Equilibrium

1) A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature.

The volume of the container is suddenly increased.

a) What is the initial effect of the change on vapour pressure?

b) How do rates of evaporation and condensation change initially?

c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

Solution

a) Due to increase in the volume, there will be decrease in the vapour pressure of the liquid because vapours now would be disturbed over larger space, that is due to increase in volume.

b) The rate of evaporation will increase due to the increase in the surface area of liquid and hence more space is available for the molecules to escape. The rate of condensation will decrease due to decrease in the number of vapour molecules in unit volume with increase in volume.

c) Finally, the equilibrium will be restores when the rates of forward and backward reactions become equal . The vapour pressure will remain same as it is dependent on the temperature and not on the volume of the container.

2) What is KC for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60 M, [O2]=0.82 M and [SO3]=1.90 M?

2SO2 (g) + O2 ⇌ 2SO3(g)

Solution

2SO2 (g) + O2 ⇌ 2SO3(g)

The equilibrium constant , KC , for the reaction is

= [1.90 M]2/ [0.60 M]2 [0.82 M]

= 3.61 M2/0.36 x 0.82 M3

= 12.229 M-1

KC =12.229 M-1

3) At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms

I2 (g) ⇌ 2I(g)

Calculate Kp for the equilibrium.

Solution

Given

P, total pressure at equilibrium = 105 Pa

P1 , partial pressure of iodine atoms= 40% of 105

= 40/100 x 105

= 4 x 104 Pa

So, PI2 , partial pressure of iodine molecules= 60 % of 105 Pa

= 60/100 x 105

= 6 x 104 Pa

So, Kp = p21/pI2

= (4x 104 Pa)2/ 6 x 104 Pa = 16 / 6 x 104

Kp = 2.67 x 104 Pa

4) Write the expression for the equilibrium constant , Kc for each of the following reactions:

i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g)

ii) 2Cu[NO3]2 (s) ⇌ 2CuO (s) + 4NO2(g) + O2 (g)

iii) CH3COOC2H5(aq) + H2O ⇌ CH3COOH (aq) +C2H5OH(aq)

iv) Fe3+ (aq) + 3OH- (aq) ⇌ Fe (OH)3(s)

v) I2 (s) + 5F2 ⇌2IF5

Solution

i) Kc = [NO(g)]2 [Cl(g)]/ [NOCl(g)]2

ii) Kc =[NO(g)]4 [O2(g)]

Here, [(Cu(NO3)2(s)] = [CuO(s)] = 1

iii) Kc = [CH3COOH (aq)] [C2H5OH (aq)]/ [CH3COOC2H5(aq)]

Here, [H2O (l)]=1

iv) Kc = 1/[ Fe3+ (aq)] [OH- (aq) ]3 Here, [Fe (OH)3(s)]=1

v) Kc = [IF5]2 / [F2(g)]

5

Kc = 1/ [F2(g)]5 Here, [I2(s)]=1 [IF(l)]= 1

5) Find out the value of Kc for each of the following equilibria from the value of KP:

i) 2NOCl(g) ⇌ 2NO (g) + Cl2(g) ; Kp=1.8 x 10-2 at 500 K

Solution

Given,

Kp = 1.8 x 10-2

T= 500 K

For reaction , ∆ng =3-2 =1

We know,

Kp =KC. (RT) ∆ng

∴ KC = Kp / ((RT) ∆ng

= 1.8 x 10-2 atm/ (0.0821 L atm mol-1 K-1 x 500 K)1

= 1.8 /41 x 10-2

= 180/41 x 10-4

Kp = 4.4 x 10-4

ii) CaCO3(s) ⇌ CaO (s) + CO2 (g) ; Kp =167 at 1073 K

Solution

KC = 167 atm

T = 1073 K

For reaction , ∆ng = 1+0 -0 =1

So,

Kp = KC .RT

KC = Kp / RT

= 167 atm/ (0.0821 L atom mol-1K-1 x 1073 K )-1

= 167 / 0.0821 x 1073 mol L-1

KC = 1.8957 mol L-1

6) For the following equilibrium , Kc = 6.3 x 1014 at 1000 K

NO (g) + O3 (g) ⇌ NO2(g) + O2

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is KC , for the reverse reaction?

Solution

For a reaction at its equilibrium , the value of K’C of reverse direction is inverse of its KC

in forward direction.

K’C =1/ KC

Given.

KC =6.3 x 1014

So,

K’C =1/6.3 x 1014

=1/6.3 x 10-14

K’C = 1.59 x 10-15

7) Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Solution

Equilibrium constants of reactions are expressed in terms of molar concentration of reactants and products.

Molar concentration= Number of moles/volume

= Mass/ Molar mass x 1/volume

= Mass/ volume x 1/Molar mass

= Density / Molar mass

Because , the density of a pure liquid and a pure solid are constant and the molar mass of a substance is also constant . Hence, pure solids and pure liquids , are ignored in expressing equilibrium constant.

8) Reaction between N2 and O2 takes place as follows:

2N2(g) + O2(g) ⇌ 2N2O(g)

If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which KC= 2.0 x 10-37 , determine the composition of equilibrium mixture.

Solution

Let x moles of N2 (g) take part in the reaction. From the equation x/2 moles of O2(g) will react to form x moles of N2O (g).

2N2 (g) + O2(g) → 2N2O (g)

initially 0.482/10 0.933/10 0

At equilibrium 0.482-x/10 0.933-x/10 x/10

Let x moles of N2 reacts at the position of equilibrium

Now, from law of mass action,

Since, KC is very small, very small amount of reactanst would have reacted , i.e., x is very small , and can be neglected.

0.482-x ≈0.482

0.933 – x ≈ 0.933

Then, substituting in the above equation for KC we have

2.0 x 10-37 = (x/10)2/(0.482/10)2 (0.933/10)

= 0.01 x2/ 2.1676 x 10-4

x2 = 43.35 x 10-40

x = √(43.352 x 10-40 ) = 6.6 x 10-20

As x is very small, it can be neglected.

So,

Thus at equilibrium,

[N2]= 0.482/10 = 0.0482 M

[O2]=0.933/10 = 0.0933 M

[N 2O]=x/10 = 6.6 x 10-20 / 10

= 6.6 x 10-21 M

9) Nitric oxide with Br2 and gives nitrosyl bromide as per reaction given below:

2NO (g) + Br2 (g) ⇌ 2NOBr(g)

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium . Calculate equilibrium amount of NO and Br2 .

Solution

2NO + Br2 (g) ⇌ 2NOBr(g)

2 moles 1 mole 2 mole

2 moles of NO react with every one mole of Br2 to give 2 moles of NOBr

So, At equilibrium

nNOBr = 0.0518 mole

So, nNO = 0.0518 mole

nNO = 0.087 (initially)

nNO left at equilibrium = 0.087 – 0.0518 = 0.0352 mol

10) At 450 K , KP = 2.0 x 1010 / bar for the given reaction at equilibrium.

2SO2 (g) + O2(g) ⇌ 2SO3(g)

What is Kc at this temperature?

Solution

Given,

KP = 2.0 x 1010 / bar

And for reaction,

2SO2 (g) + O2(g) ⇌ 2SO3(g)

∆ng =2-3 = -1

So, KP = Kc .RT ∆ng

∴ Kc = KP / RT ∆ng

Kc = KP / RT -∆ng

Substituting we have,

Kc = (2.0 x 1010 bar-1) x [ ( 0.0821 L bar K-1 mol-1) x (450 K) ]

= 7.47 X 1011 mol-1 L = 7.47 X 1011 M-1

Kc =7.47 X 1011 M-1

11) A sample of HI (g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI (g) is 0.04 atm. what is Kp for the given equilibrium?

2HI ⇌ H2 (g) + I2(g)

Solution

2HI(g) ⇌ H2 (g) + I2(g)

Given,

PHI =0.04 atm

PH2 =0.08 atm

PI2=0.08 atm

So,

Kp = PH2 x PI2/ P2

HI

= (0.08 atm) (0.08 atm)/(0.04 atm)2

= 64 x 10-4 atm2/ 16 x 10-4 atm2

= 4

Kp = 4

12) A mixture of 1.57 mol of N2 , 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction at 500 K . At this temperature , the equilibrium constant, KC for the reaction

N2 (g) + 3H2(g) ⇌ 2NH3(g) is 1.7 x 102

Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction?

Solution

Concentration of N2 , [N2]= 1.57 /20 M

Concentration of H2 , [H2]= 1.92 /20 M

Concentration of NH3, [NH3]= 8.13 /20 M

The reaction quotient, QC , of the reaction is QC = [NH3]2/ [N2] [H2]

3

= (8.13/20)2 ⧸(1.57/20) (1.92/20)3

= (8.13/20)2 ⧸(1.57/1.92) x 400

= 66.0969/11.12 x 400

= 2377.58

QC = 2.377 x 103

As QC ≠ Kc it shows that the reaction is not in a state of equilibrium.

Since, QC > Kc , the reaction is not at equilibrium and the reaction will proceed in backward reaction.

13) The equilibrium constant expression for a gas reaction is ,

Kc = [NH3]4 [O2]

5/ [NO]4[H 2O]6

Write the balanced chemical equation corresponding to this expression.

Solution

The balanced chemical equation corresponding to the equilibrium expressions is

4NO (g) + 6H2O⇌ 4NH3(g) + 5O2(g)

14) One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40 % of water (by mass) reacts with CO according to the equation,

H2O (g) + CO(g) ⇌ H2 (g) + CO2(g)

Calculate the equilibrium constant for the reaction.

Solution

Number of moles of water present initially = 1 mole

% of water reacted= 40 %

Number of moles of water reacted = 1x 40/100 = 0.40 mol

15) At 700 K, equilibrium constant for the reaction:

H2 (g) + I2(g) ⇌ 2HI (g)

is 54.8 . If 0.5 mol L-1 of HI (g) is present at equilibrium at 700 K , what are athe concentration of H2 (g) and I2(g) assuming that we initially started with HI (g) and allowed it to reach equilibrium at 700 K?

Solution

The equilibrium constant KC is given as KC = [HI]2 / [H2] [I2]

The reverse reaction can be written as

HI (g) ⇌ H2 (g) + I2(g)

For reverse reactions

K’C =1/KC = 1/54.8

K’C = [H2(g)] [I2(g)]⧸ [HI(g)]2

Let , [H2] and [I2]= x

Then,

1/54.8 = [x ] [x ]/ [0.5 mol L-1 ]2

x = (0.25/54.8 mol2 L-2 )1/2

= 0.0675 mol L-1

[H2] and [I2]= 0.0675 mol L-1 each

16) What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2ICl (g) ⇌ I2 (g) + Cl2 (g) ; Kc = 0.14

Solution

Let the molar concentration of I2 (g) and Cl2 (g) is x mol L-1 at equilibrium

2ICl (g) ⇌ I2 (g) + Cl2 (g)

Initially 0.78 M 0 0

At equilibrium (0.78-2x) x x

Kc = [ I2 ] [Cl2 ] / [ICl]2

= x X x / (0.78 – 2x )2

Kc = x2 /(0.78- 2x)2

0.14 = x2 /(0.78- 2x)2

x /(0.78- 2x) = (0.14)1/2 = 0.374

So, x= 0.374 x 0.78 – 0.374 x 2x

x = 0.292 – 0.748 x – x= 0

1.748 x = 0.292

x= 0.292/1.748= 0.167

[ICl] = (0.78-2x) = (0.78 – 0.334)

= 0.446 M

[ I2 ] = x = 0.167 M

[Cl2 ] = x= 0.167 M

17) Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C2H6 (g) ⇌ C2H4 (g) + H2(g)

Solution

C2H6 (g) ⇌ C2H4 (g) + H2(g)

Initially 4 atm 0 0

At equilibrium 4-x atm x atm x atm

Here, x is the change in pressure of C2H6 at equilibrium

Now,

KP = PC2H4 x PH2 / PC2H6

= 0.04 = x X x/4-x

= 0.16 -0.04 x = x2

x2 + 0.04 x – 0.16 =0

x= -0.04 ± √ [(0.04)2 + 4 x 1 x 0.16] / 2 x 1

= -0.04 ± √ ( 0.0016 + 0.64) / 2

= -0.04 ± 0.801/ 2

x= -0.4205 (not possible)

x = 0.3805

So,

PC2H6 = 4-x

4 – 0.3805

PC2H6 = 3.6195 atm

18) Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as :

CH3COOH (l) + C2H5OH(l) ⇌ CH3COO C2H5 (l) + H2O(l)

i) Write the concentration ratio (reaction quotient), Qc , for this reaction (note: water is not in excess and is not a solvent in this reaction)

ii) At 292 K , if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

Solution

i) The concentration ratio , Q, for the reaction is given as

Q= [CH3COO C2H5 ] +[ H2O]/ [ CH3COOH] [C2H5OH ]

ii) Initial concentration of CH3COOH = 1.0 mol

Initial concentration of C2H5OH = 0.18 mol

At equilibrium ,

concentration of CH3COOH = 0.171 mol

So, concentration of H2O =0.171 mol

At equilibrium

concentration of CH3COOH = 1-0.171 =0.829 mol

concentration of C2H5OH = 0.18 -0.171 = 0.009 mol

Now, by applying law of mass action

KC= [CH3COO C2H5 ] [ H2O]/ [ CH3COOH] [C2H5OH ]

= 0.171 X 0.171/ 0.829 X0.009

= 171 X 171/ 829 X 9 = 3.92

KC= 3.92

iii) Initially

concentration of CH3COOH = 1.0 mol

concentration of C2H5OH = 0.5 mol

At equilibrium

concentration of CH3COO C2H5 = 0.214 mol

concentration of H2O = 0.214 mol

So, at equilibrium

concentration of CH3COOH = 1-0.214 = 0.786 mol

concentration of C2H5OH = 0.5 - 0.214 = 0.286 mol

Now, by applying law of mass action,

QC = [CH3COO C2H5 ] [ H2O]/ [ CH3COOH] [C2H5OH ]

= 0.214 x 0.214/ 0.786 x 0.286

= 214 x 214 / 786 x 786= 0.204

QC = 0.204

Since, QC < KC the equilibrium has been reached . The reactants continue to take part in the reaction to form products.

19) A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained , concentration of PCl5 was found to be 0.5 x 10-1 mol L-1 .

If value of KC is 8.3 x 10-3 , what are the concentrations of PCl3 and Cl2 at equilibrium?

PCl5 (g) ⇌ PCl3 + Cl2

Initially x moll-1 0 0

At equilibrium 0.05 mol L-1 (X -0.05) (X-0.05)

Given , KC = 8.3 X 10-3 = 0.0083

Now, KC =[ PCl3 ] [Cl2]/ [PCl5 ]

0.0083 = (x-0.05) (x-0.05)/0.05

(x-0.05)2 =0.0083 x 0.05

= 4.15 x 10-4

(x – 0.05) = √ (4.15 x 10-4)

= 2.037 x 10-2

= x – 0.05 = 0.02037

∴ x= 0.02037 + 0.05

= 0.07037 mol L-1

So, at equilibrium , [ PCl3 ]= 0.07037 – 0.05

= 0.02037 mol L-1

[Cl2] = 0.07037 – 0.05 = 0.02037 mol L-1

20) One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2 .

FeO (s) + CO (g) ⇌ Fe (s) + CO2 (g) ; KP =0.265 atm at 1050 K

What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are : Pco = 1.4 atm and CO2 p= 0.80 atm?

Solution

FeO (s) + CO (g) ⇌ Fe (s) + CO2 (g)

Initially 0 1.4 atm 0 0.80 atm

At equilibrium 0 1.4 –x atm 0 (0.80 +x) atm

Now, KP = pCO2 / Pco

KP = 0.80 +x/ 1.4 –x

0.265 = 0.80 +x/1.4 –x

= 0.371 – 0.265 x = 0.80 +x

= 1.265 x =-0.429

= x= -0.429/1.265

x= -0.34 atm

so ,at the equilibrium

[pco]= 1.4 – x = 1.4 +0.34= 1.74 atm

[pco2]=0.46 atm

21) Equilibrium constant , KC for the reaction

N2 (g) + 3H2(g) ⇌ 2NH3 (g) at 500 K is 0.061

At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L-1 N2 , 2.0 mol L-1 H2 and 0.5 mol L-1 NH3 . Is the reaction at equilibrium ? If not in which direction does the reaction tend to proceed to reach equilibrium?

Solution

Given, KC = 0.061

At equilibrium,

[NH3] = 0.5 M

[N2] =3.0 M

[H2] =2.0 M

The reaction quotient QC

QC = [NH3]2 /[N2] [H2]

3

= (0.5)2 / (3.0)(2.0)3 = 0.25/ 3 x 8 = 0.0104

QC =0.0104

Since, QC < KC , the reaction mixture is not in equilibrium and the reaction will proceed in the forward direction to reach equilibrium.

22) Bromine monochloride , BrCl decomposes into bromine and chlorine and reaches the equilibrium:

2BrCl(g) ⇌ Br2 (g) + Cl2(g)

for which KC = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 x 10-

3 mol L-1 , what is its molar concentration in the mixture at equilibrium?

Solution

let x be the moles of BrCl which decompose to attain equilibrium.

2BrCl(g) ⇌ Br2 (g) + Cl2(g)

Initially 0.0033 M 0 0

At equilibrium (0.0033-x) x/2 x/2

Now, KC = [Br2] [Cl2]/ [BrCl]2

32= x/2 X x/2⧸(0.0033-x)2

= (x/2)2 ⧸ (0.0033 –x)2

5.656= x/2⧸ 0.0033 –x

0.0186 – 5.656 x = x/2

0.0186 = x/2 + 5.656 x = x+11.31x/2

0.0371 = 12.31 x

x= 0.003 M

Molar concentration of BrCl at equilibrium

= 0.0033 – 0.003

= 0.0003

= 3.0 x 10-4 M

Molar concentration of BrCl at equilibrium= 3.0 x 10-4 M

23) At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55 % CO by mass

C(s) + CO2 (g) ⇌ 2CO(g)

Calculate KC for this reaction at the above temperature.

Solution

Let total mass of gaseous mixture= 100 g

So, the mass of CO in mixture = 90.55 g

Mass of CO2 in mixture= (100-90.55) = 9.45 g

And , nco = 90.55 / 28 =3.23 mol

nco2 =9.45/44 = 0.215 mol

∴ mole fraction of CO = XCO2 = 3.23 / 3.23 + 0.215 = 0/937

Mole fraction of CO2 = XCO2

= 1-0.937

= 0.063

Now, Partial pressure of CO = PCO

= XCO x total pressure

= 0.937 atm

PCO2 = 0.063 atm

So, KP = PCO 2 / PCO2

=(0.938)2 / 0.063 = 0.8798/0.063 =13.96 atm

Kp = 13.96 atm

Calculation of Kc

(C(s) + CO2 (g) →2CO(g) )

Hence, ∆ng = 2-1 =1

T= 1127 K

Kp = KC(RT) ∆ng

∴ KC = Kp / (RT) ∆ng

= 13.96 atm/ (0.082 x 1127)1

KC = 0.151 molL-1

24) Calculate a) ∆Gθ and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K

NO (g) + ½ O2 (g) ⇌ NO2 (g)

where,

∆fGθ (NO2)= 52.0 kJ/mol

∆fGθ (NO)= 87.0 kJ/mol

∆fGθ (O2)= 0 kJ/mol

Solution

a) For the reaction,

= 52.0 – (87+0)

= -35 kJ/mol

Hence, ∆fGθ of O2 =0

(as it is in elemental state)

Now, ∆Gθ = -2.303 RT log Kc

b)

= - -35 x 103 / 2.303 x 8.314 x 298

log Kc = +6.134

So, Kc= Antilog (6.314)

= Antilog (6+ 0.134)

= 1.36 x 106

Kc= 1.36 x 106

25) Does the number of moles of reaction products increase , decreases or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

a) PCl5(g) ⇌ PCl3 (g) + Cl2(g)

b) CaO (s) + CO2(g) ⇌ Ca CO3 (s)

c) 3Fe (s) + 4H2O(g) ⇌Fe3O4(s) + 4H2(g)

Solution

i) PCl5(g) ⇌PCl3 (g) + Cl2 (g)

1 vol 2 vols

Since there is increase in volume in forward direction the reaction will progress in forward direction, hence , increase in number of moles of products.

ii) CaO (s) + CO2 (g) ⇌ CaCO3(s)

0 vol 1 vol 0 vol

Since there is decrease in volume in forward direction the number of mole of products will decreases with increase in volume.

iii) 3Fe (s) + 4H2O ⇌ Fe3O4(s) + 4H2(g)

0 vol 4 vol 0 vol 4 vol

Since, there is no change in volume of reaction there will be no change in the number of moles as the change in pressure will have no effect.

26) Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.

i) COCl2(g) ⇌ CO (g) + Cl2(g)

ii) CH4(g) + 2S2 (g) ⇌ CS2(g) + 2H2S(g)

iii) CO2(g) + C(s) ⇌ 2CO (g)

Solution

i) COCl2(g) ⇌ CO (g) + Cl2(g)

Here, increase in pressure will favour backward reaction

ii) CH4(g) + 2S2 (g) ⇌ CS2(g) + 2H2S(g)

Since there is no change in the volume , increase in the pressure will have no effect on equilibrium.

iii) CO2(g) + C(s) ⇌ 2CO (g)

Since there is increase in volume of the reaction the increase in pressure will shift the equilibrium in backward direction.

b) iv) 2H2(g) + CO (g) ⇌ CH3OH (g)

v)CaCO3(s) ⇌ CaO (s) + CO2(g)

vi) 4 NH3 (g) + 5O2(g) ⇌ 4NO (g) + 6H2O (g)

Solution

iv) 2H2(g) + CO (g) ⇌ CH3OH (g)

The increase in pressure will cause the equilibrium to shift in forward direction.

v)CaCO3(s) ⇌ CaO (s) + CO2(g)

Increase in pressure will cause the equilibrium to shift in backward direction.

vi) 4 NH3 (g) + 5O2(g) ⇌ 4NO (g) + 6H2O (g)

Increase in pressure will favour backward direction.

27) The equilibrium constant for the following reaction is 1.6

H2(g) +Br2(g) ⇌ 2HBr(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024k.

Solution

H2(g) +Br2(g) ⇌ 2HBr(g)

Initially 0 0 10 bar

At equilibrium x bar x bar (10-x )bar

Now, KP = P2 HBr/ PH2 x PBr2

1.6 x 105 = (10-x)2/ x X x = (10-x)2/x2

or. 10-x/ x= √ (1.6 x 105)

= 10-x/ x= 400

= 401 x = 10

x=10/401 = 0.02493

= 2.493 x 10-2

= 2.5 x 10-2 bar

PH2= 2.5 X 10-2 bar

PBr2 = 2.5 x 10-2 bar

PHBr = 10- 2.5 x 10-2 bar

= 9.975 bar

PHBr = 9.975 bar

28) Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

CH4 (g) + H2O(g) ⇌ CO (g) + 3H2(g)

a) Write as expression for Kp for the above reaction.

b) How will the values of Kp and composition of equilibrium mixture be affected by

i) increasing the pressure ii) increasing the temperature iii) using a catalyst?

Solution

a) CH4 (g) + H2O(g) ⇌ CO (g) + 3H2(g)

Kp =

p= partial pressure of gas

b)i) Since there is increase in the volume of reaction, increase in pressure will shift the equilibrium in backward direction. More reactants would be formed, hence, the value of Kp will decrease.

ii) Since reaction is endothermic , increase in temperature will favour forward shift in reaction , i.e. more products will be formed. hence the value of Kp will increase.

iii) Addition of catalyst will not change the position of equilibrium as it will influence both the forward and backward reaction to same extent . Hence, Kp will remain unchanged.

29) Describe the effect of:

a) addition of H2 b) addition of CH3OH c) removal of CO d) removal of CH3OH on the equilibrium of the reaction:

2H2(g) + CO (g) ⇌ CH3OH(g)

Solution

2H2(g) + CO (g) ⇌ CH3OH(g)

a) Addition of H2 will shift equilibrium in forward direction due to increase in concentration of reactant.

b) Addition of CH3OH will shift the equilibrium in backward direction due to increase in concentration of products.

c) Removal of CO will shift the reaction in backward direction due to decrease in concentration of reactant.

d) Removal of CH3OH will shift the equilibrium in forward direction due to decrease in concentration of products.

30) At 473 K , equilibrium constant KC for decomposition of phosphorus pentachloride , PCl5 is 8.3 x 10-3 . I f decomposition is depicted as,

PCl5 ⇌ PCl3 (g) + Cl2 (g) ∆rθH= 124.0 kJ mol-1

a) write an expression for KC for the reaction.

b) what is the value of KC for the reverse reaction at the same temperature?

c) what would be the effect on KC if i)more PCl5 is added

ii) pressure is increased iii) the temperature is increased?

Solution

PCl5 ⇌ PCl3 (g) + Cl2 (g)

a)

b)

It is because, the equilibrium constant of a reaction in reverse direction is reciprocal of that in forward direction.

c)i)No change in value of KC on addition of PCl5 (s) as it does not involve in the equilibrium expression.

ii) As pressure is increased , the value of KC is not changed as pressure is not involved in the equation KC.

iii) Since the reaction is endothermic, an increase in temperature will favour forward reaction and hence the value of KC will increase.

31) Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2 . In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

CO(g) + H2 O (g) ⇌ CO2 (g) + H2 (g)

If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and steam such that PCO = PH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium ? KP = 10.1 at 400 °C

Solution

CO(g) + H2 O (g) ⇌ CO2 (g) + H2 (g)

Initially 4 bar 4 bar 0 0

At equilibrium (4-x) (4-x) x bar x bar

Now,

= 10.1 = x X x/ (4-x) (4-x)

= 10.1 = x2 / (4-x)2

= x/4-x = √10.1 = 3.179

= 12.712 – 3.179 x=x

= 4.179 x = 12.712

x = 12.712/4.179 = 3.042 bar

Then, the partial pressures of H2 and at equilibrium is 3.042 bar.

32) Predict which of the following reaction will have appreciable concentration of reactants and products:

Solution

Larger the value of equilibrium constant (KC or KP), larger the concentration of products while less value of KC indicates low concentration of products. For reaction (c) ,

equilibrium has moderate value, so appreciable concentration of reactant and products will have appreciable concentration.

33) The value of KC for the reaction 3O2 (g) ⇌ 2O3 (g) is 2.0 x 10-50 at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6 x 10-2 , what is the concentration of O3?

Solution

3O2 (g) ⇌ 2O3 (g)

Given,

KC = [O3]2/[O2]

3

2.0 x 10-50 = [O3]2/[1.6 x 10-2)3

[O3]2 = (2.0 x 10-50 ) x (.6 x 10-2 )

[O3]2 = 8.192 x 10-56

∴[O3] = (8.192 x 10-56)1/2

∴[O3] = 2.86 x 10-28 M

34) The reaction , CO(g) + 3H2 (g) ⇌ CH4(g) + H2O(g)

is at equilibrium at 1300 K in a 1 L flask . It also contain 0.30 mol of CO, 0.10 mol of H2

and unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium , KC for the reaction at the given temperature is 3.90.

Solution

CO(g) + 3H2 (g) ⇌ CH4(g) + H2O(g)

Equilibrium concentration : 0.30 mol 0.10 mol x mol 0.02 mol

Hence, X is the number of moles CH4.

3.90 = x X 0.02/ (0.30)(0.10)3

= 0.02 x/ 0.30 x (0.10)3

x = 0.30 x (0.10)3 x 3.90 / 0.02

= 0.30 x 0.001 x 3.90 / 0.02

= 0.00117/ 0.02

= 0.0585 M

So , the concentration of CH4 at equilibrium is 0.0585 M.

35) What is meant by the conjugate acid-base pair ? Find the conjugate acid / base for the following species:

HNO2 , CN-, HClO4 , F-, OH- , CO3

2- and S2-.

Solution

The species obtained by removing a proton from an acid is called its conjugate base and the species obtained by adding a proton to a base is called its conjugate acid.

Acid Conjugate base

HNO2 NO2-

HClO4 ClO4-

Base Conjugate acid

CN- HCN

OH- H2O

CO32 HCO-

3

S2- HS-

36) Which of the followings are Lewis acids? H2O , BF3 , H+ , and NH4

+

Solution

An electron pair acceptor is a Lewis acid. As they are electron deficient, they accept electrons.

Among given examples

BF3 , H+ , and NH4

+ ions are Lewis acids.

37) What will be the conjugate bases for the Bronsted acids: HF,H2SO4, and HCO3?

Solution

The species obtained by loss of a proton (H+) from an acid is its conjugate base.

Conjugate base = Acid - H+

Acid Conjugate base

HF F-

H2SO4 HSO4-

HCO3- CO3

2-

38) Write the conjugate acids for the following Bronsted bases : NH2-, NH3 and HCOO-.

Solution

Base Conjugate acid

NH2- NH3

NH3 NH4+

HCOO- HCOOH

39) The species : H2O, HCO3-, HSO4

- and NH3 can act both as Bronsted acids and bases. For each case given the corresponding conjugate acid and base.

Solution

The species obtained by removing a proton from an acid is its conjugate base and the species obtained by adding a proton to a base is its conjugate acid.

40) Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/ base : a) OH- b) F- c) H+ d) BCl3

Solution

Lewis acids are substance which can accept an electron pair while lewis bases are substance which can donate an electron pair.

So, (a) OH- b) F- ion are lewis bases as they are electron rich species and can donates electron pair.c) ) H+ d) BCl3 are lewis acids as they can accept an electron pair.

41) The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10-3 M. what is its pH ?

Solution

Given.

[H+]= 3.8 x 10-3 M

pH = -log [H+]

= -log 3.8 x 10-3

=-(log10-3 + log 3.8)

= -(-3 + 05798)

=-(-2.4202)

pH = 2.4202

42) The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

Solution

pH = -log [H+]

log [H+]= - pH

= -3.76

= log [H+]= Antilog (-3.76)

Antilog (-4 + 0.24)

[H+]= 1.738 x 10-4 M

43) The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 x 10-4 , 1.8 x 10-4 and 4.8 x 10-9 respectively. Calculate the ionisation constants of the corresponding conjugate base.

Solution

We have,

KW = Kb x Ka

Here, KW = ionic product of water

= 1.0 x 10-14

Ka = ionisation constant of acid

Kb = ionisation constant of base

So, for HF

Kb = KW / Ka = 10-14/6.8 X 10-14

Kb = 1.47 X 10-11

For HCOOH Kb = KW / Ka = 10-14/1.8 X 10-14

Kb = 5.55 X 10-11

Fr HCN , Kb = KW / Ka = 10-14/ 4.8 X 10-14

= 2.08 X 10-11

Kb = 2.08 X 10-11