate of forward reaction is equal to the rate of backward reaction. (d) The concentrations of both reactants and products remain unchanged

but not zero. Thus, there are no observable changes of the reaction mixture.



Chapter 42 Equilibrium constant Class Practice A42.1 The answer is (e) because the amount/concentration of H2, I2 and HI become constant at (e) and afterwards. (Three H2 molecules, three I2 molecules and fourteen HI molecules) A42.2

1. (a) 2eqm2

eqm22eqm2

cS(g)]H[

](g)S[](g)[HK

(b) eqm2eqm

eqm2c (g)][ClCO(g)][

](g)[COClK

2. (a) 2NO(g) + O2(g) 2NO2(g) (b) 2CH4(g) C2H2(g) + 3H2(g) A42.3

(a) (i) Kc1 = 3eqm2eqm2

2eqm3

(g)][H(g)]N[

(g)]NH[

(ii) Kc2 = 2eqm3

3eqm2eqm2

(g)]NH[

(g)][H(g)]N[

(b) Kc1 = c2K

1

(c) Kc1: mol2 dm6

Kc2: mol2 dm6

A42.4 (a) and (b) are heterogeneous equilibria. (a) Kc = [CO2(g)]eqm

(b) Kc= [Ag+(aq)]eqm[Cl(aq)]eqm

A42.5 (b), (c), (d), (a)



A42.6

1. Kc = 2eqm4

3eqm2eqm22

(g)]CH[

(g)][H(g)]HC[

Substituting the equilibrium concentration values into the expression of equilibrium constant, we have

Kc = 23

333

)dm mol 02030(

)dm mol 11200)(dm mol 04510(

.

..

= 62dmmol 1540 . 2.

CO(g) + 2H2(g) CH3OH(g) Concentration (mol dm3)

Initial 0.50 1.00 0 Change 0.15 – 0.50

= –0.35–2 0.35 = –0.70 +0.35

Equilibrium 0.15 0.30 0.35

Kc= 2eqm2eqm

eqm3

(g)]][HCO(g)[

OH(g)]CH[

Substituting the equilibrium concentration values into the expression of equilibrium constant, we have

Kc= 233

3

)dm mol 30.0)(dm mol 15.0(

dm mol 35.0

= 62 dm mol 93.25 3.

2HI(g) H2(g) + I2(g) Concentration (mol dm3)

Initial

5

4= 0.8

0 0

Change –2x +x +x Equilibrium 0.8 – 2x x x

2eqm

eqm2eqm2c

][HI(g)

](g)I[(g)]H[K

Substituting the equilibrium concentration values into the equilibrium constant expression gives

23

33

)dm mol )28.0((

)dm mol )(dm mol (016.0

x

xx

Take the square root of both sides of the equation to obtain

)28.0(

)(126.0

x

x

Solving for x, x = 0.0808



Thus, [H2(g)]eqm = 0.0808 mol dm3

[I2(g)]eqm = 0.0808 mol dm3

[HI(g)]eqm = 0.8 – 2(0.0808) mol dm3

= 0.638 mol dm3



Chapter 42 Equilibrium constant Chapter Exercise

1. ba

dc

eqmeqm

eqmeqm

[B][A]

[D][C], concentrations, Equilibrium

2. temperature, initial, equation 3. omogeneous 4. eterogeneous, concentrations 5. osition 6. xtent, ate 7. large, reactant 8. C 9. D 10. B 11. D 12. A 13. A 14. B 15. A 16. C 17. B 18. B 19. C

20. (a) Kc = eqm2

4eqm

2eqm2

2eqm2

(g)][O[HCl(g)]

O(g)][H(g)][Cl

(b) Kc = 2eqm2eqm (g)][H[CO(g)]

1

(c) Kc = eqm2

2eqm2

2eqm2

(g)][S(g)][H

S(g)][H

(d) Kc = eqm2eqm

eqm2eqm2

O(g)][H[CO(g)]

(g)][H(g)][CO

(e) Kc = eqm23

eqm2

(aq)])[Ca(HCO

(g)][CO



21. (a) Kc = eqm

24eqm

2 (aq)][SO(aq)]Ba[

1

Unit of Kc = 6233

dmmol )dm mol)(dm (mol

1

(b) Kc = eqm2

2eqm

2eqm

(g)][ClNO(g)][

NOCl(g)][

Unit of Kc = 31323

23

dmmol )dm mol()dm (mol

)dm (mol

(c) Kc = eqm2

2eqm

eqm22eqm

(g)][FHCl(g)][

](g)Cl[HF(g)][

Unit of Kc = )dm (mol)dm (mol

)dm (mol)dm (mol323

323

i.e. no unit

(d) Kc = eqm4

4eqm

(g)]Ni(CO)[

CO(g)][

Unit of Kc = 933

43

dm mol dm mol

)dm (mol

22. Kc = 322

33

2eqm

eqm2eqm2 10 2.07 )10(2.2

)10)(1.010 (1.0

[HI(g)]

(g)][I(g)]H[

23. Kc = eqm2

2eqm2

2eqm3

(g)][O(g)][SO

(g)][SO= 10.1dmmol

50.175.1

25.2 312

2

mol1 dm3

24.

2CO2(g) 2CO(g) + O2(g) Concentration (mol dm3)

Initial

1

001.0= 0.001

0 0

Change 0.001 – 1 104 +1 104 +

2

1 1 104

Equilibrium 9 104 1 104 5 105

Kc = 2eqm2

eqm22eqm

][CO

][O[CO] = 24

524

)109(

)105()101(

mol dm3 = 6.2 107 mol dm3



25. (a) Let x be the number of moles per dm3 of H2 that react. H2(g) + Br2(g) 2HBr(g)

Concentration (mol dm3)

Initial

00.2

0.4= 2.0

00.2

0.4= 2.0

0

Change –x –x +2x Equilibrium 2.0 – x 2.0 – x 2x

Since Kc = eqm2eqm2

2eqm

(g)][Br(g)][H

[HBr(g)] = 2

2

)0.2(

)2(

x

x

=12.0

Solving for x, x = 1.268 At equilibrium, [H2]eqm = [Br2]eqm = 2 – 1.268 mol dm3 = 0.732 mol dm3 [HBr]eqm = 2 1.268 mol dm3 = 2.536 mol dm3 (b) Let x be the number of moles per dm3 of H2 that react.

H2(g) + Br2(g) 2HBr(g) Concentration (mol dm3)

Initial

00.2

0.6= 3.0

00.2

0.4= 2.0

0

Change –x –x +2x Equilibrium 3.0 – x 2.0 – x 2x

Since Kc = eqm2eqm2

2eqm

(g)][Br(g)][H

[HBr(g)]

= )0.2)(0.3(

)2( 2

xx

x

= 12.0

Solving for x, x = 1.5 or 6.0 (rejected) At equilibrium, [H2]eqm = 3.0 – 1.5 mol dm3 = 1.5 mol dm3 [Br2]eqm = 2.0 – 1.5 mol dm3 = 0.5 mol dm3 [HBr]eqm = 2 1.5 mol dm3 = 3.0 mol dm3 26. Let x be the number of moles per dm3 of COCl2 that react.

CO(g) + Cl2(g) COCl2(g) Concentration (mol dm3)

Initial 0 0

0.8

20.0= 0.025

Change +x +x –x Equilibrium x x 0.025 – x

Kc = eqm2eqm

eqm2

(g)][Cl[CO(g)]

(g)][CoCl

Substituting the equilibrium concentrations into the expression,

4.7 109 = 2

025.0

x

x

Since Kc is large, 0.025 – x 0.025

4.7 109 = 2

025.0

x



Solving for x, x = 2.3 106 Equilibrium concentrations are: [CO(g)]eqm = [Cl2(g)]eqm = 2.3 106 mol dm3

[COCl2(g)]eqm = 0.025 mol dm3 27. Let x mol dm3 be the initial concentration of N2O4 and y be the number of moles

per dm3 of N2O4 that react. N2O4(g) 2NO2(g)

Concentration (mol dm3) Initial x 0 Change y +2y Equilibrium x y = x 1.5 103 2y = 3.0 103

Kc = eqm42

2eqm2

(g)]O[N

(g)][NO

0.2 = )105.1(

)100.3(3

23

x

Solving for x, x = 0.00155 Initial concentration of N2O4(g) is 0.00155 mol dm3.

28. Kc = eqm5

eqm2eqm3

][PCl

(g)][Cl(g)][PCl

33.3 = 3

1eqm3

101.29

)10(1.87(g)][PCl

[PCl3(g)]eqm = 0.230 mol dm3




A43.1 (a) (i) The equilibrium position shifts to the right. (ii) The equilibrium position shifts to the left. (iii) The equilibrium position shifts to the left. (b) A: H2(g) is added to the equilibrium mixture. B: HI is removed from the equilibrium mixture. A43.2

1. The reaction quotient, (aq)](aq)][SCN[Fe

(aq)][FeSCNQ

3

2

c

[FeSCN2+(aq)] = 0.2

1020.0mol dm3 = 0.0510 mol dm3

[Fe3+(aq)] = 0.2

0320.0mol dm3 = 0.0160 mol dm3

[SCN(aq)] = 0.2

0600.0mol dm3 = 0.0300 mol dm3

3133

3

c dmmol 106.25)dm mol )(0.0300dm mol (0.0160

dm mol 0.0510Q

Since Qc < Kc, more products (FeSCN2+) will be formed and more reactants (Fe3+ and SCN) will be consumed until Qc = Kc. Hence, the equilibrium position shifts to the right.

2. (a) Let 2x be the number of moles per dm3 of HI that decompose. H2(g) + I2(g) 2HI(g) Concentration (mol dm3) Initial 0 0

10

40.0= 0.040

Change +x +x –2x Equilibrium x x 0.040 – 2x Substituting equilibrium concentration values into the equilibrium law

expression,

0.00443 ,for Solving

)2 (0.04049.5

(g)][I(g)][H

[HI(g)]K

2

2

eqm2eqm2

2eqm

c

xxx

x

At equilibrium, [H2(g)]eqm = 0.00443 mol dm3 [I2(g)]eqm = 0.00443 mol dm3 [HI(g)]eqm = 0.040 – 0.00443 2 mol dm3 = 0.0311 mol dm3 (b) After adding 0.10 mol of HI to the reaction vessel, [HI(g)] = 0.0311 +

10

10.0mol dm3 = 0.0411 mol dm3



86.07 )dm mol )(0.00443dm mol (0.00443

)dm mol (0.0411

(g)](g)][I[H

[HI(g)]Q

33

23

22

2

c

Since Qc > Kc, more product (HI) will be consumed and more reactants (H2 and I2) will be formed until Qc = Kc. Hence, the equilibrium position shifts to the left.

(c) Let 2y be the number of moles per dm3 of HI that decompose. H2(g) + I2(g) 2HI(g) Concentration (mol dm3) Initial 0.00443 0.00443 0.0311 + 0.01

= 0.0411 Change +y +y –2y Equilibrium 0.00443 + y 0.00443 + y 0.0411 – 2y Substituting the new equilibrium concentration values into the equilibrium

law expression,

2

2

eqm2eqm2

2eqm

c

) (0.00443

)2 (0.0411549

(g)][I(g)][H

[HI(g)]K

y

y.

Solving for y, y = 0.00110 At equilibrium, [H2(g)]eqm = [I2(g)]eqm = 0.00443 + 0.00110 mol dm3 = 0.00553 mol dm3 [HI(g)]eqm = 0.0411 – 2 0.00110 mol dm3 = 0.0389 mol dm3 (d) No, the concentration of HI (0.0389 mol dm3) at the new equilibrium

position is still higher than that before the disturbance (0.0311 mol dm3). A43.3 (a) The colour of the mixture becomes bluer, i.e. the equilibrium position shifts to

the right (the product side). As the forward reaction is an endothermic change, an increase in temperature favours endothermic reaction causing an increase in the amount of the blue CoCl4

2(aq) ions. (b) The colour of the mixture becomes more pink, i.e. the equilibrium position shifts

to the left (the reactant side). As the backward reaction is an exothermic change, a decrease in temperature favours exothermic reaction causing an increase in the amount of the pink Co2+(aq) ions.

A43.4 (a) When potassium dichromate is acidified, there is a high concentration of acid,

H+(aq). According to the Le Châtelier’s Principle, the high concentration of H+(aq) shifts the equilibrium position of the system to the right. This leads to the formation of more orange Cr2O7

2(aq) ions. So, acidified potassium dichromate solution is orange in colour.

(b) When sodium hydroxide pellets are added, it neutralizes the acid present. The concentration of H+(aq) is lowered. According to the Le Châtelier’s Principle, the system counteracts the change (removal of H+(aq)) in a direction to replace the removed H+(aq). This is done by shifting the equilibrium position to the left. As more Cr2O7

2(aq) ions are converted to CrO42(aq) ions, the colour of the

solution changes from orange to yellow.



A43.5 (a) A decrease in volume causes an increase in pressure of the gaseous equilibrium

mixture. According to the Le Châtelier’s Principle, the equilibrium position of the system will shift to the side with a decrease in the number of moles of gases. However, there are equal numbers of moles of gases on both sides of the equation. So, a change in volume has no effect on the equilibrium position for this reaction.

(b) It can be concluded that a change in volume has no effect on the equilibrium position for reactions having the same number of moles of gases on both sides of the equation.




Chapter Exercise 1. osition, equilibrium constant, Kc 2. reaction quotient, equilibrium constant

(a) equilibrium, does not (b) product, right (c) reactant, left

3. onstant, osition 4. endothermic, exothermic 5. oncentration, emperature, counteract 6. Le Châtelier’s Principle, shift 7. yield 8. A 9. D 10. A 11. C 12. C 13. C 14. D 15. C 16. B 17. A 18. A 19. D 20. (a) Shift to the right/product side (b) Shift to the left/reactant side (c) Shift to the left/reactant side (d) No shift in equilibrium position (e) Shift to the right/product side 21. (a) Fe3+(aq) is yellow, SCN(aq) is colourless and FeSCN2+(aq) is dark red.

When the mixture becomes darker red, equilibrium position has shifted to the product side. When the mixture becomes paler yellow, equilibrium position has shifted to the reactant side.

(b) CrO42(aq) is yellow and Cr2O7

2(aq) is orange. When the mixture becomes orange, equilibrium position has shifted to the product side. When the mixture becomes yellow, equilibrium position has shifted to the reactant side.

(c) NO2(g) is brown and N2O4(g) is colourless. When the mixture becomes pale brown or yellow, equilibrium position has shifted to the product side. When the mixture becomes darker brown, equilibrium position has shifted to the reactant side.



22. Qc = (g)](g)][HH[C

(g)]H[C

222

42 =4.002.0

102.3 4

mol1dm3 = 0.04 mol1dm3

Since Qc < Kc, the equilibrium position shifts to the product side, i.e. to the right.

23. Qc = (g)](g)][H[CO

O(g)][CO(g)][H

22

2 =0.20.4

0.20.4

= 1.0

Since Qc > Kc, the reaction will proceed to the reactant side, i.e. to the left. 24. Description (1) applies to Figure 24(b). When H2(g) is added, there is a sudden

increase of [H2(g)]. When [H2(g)] increases, the equilibrium is disturbed. Qc is decreased, so the equilibrium position will shift to the right. This will decrease [I2(g)] and increase [HI(g)].

Description (3) applies to Figure 24(c). The graph shows a decrease in [HI(g)] and an increase in both [H2(g)] and [I2(g)]. That is, the equilibrium position has shifted to the left which is an endothermic process. Increase in temperature favours the process. Furthermore, there is no sudden change in the concentrations of all substances, so the pressure of the system has not changed.

25. Low temperatures favour exothermic reaction, i.e. the combination of brown

nitrogen dioxide, NO2(g), to form colourless dinitrogen tetroxide, N2O4(g). That explains the decrease in colour intensity of the brown haze.

High temperatures favour endothermic reaction, i.e. the decomposition of colourless N2O4(g) back to brown NO2(g). So the colour intensity of brown haze increases.

26. Hydroxide ions from sodium hydroxide combine with the hydrogen ions in the

system. H+(aq) + OH(aq) H2O(l) As the concentration of hydrogen ions in the system decreases, the reaction

quotient Qc = (aq)]O[Cr

(aq)][H(aq)][CrO2

72

2224

will be smaller than Kc. As Qc is smaller

than Kc, the equilibrium position shifts to the right forming more CrO42(aq)

which is yellow. 27. A higher concentration of O2 will shift the equilibrium position towards the

products. Thus, CO is released from the haemoglobin and more HbO2 will form.



28. (a) According to the Le Châtelier’s Principle, when there is an increase in [CO(g)], the system shifts in a direction to lower the [CO(g)]. This is done by shifting the equilibrium position to the left which can help to remove some of the added CO.

(b) According to the Le Châtelier’s Principle, when there is an increase in [H2O(g)], the system shifts in a direction to lower the [H2O(g)]. This is done by shifting the equilibrium position to the right which can help to remove some of the added H2O.

(c) Increasing the reaction volume is equivalent to a decrease in pressure. According to the Le Châtelier’s Principle, the system shifts in a direction to increase the pressure of the system again. This is done by producing more molecules. Thus, the equilibrium position shifts to the right as there are larger number of moles of molecules on the right hand side.

(d) According to the Le Châtelier’s Principle, when there is an increase in temperature, the system will shift in a direction to absorb the heat added in order to lower the temperature of the system again. This is done by shifting the equilibrium position to favour the endothermic change, i.e. to the right.



Part X Chemical equilibrium Part Exercise 1. C 2. B 3. C 4. D 5. A 6. C 7. B 8. A 9. B 10. COCl2(g) CO(g) + Cl2(g) Concentration (mol dm3)

Initial 1.0 0 0 Change –0.028 +0.028 +0.028 Equilibrium 1.0 – 0.028 =

0.972 0.028 0.028

0.972

(0.028)

(g)][COCl

(g)][Cl[CO(g)]K

2

eqm2

eqm2eqmc mol dm3 = 8.07 104 mol dm3

11. Kc = [Mg2+(aq)]

eqm[F(aq)] 2

eqm = (2.6 104)(5.2 104)2 M3 = 7.03 1011 M3

12. (a) The forward reaction rate is equal to the backward reaction rate. Both reactants and products are present and their concentrations remain

unchanged.

(b) 2eqm2

eqm22eqm2

c S(g)][H

(g)][S(g)][HK

(c) mol dm3/M (d) Let x be the number of moles per dm3 of H2S that decompose.

2H2S(g) 2H2(g) + S2(g) Concentration (mol dm3)

Initial

1

20.0= 0.20

0 0

Change –2x +2x +x Equilibrium 0.20 – 2x 2x x

2eqm2

eqm22eqm2

c S(g)][H

(g)][S(g)][HK = 2

2

)2 (0.20

)()2(

x

xx

Since equilibrium constant is very small, x is expected to be very small.

Thus, 0.20 – 2x is almost 0.20.



So, 4.2 106 = 2

3

(0.20)

4x

x = 3.5 103 [S2(g)]eqm = 3.5 103 mol dm3 [H2(g)]eqm = 2 3.5 103 mol dm3 = 7.0 103 mol dm3 (e) The equilibrium position shifts to the left. 13. (a)

PCl5(g) PCl3(g) + Cl2(g) Concentration (mol dm3)

Initial 1.0 0 0 Change 0.54 – 1.0

= –0.46+0.46 +0.46

Equilibrium 0.54 0.46 0.46 The changes in concentration of PCl3(g) and Cl2(g) are plotted with time and

their equilibrium concentration = 0.46 mol dm3. (b) PCl5(g) PCl3(g) + Cl2(g) Concentration (mol dm3)

Initial 0 0.50 0.30 Change +0.14 0.36 – 0.50

= –0.14–0.14

Equilibrium 0.14 0.36 0.16 The changes in concentration of PCl5(g) and Cl2(g) are plotted with time and

their equilibrium concentrations = 0.14 mol dm3 and 0.16 mol dm3 respectively.

PCl3 and Cl2

Con

cent

rati

on o

f P

Cl 5

(g)

Time



14. (a)

Temperature (K)

[N2]eqm [H2]eqm [NH3]eqm Kc

500 0.115 0.105 0.439 1448 575 0.110 0.249 0.128 9.6 775 0.120 0.140 0.00439 0.0584

(b) The forward reaction is exothermic. From the table, it is noticed that increasing temperature decreases the value of Kc, i.e. the forward reaction is not favoured. Since increasing temperature favours endothermic reaction, the forward reaction must be exothermic.

15. (a) Kc= 2eqm3

2eqm42

eqm24eqm2

(g)][O(g)]H[C

(g)][OO(g)][CH

(b) It has the same form as equilibrium constant, Kc, but Qc is calculated from concentrations at any particular moment, not necessarily at equilibrium. Qc is used to predict the shift in equilibrium position.

(c) (i) As Qc = 2

32

42

24

2

(g)][O(g)]H[C

(g)][OO(g)][CH, an increase in [C2H4(g)] will make Qc

smaller than Kc. Thus, the equilibrium position shifts to the product side, i.e. to the right.

(ii) As Qc = 2

32

42

24

2

(g)][O(g)]H[C

(g)][OO(g)][CH, a decrease in [O3(g)] will make Qc

greater than Kc. Thus, the equilibrium position shifts to the reactant side, i.e. to the left.

(iii) As Qc = 2

32

42

24

2

(g)][O(g)]H[C

(g)][OO(g)][CH, a decrease in [O2(g)] will make Qc

smaller than Kc. Thus, the equilibrium position shifts to the product side, i.e. to the right.

Cl2

PCl5

Time

Con

cent

rati

on o

f P

Cl 3

(g)



16. At A, there should be a sudden decrease in pressure or an increase in volume of

the system. This change favours the formation of more gas molecules by shifting the equilibrium position to the left producing more C2H4 and H2.

At B, there is an increase in temperature of the system. The lines show that both [C2H4(g)] and [H2(g)] increase while [C2H6(g)] decreases. The equilibrium position of the system has shifted to the left which is an endothermic process. An increase in temperature favours endothermic reaction.

At C, C2H6(g) is added to the system. This makes the reaction quotient greater than the equilibrium constant, i.e. Qc > Kc, the equilibrium position shifts to the reactant side. This lowers [C2H6(g)] and increases [C2H4(g)] and [H2(g)].

17. (a) Addition of hydrochloric acid increases the concentration of Cl(aq) ions,

reaction quotient Qc = (aq)][CuCl

(aq)](aq)][ClO)[Cu(H2

4

4242

will be greater than the

equilibrium constant, Kc. Thus, the equilibrium position shifts to the left producing more CuCl4

2(aq), so the solution becomes yellow in colour. (b) Addition of silver nitrate solution removes Cl(aq) ions in the system by

precipitation with Ag+(aq) ions. Ag+(aq) + Cl(aq) AgCl(s)

As [Cl(aq)] decreases, reaction quotient Qc = (aq)][CuCl

(aq)](aq)][ClO)[Cu(H2

4

4242

will be smaller than the equilibrium constant, Kc. Thus, the equilibrium position shifts to the right producing more Cu(H2O)4

2+(aq), so the solution becomes blue in colour.

18. Addition of HCl(aq) removes OH(aq) in the solution.

H+(aq) + OH(aq) H2O(l) As OH(aq) is removed, the reaction quotient, Qc = [Fe3+(aq)][OH(aq)]3,

becomes smaller than the equilibrium constant, Kc. Thus, the equilibrium position shifts to the right making more Fe(OH)3 to dissolve. Then [Fe3+(aq)] and [OH(aq)] increase and eventually Qc becomes the same as Kc.



19. SO2Cl2(g) SO2(g) + Cl2(g) Concentration (mol dm3)

Initial 0.020 0 0 Change 1.2 102 +1.2 102 +1.2 102 Equilibrium 0.020 1.2 102

= 8 103 1.2 102 1.2 102

eqm22

eqm2eqm2c (g)]Cl[SO

(g)][Cl(g)][SOK =

)108(

)102.1)(102.1(3

22

mol dm3 = 0.018 mol dm3

20. eqm2eqm

eqm2c (g)][Cl[CO(g)]

(g)][COClK

2.9 1010 = )10)(7.310 (1.8

(g)][COCl65

eqm2

[COCl2(g)]eqm = 3.81 mol dm3 No. of moles of COCl2 = 3.81 6 mol = 22.86 mol Mass of COCl2 = 22.86 (12.0 + 16.0 + 35.5 2) g = 2263 g 21. (a) The equilibrium position does not shift to any direction, so there is no

change in yield of CH4. (b) The equilibrium position shifts to the right, so the yield of CH4 increases. (c) The equilibrium position shifts to the right, so the yield of CH4 increases. (d) The equilibrium position shifts to the left, so the yield of CH4 decreases.

hkdse chemistry – book 4a ans

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