8 Chapter Chemical Equilibrium Text Book Exercise

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CHAPTER 8

CHEMICAL EQUILIBRIUM

TEXT BOOK EXERCISE

Q1. Multiple Choice questions(i) For which system does the equilibrium constant, Kchas units of

(concentration)-1

(a) N2+ 3H2 2NH3

(b) H2+ I2 2HI

(c) 2NO2 N2O4

(d) 2HF H2+ F2(ii) Which statement about the following equilibrium is correct

2SO2+ O2 2SO3

(a) The value of Kpfalls with a rise in temperature

(b) The value of KPfalls with increasing Pressure

(c) Adding V2O5catalyst increases the equilibrium yield of

SO3

(d) The value of Kpis equal to Kc

(iii) The pH of 10-3

mol/dm3

of an equous solution of H2SO4is(a) 3 (b) 2.7 (c) 2 (d) 1.5

(iv) The solubility product of AgCI is 2 x 1010

mol2dm

-6. The

maximum concentration of Ag+ ions in the solution is

(a) 2 x 10-10

mol dm-3

(b) 1.41 x 10-10

mol dm-3

(c) 1 x 10-10

mol dm-3

(d) 4 x 10-20

mol dm-3

(v) An ecxcess of aqueous silver nitrate is added to aqueous bariumchloride and precipitate is removed by filtration. What are the

main ions in the filtrate?

(a) Ag+and NO3

-1only

(b) Ag+and Ba

2+and NO3

-1

(c) Ba2+

and NO3-1

only


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(d) Ba2+

and NO3-1

and CI-1

Ans. i)c ii)a iii)b iv)b v)b

Q2. Fill in the blanks(i) Law of mass action states that the rate at which a reaction

proceeds is directly proportional to the product of the active

masses of the ______.

(ii) In an exothermic reversible reaction, _________in temperature

will shift the equilibrium towards the forward direction.

(iii) The equilibrium constant for the reaction 2O3 3O2is 1055

at

25oC, it tells that ozone is _______at room temperature.

(iv) In a gas phase reaction, if the number of moles of reactants are

equal to the number of moles of the products, KCof the reaction is______to the Kp.

(i) Buffer solution is prepared by by mixing together a weak

base and its salt with _______ or a weak acid and its salt with

________.

Ans. i)reactants ii)decrease iii)unstable

iv)equal v)strong acid, strong base

Q3. Label the sentences True or False

(i) When a reversible reaction attains equilibrium bothreactants and products are present in a reaction mixture.

(ii) The Kcof the reaction

A+B C+D is given by

Kc=

Therefore it is assumed that [A] = [B]=[C]=[D]

(iii) A catalyst is a compound, which increases the speed of the

reaction and consequently increases the yield of the product.

(iv) Ionic product Kwof pure water at 25oC is 10-14dm-6and is

represented by an expression

Kw=[H+][OH

-]=10

-14mol

2dm

-6

(v) AgCI is a sparingly soluble ionic solid in water. Its solution

produces excess of Ag+

and CI-ions.


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Ans. i)True ii)False

iii)False iv)True v)False

Q4. (a) Explain the terms reversible reaction and state of

equilibrium

See Section 8.1 and 8.1.2

(b) Define and explain the law of mass action and drive the

expression for the equilibrium constant KC.See Section 8.1.3

(c) Write KCfor the following reactions

(i) Sn2+(aq) + 2Fe3+(aq) Sn4+(aq)+ 2Fe2+(aq)(ii) Ag+(aq) +Fe

2+(aq) Ag(s)+ Fe

3+(aq)

(iii) N2(g)+ O2(g) 2NO(g)

(iv) 4NH3(g) 5O2(g) 4NO(g)+6H2O(g)

(v) PCI5(g) PCI3(g) +CI2(g)

(i) Kc=

(ii) Kc=

(iii) Kc=

(iv) Kc=

(v) Kc=

Q5. (a) Reversible reactions attain the position of equilibrium,

which is dynamic in nature and not static. Explain it.See Section 8.1.1

(b) Why do the rates of forward reactions slow down when

a reversible in nature and not static. Explain it.

See Section 8.1.1Q6. When a graph is plotted between time on X-axis and the

concentration of reactants and products on Y-axis for a

reversible reaction, the curve becomes parallel to time axis at a

certain stage.

(a) At what stage the curves become parallel?


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(b) Before the curves become parallel, the steepness of

curves falls? Give reasons.

(c) The rate of decrease of concentrations of the reactantsand rate of increase of concentrations of any of products

may or may not be equal for various types of reactions

before the equilibrium time. Explain it.See section 8.1.1

Q7. (a) Write down the relationship of different types of

equilibrium constants. i.e. Kcand Kpfor the following general

reactions.

aA + bB cC +dD

See section 8.1.2(b) Decide the comparative magnitudes of Kc, Kpfor the

following reactions.Synthesis of NH3

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

KCis given by

Kc=

Kpis given byKp=

For this reaction, change in number of moles is given by

n =number of mole of product-number of moles of reactants

=2(1+3)=-2

Hence

Kp=Kcx (RT)-2

Or Kp= Kcx

Thus if T is such that RT > 1, then Kp< Kc

If T is such that RT< 1, then Kp> Kc

Dissociation of PCI5

PCI5(g) PCI3(g)+ CI2(g)


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Kcis given by according to law of Mass of Action

Kc=

Kpis given by

Kp=

For this reaction, change in number of moles is given by

n =number of mole of product-number of moles of reactants

=21 =1

Kp=Kcx (RT)

Thus If T is such that RT > 1, then Kp> Kc

If T is such that RT< 1, then Kp


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(b) The change of temperature disturbs both the

equilibrium position and the equilibrium constant of a

reaction.(c) The solubility of glucose in water is increased by

increasing the temperature.

Q11. (a) What is ionic product of water? How does this value

vary with the change in T? Is it true that this value is 75 times

when the T of water increased form 0oC to 100

oC.

Ionic Product of water is given by the equation

Kw[H+][OH

-]

Value of Kwincreases with increase in temperature. It is

because increase in temperature increase the ionization of H2O.Thus, more H

+or OH

-ions are produced. Hence value of Kw

increase.

e.g.

At 25oC Kw=1 x 10

-14

and At 100

oC Kw=7.5 x 10

-14

Further

At 0oC (Kw)o =0.1 x 10

-14_____(1)

At 100o

C (Kw)100=7.5 x 10-14

_____(2)Divide eq (2) by eq (1)

= =75

or (KW) 100=75 x (Kw)0

Hence, KW at 100oC is 75 times more than at 0

oC

(b) What is the Justification for the increase of ionic

product with temperature?

Value of Kwincrease with increase in temperature. It isbecause increase in temperature increase the ionization of

H2O. Thus, more H+

or OH-ions are produced. Hence value

Kwincreases.

(c) How do you prove that at 25O

C in 1 dm3of water, there

are 10-7

moles of H3O+?


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At 25OC

Kw=[H3O+][OH

-14____(1)

Since ionization of water gives equal no. of H3O

+

and OH

-

ions

therefore

[H3O+]=[OH

-]

Hence, eq (1) can be written as

Kw=[H3O+][H3O

+]=10

-14

Or [H3O+]

2=10

-14

Taking square root on both sides

[H3O+]

2=10

-7mol/dm

3

Hence, at 25oC, water has 10-7mole/dm3of H3O+ions.Q12. (a) Define pH and pOH. How are they related with pKw.

(b) What happens to the acidic and basic properties of

aqueous solutions when pH varies form 0 to 14.

(c) Is it true that the sum of pHaand pKbis always equal to

14 to all temperatures for any acid? If not why?

Q13. (a) What is Lowry bronsted idea of acids and bases?

Explain conjugate acids and bases.

(b) Acetic acid dissolves in water and gives proton to water.But when dissolves in H2SO4, it accepts proton. Discuss

the role of acetic acid in both cases.

CH3COOH + H2O CH3COO-+ H3O

+

However, H2SO4is a stronger acid than acetic acid, therefore,

H2SO4donates proton and acts as an acid while acetic acid accepts

proton and acts as a base.

H2SO4+ CH3COOH HSO4-+ CH3COOH2

+

Q14. In the equilibrium PCI5(g) PCI3(g) + CI2(g) H=90 kJ/mol

(a) The position of equilibrium

(b) Equilibrium constant?

If (i) Temperature is increased

(ii) Volume of the container is decrease.


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(iii) Catalyst is added

(iv) CI2is added

See section 8.2Explain your answer.

Q15. Synthesis of NH3by Habers process is an exothermic

reaction.

N2 + 3H2 2NH3 H=92.46 kJ/ mol

(a) What should be the possible effect of change of

temperature at equilibrium stage?

(b) How does the change of pressure or volume shifts the

equilibrium position of this reaction?

(c) What is the role of the catalyst in this reaction?(d) What happens to equilibrium position of this reaction

if NH3is removed form the reaction vessel form time to time.See Section 8.2.1

Q16. Sulphuric acid is the king of chemicals. It is produced by the

burning of SO2to SO3through an exothermic reversible

process.

(a) Write the balanced reversible reaction

(b) What is the effect of pressure change on this reaction?(c) Reaction is exothermic but still the temperature of

400-500oC is required to increase the yield of SO3. Give

reasons.See Section 8.2.2

Q17. (a) What are buffer solutions? Why do we need them in

daily life?See Section 8.2

(b) How does the mixture of sodium acetate and acetic acidgive us the acidic buffer?See Section 8.7

(c) Explain that a mixture of NH4OH and NH4CI gives us

the basic buffer?See Section 8.7


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(d) How do you justify that the greater quantity of

CH3COONa in acetic acid decrease the dissociating

power of acetic acid and so the pH increases.CH3COOH is a weak acid and ionizes very small, while

CH3COONa is a strong electrolyte and it ionizes in water to

greater extent and provides acetate ions.

CH3COOH +H2O CH3COO-+ H3O

+

CH3COONa CH3COO-+ Na

+

Thus CH3COONa decreases the ionization of

CH3COOH due to common CH3COO-ion and pH of solution

increases.

(d) Explain the term buffer capacity.See Section 8.7.1

Q18. (a) What is the solubility product? Derive the

solubility product expression for sparingly soluble compounds,

AgCI, Ag2CrO4and PbCI2.See Section 8.8

(b) How do you determine the solubility product of a

substance when its solubility is provided in grams/100 g

of water?See Section 8.8

(c) How do you calculate the solubility of a substance from

the value of solubility product.

See Section 8.8

Q19. Kcvalue for volume for the following reaction is 0.076 at

520oC

2HI H2 + I2

Equilibrium mixture constants [HI] =0.08 M, [H2] =0.01 M.To this mixture more HI is added so that its new concentration is

0.096 M. what will be the concentration of [HI], [H2] and [I2] when

equilibrium is re-established.

Solution

2HI H2 + I2


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Equilibrium conc. 0.08 0.01 0.01

(mol/dm3)

Initial conc.

0.096 0.01 0.01After adding more HI(mol/dm3)

Equilibrium conc.

0.0962x 0.01 0.01When equilibrium is re-established

(mol/dm3)

According to law of mass action

Kc=

Kc= =0.016

= =0.016

Taking square root on both sides

= =0.126

0.01 + x=0.126 (0.0962x)0.01 + x =0.01210.252 x

x + 0.252 x=0.01210.01

1.252 x=0.0021

x ==

x=0.00168 mol/dm-3

Thus

Concentrations when equilibrium is re-established are[H2] =0.01 + x =0.01 +0.00168 =0.01168 mol dm-3

[I2] =0.01 + x =0.01 +0.00168 =0.01168 mol dm-3

[HI] =0.096 - 2 x =0.096 -2x0.00168 =0.0926 mol dm-3

Q20. The equilibrium constant for the reaction between acetic and

ethyl alcohol is 4. A mixture of 3 moles of acetic acid and one


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mole of C2H5OH is allowed to come to equilibrium. Calculate

the amount of ethyl acetate at equilibrium stage in number of

moles and grams. Also, calculate the masses of reactants leftbehind.

Solution

CH3COOH + C2H5OH CH3COOC2H5+H2OInitial conc.

3 1 0 0(mol/dm3)

Equilibrium conc.

3- x 1x x x(mol/dm3)

According to law of mass action

Kc=

Kc=

X2

=4(3x) (1x)

X2

=4(33x) (1x)

X2

=4(34x + x2)

X2

=1216x + 4 x2)

Or 1216x + 4x

2

x

2

=03x216x + 12=0

It is quadratic equation and can be solved by using quadric

formula

Here a = 3 , b = 12 , c = 12

Thus

x=

x=

x=

x=


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x=

Either x= or x=

x= 4.43 mol dm-3

or x= 0.9 mol dm-3

x= 4.43 is not possible as it is greater than the initial

concentrations of reactants, therefore , x= 0.9 mol dm-3

Therefore

Moles of ethyl acetate =x=0.9 moles

Mass of ethyl acetate = 0.9x88=79.46g.

Moles of water =x=0.9 moles

Mass of water =0.9 x 18=16.2 g.Moles of acetic acid = 3x =30.9 =2.1 moles

(left behind)

Moles of ethyl alcohol = 1x =10.9= 0.1 moles

(Left behind)

Mass of ethyl alcohol =0.1 x 46=4.6g.

(left behind)

Q21. study the equilibrium

H2O + CO H2+ CO2

(a) Write an expression of Kp

Kp=

(b) When 1 mole of steam and 1 mole of CO are allowed to

reach equilibrium, 33.3% of the equilibrium mixture is

hydrogen. Calculate the value of Kp. state the units of Kp.

Solution

H2O + CO H2+ CO2Initial conc.

1 1 0 0(mol/dm3)

Equilibrium conc. 1x 1x x x

(mol/dm3)

Total no. of moles at equilibrium =1x + 1x + x + x=2

Hence


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% of H2=

33.3=

or no of moles of H2 =0.67 moles

Hence

At equilibrium

Moles of H2=x=0.67 moles

Moles of CO2=x=0.67 moles

Moles of H2O =1x = 1 - 0.67= 0.33 moles

Moles of CO =1x = 1 - 0.67= 0.33 moles

Hence, Kcis given as

Kc=

Kc=

Since Kp=Kc(RT) and n =n products -nrectants =0, therefore

Kp=Kc=4

Q22. Calculate the pH of

(a) 10-4moles/ dm3of HCI

HCI ionizes as

HCI H+

+ CI-

Since HCI is a strong acid, and it is 100% dissociated. Hence 10-

4mol/dm

3of HCI produces 10

-4mol/dm

3of H

+ions.

Thus

[H+] = 10

-4mol/dm

3

So

pH = - log [H+]pH = - log [10

-4]

pH = 4

(b) 10-4

moles/dm3of Ba(OH)2

Ba(OH)2 ionizes as

Ba(OH)2 Ba+2

+ 2 OH-


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Since Ba(OH)2is a strong base it is 100% dissociated.

Hence 10-4

mol/dm3of Ba(OH)2produces 2 x 10

-4mol/dm

3of OH

-ions.

Thus [OH-]= 2 x 10

-4mol/dm

3

So

pOH= - log[OH-]

pOH= - log[2 x 10-4

]

pOH= 3.699

Since

pH + pOH=14

Therefore

pH= 14pOH=143.699 =10.301

(c) 1 mol/dm3of H2X, which is 50% dissociated

H2X ionizes as

H2X 2H+

+ X-

1mole of H2X produces 2 moles of H+ions if 100% dissociated

However, since H2X is 50% dissociated therefore 1 mole of H2X

produces 1 mole of H+ion

Thus[H

+]=1 mol/dm

3

So

pH= -log [H+]

pH= -log [1]

pH= 0

(d) 1 mol/dm3of NH4OH which is 1% dissociated

NH4OH ionizes as

NH4OH NH4+

+ OH

-

It shows that 1 mole of NH4OH produces 1 mole of OH-

ions.

NH4OH is only 1% dissociated

Hence


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% dissociation = x 100

1 = x100

mol of OH- x1=0.01 mol/dm

3

Thus

[OH-] =0.01 mol/dm

3

So

pOH = - log [OH-]

pOH = - log [0.01]

pOH = 2Since

pH + pOH =14

Therefore

pH=14pOH

=142=12

Q23. (a) Benzoic aicd C6H5COOH is weak mono-basic acid

(Ka=6.4 x10-5

mol/dm3). What is the pH of a solution

containing 7.2 g of sodium benzoate in one dm3of o.02

mol/dm3benzoic acid.Mass of sodium benzoate =7.2 g /dm

3

Formula of

sodium

benzoate is

C6H5COONa

Mol. Mass of sodium benzoate =144 g/mol

Moles of sodium benzoate = =0.05 mol/dm3

Moles of benzoic acid =0.02 mol/dm3

Ka of benzoic acid 6.4 x 10-5

mol/dm3

Thus pKa= - log Ka= - log (6.4 x10-5

) =4.2

Hence, according to Hendersons eq.


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pH =pKa + log

pH =4.2 + logpH = 4.2 + 0.39

pH =4.59

(b) A buffer solution has been prepared by mixing 0.2 M

CH3COONa and 0.5 M CH3COOH in 1 dm3of solution.

calculate the pH of solution. pKaof acid is 4.74 at 25oC.

How the value of pH will change by adding 0.1 mole 0.1

mole of NaOH and 0.1 mole mol HCI respectively.

Solution[CH3COOH] =0.5 M

[CH3COONa] =0.2 M

pKa of CH3COOH =4.74

pH =?

Since pH =pKa+ log

or pH =pKa+ log

pH =4.74 + log

pH =4.74 - 0.4

pH =4.34

When 0.1 mole of NaOH is addedNaOH is strong base. It dissociates completely. Therefore, it

produces 0.1 moles of OH-ions. Thus, 0.1 moles of OH

-ions

reacts with 0.1 moles of CH3COOH. Hence, out of 0.5 moles of

CH3COOH, 0.4 moles of CH3COOH are behind.On the other hand, due to slat formed by the neutralization

reaction, conc. of salt (CH3COONa) is increased from 0.2 moles

to 0.3 moles.

Hence, new conc. will be


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[CH3COOH] =0.4 M [CH3

COONa] =0.3M

Thus pH =pKa+ log

pH =4.74 + log

pH =4.74 - 0.12

pH =4.62

Addition of 0.1 mole of HCIHCI is a strong acid. It dissociates completely. Therefore, it

produces 0.1 moles of H+ions. Thus, 0.1 moles of H

+ions react

with 0.1 moles of CH3COO-

ions. Hence, out of 0.2 moles of salt(CH3COO

-Na

+), 0.1 moles of salt are left behind.

On the other hand, conc. of acid (CH3COOH) is increased

form 0.5 moles to 0.6 moles.

Hence, new conc. will be

[CH3COOH] =0.6 M [CH3

COONa] =0.1M

Thus pH =pKa+ log

pH =4.74 + log

pH =4.74 - 0.78

pH =3.96

(See Section 8.7.1 for complete understanding of this numerical)

Q24. Solubility of CaF2in water at 25O

C is found to be 2.05 x 10-4

mol dm-3

. What is the value of Kspat this T.Solubility of CaF2= 2.05 x 10

-4mol dm

-3

According to balanced chemical eq.CaF2(aq) Ca

2+(aq) + 2F

-(aq)

At initial stage 2.05 x 10-4

0 0

(mol/dm3)

After solubility 0 2.05 x 10-4

2x2.05 x 10-4

Hence Ksp=[Ca+2

][F-]


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Ksp=[0.05 x 10-4

][2 x 2.05 x 10-4

]2

Ksp=3.446 x 10-11

mol3dm

-9

Q25. The solubility product of Ag2CrO4is 2.6 x 120

-2

at 25

o

C.Calculate the solubility of the compound.Kspof Ag2CrO4=2.6 x 10

-2

We know

Ag2CrO4(aq) 2Ag+

CrO42-

(aq)Initial stage (mol/dm3)

Ag2CrO4 0 0At equilibrium (mol/dm3)

Ag2CrO4 2S S

Hence

Ksp =[Ag+]2[CrO42-]Ksp=[2S]

2[S] =2.6 x 10

-2

4[S]3=2.6 x 10

-2

[S]=

or [S]=0.1866 mol/dm3

Hence at equilibrium

[Ag+] =2 x 0.1866 mol/dm

3=0.3732 mol/dm

3

and [CrO42-

] =0.1866 mol/dm3

Since

1 mole of Ag2CrO4gives 1 mole of CrO42-

ions, hence

Solubility of Ag2CrO4=[CrO42-

]=0.1866 mol/dm3

8 Chapter Chemical Equilibrium Text Book Exercise

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