Equilibrium

NCERT Solution

Part 2

Question 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): K_c = 0.14

Answer: Let us assume concentration of each product is x

`K_c=([I_2][Cl_2])/([ICl]^2)`

Or, `0.14=(x^2)/((0.78-x)^2)`

Or, `c/(0.78-x)=sqrt(0.14)=0.374`

Or, `x=0.374(0.78-2x)`

Or, `x=0.292-0.748x`

Or, `1.748x-0.292`

Or, `x=(0.292)/(1.748)=0.167`

So, [ICl] = 0.78 – 0.167 = .613

[I₂] = [Cl₂ = 0.167 M L^-1

Question 17: K_p = 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)

Answer: The equilibrium in the reaction is:

C₂H₆ (g) ⇌ C₂H₄ (g) + H₂ (g)

`K_=(p_(C_2H_4xx\p_(H_2)))/(p_(C_2H_6))`

Or, `0.04=(p^2)/(4-p)`

Or, `P^2=0.04(4-p)=0.16-0.04p`

Or, `p^2+0.04p-0.16=0`

Using the formula for root of quadratic equation:

`p=((-0.04)±sqrt(0.0016-4(-0.16)))/2`

`=((-0.04)±sqrt(0.6416))/2`

`=((-0.004)±0.8)/2=(0-76)/2=0.38`

Equilibrium pressure of C₂H₆ = 4- 0.38 = 3.62 atm

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

CH₃COOH (l) + C₂H₅OH (l) ⇌ CH₃COOC₂H₅ (l) + H₂O (l)

(a) Write the concentration ratio (reaction quotient) Q_c, for this reaction (Note: water is not in excess and is not a solvent in this reaction).

Answer: The concentration ratio Q_c is as follows:

`Q_c=([CH_3CO\OC_2H_5][H_2O])/([CH_2CO\H][C_2H_5OH])`

(b) At 293 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.

Answer:

`K_c=([CH_3CO\OC_2H_5][H_2O])/([CH_2CO\H][C_2H_5OH])`

`=(0.171xx0.171)/(0.829xx0.009)-3.92`

(c) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Answer:

`Q_c=([CH_3CO\OC_2H_5][H_2O])/([CH_2CO\H][C_2H_5OH])`

`=(0.214xx0.214)/(0.786xx0.286)=0.204`

As Q_c < K_c equilibrium has not been reached.

Question 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 × 10^-1 mol L^-1. If value of K_c is 8.3 × 10^-3 what are the concentrations of PCl₃ and Cl₂ at equilibrium?

PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

Answer:

`K_c=([PC\l_3][Cl_2])/([PC\l_5])`

Or, `0.0083=((x-0.05)xx(x-0.05))/(0.05)`

Or, `(x-0.05)^2=0.0083xx0.05=4.15xx10^(-4)`

Or, `x-0.05=sqrt(4.15xx10^(-4))`

`=2.037xx10^(-2)=0.02`

Or, `x=0.05+0.02=0.07` M

Molar concentration of PCl₃ at eql = 0.07 – 0.05 = 0.02 M L^-1

Molar concentration of Cl₂ at eql = 0.07 – 0.05 = 0.02 M L^-1

Question 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 at CO₂.

FeO (s) + CO (g) ⇌ Fe (s) + CO₂ (g): K^p = 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: p_CO = 1.4 atm and p_CO2 = 0.80 atm?

Answer: Initial pressure of CO = 1.4 atm and that of CO₂ = 0.8 atm

`Q_p=(p_(CO_2))/(p_(CO))`

`=(0.8)/(1.4)=0.571`

Here, Q_p > K_p

So, reaction will proceed in backward direction to attain equilibrium. So, p_CO2 will decrease and p_CO will increase to attain equilibrium.

Let us assume that p atm is the reduction in partial pressure of CO₂. So, partial pressure of CO will increase by same magnitude, i.e. p atm

p_CO2 = 0.8 – p atm

p_CO= 1.4 + p atm

`=(0.8-p)/(1.4+p)`

`K_p=(p_(CO_2))/(p_(CO))`

Or, `0.265=(0.8-p)/(1.4+p)`

Or, 0.371 + 0.265p = 0.8 – p

Or, 1.265p = 0.8 – 0.371 = 0.429

Or, `p=(0.429)/(1.265)=0.339` atm

At equilibrium:

p_CO = 1.4 = 0.339 = 1.739 atm

p_CO2 = 0.8 – 0.339 = 1.461 atm

Question 21: Equilibrium constant K_c for the reaction

N_{2 (g) + 3H2 ⇌ 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?

Answer: We need to find Q_c for this reaction.

`Q_c=([NH_3]^2)/([N_2][H_2]^3)`

`=(0.5^2)/(3xx2^3)=(0.25)/(24)=0.104`

Here, Q_c < K_c

So, reaction is not at equilibrium. It will proceed towards right.

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

2BrCl (g) ⇌ Br₂ (g) + Cl₂ (g)

For which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium?

Answer: Let us assume that x mole of BrCl is utilized to attain equilibrium.

`32=((x/2)^2)/(0.0033-x)^2`

Or, `5.656=(x/2)/(0.0033-x)`

Or, `x/(0.0033-x)=11.31`

Or, `0.037-11.31x=x`

Or, `12.31x=0.037`

Or, `x=(0.037)/(12.31)=0.003` M L^-1

Question 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 ⇌ 2CO (g)

Calculate K_c for this reaction at the above temperature.

Answer: Let us assume that total mass of gaseous mixture = 100 g

Mass of CO in mixture = 90.55 g

Mass of CO₂ = 100 – 90.55 = 9.45 g

No. of moles of CO `=(90.55)/(28)=3.234` M

No. of moles of CO₂ `=(9.45)/(44)=0.215` M

p_CO `=(3.234)/(3.232+0.215)xx1` atm

`=(3.234)/(3.449)xx1 at\m=0.938` atm

p_CO2 `=(0.214)/(3.449)=0.062` atm

So, `K_p=(0.938^2)/(0.062)=14.19` atm

Now, K_c can be calculated as follows:

`K_c=(K_p)/((RT)Δ_ng)`

R = 0.0821 L atm K^-1 mol^-1, T = 1127 K, Δ_ng = 2 – 1 = 1

So, `K_c=(14.19)/(0.0821xx1127)=6.46`

Question 24: Calculate (a) ΔG^Θ 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:

Δ_fG^Θ [NO₂ = 52.0 kJ/mol

Δ_fG^Θ [NO] = 87.0 kJ/mol

Δ_fG^Θ = 0 kJ/mol

Answer: ΔG^Θ can be calculated as follows:

ΔG^Θ = Δ_fG^Θ(NO₂) - Δ_fG^Θ1/2O₂

= 52.0 – (87 + 0) = -35 kJ mol^-1

K_c can be calculated as follows:

ΔG^Θ = -2.303 RT log K_c

Or, log K_c `=-(ΔG^(Θ))/(2.303RT)`

`=((-35xx10^3)^3)/(2.303xx8.314xx298)=6.134`

K_c = Antilog 6.314 = 1.36 × 10⁶

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

1. PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)
2. CaO (s) + CO₂ (g) ⇌ CaCO₃ (s)
3. 3Fe (s) + 4H₂ (g) ⇌ Fe₃O₄ (s) + 4H₂ (g)

1. In case of (a) no. of moles of gases is more on RHS. So, number of moles of reaction products increases.
2. In case of (b) no. of moles of gases is more on LHS. So number of moles of products decreases.
3. In case of (c) no. of moles of gases is equal on both sides. So, there will be no change in this reaction.

Question 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.

(a) COCl₂ (g) ⇌ CO (g) + Cl₂ (g)

Answer: When pressure is increased then reaction will shift to the side where number of moles of gases is less. This reaction will shift backwards.

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

Answer: No change

(c) CO₂ (g) + C (s) ⇌ 2CO (g)

Answer: Shift backwards

(d) 2H₂ (g) + CO (g) ⇌ CH₃OH (g)

Answer: Shift forwards

(e) CaCO₃ (s) ⇌ CaO (s) + CO₂ (g)

Answer: Shift backwards

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

Answer: Shift backwards

Question 27: The equilibrium constant for the following reaction is 1.6 × 10⁵ at 1024 K

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 1024 K.

Answer: We know, K_p = K_c(RT)Δ_ng

In this reaction, Δ_n = 2 – 2 = 0

So, K_p = K_c = 1.6 × 10⁵

`K_p^(')=(pH_2xx\pBr_2)/(p^2HBR)`

Or, `1/(1.6xx10^5)=(p/2xxp/2)/((10-p)^2)``=(p^2)/(4(10-p)^2)`

Or, `(2(10-p))/p=sqrt(1.6xx10^5)=4xx10^2`

Or, `20-2p=4xx10^2p`

Or, `2p+400p=20`

Or, `402p=20`

Or, `p=(20)/(400)=0.050` bar

Question 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

(a) Write an expression for K_p for the above reaction.

Answer: `K_p=((p\CO)xx(p^3H_2O))/((pCH_4)xx(pH_2O))`

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

(i) Increasing the pressure

Answer: In this reaction, LHS has less no. of moles of gases. So, increasing the pressure will shift the reaction to LHS. It will increase the no. of moles of reactants. As denominator increases, the value of K_p will decrease.

(ii) Increasing the temperature

Answer: As it is an endothermic reaction, so increasing the temperature will shift the reaction to RHS.

(iii) Using a catalyst

Answer: No effect, because catalyst has equal effect on reactants and products.

Question 29: Describe the effect of:

1. Addition of H₂
2. Addition of CH₃OH
3. Removal of CO
4. Removal of CH₃OH

On the equilibrium of the reaction:

2H₂ (g) + CO (g) ⇌ CH₃OH (g)

Answer: The reaction will shift to RHS in case of (a) and (d), and to LHS in case of (b) and (c)

Question 30: At 473 K equilibrium constant K_c for decomposition of phosphorus pentachloride PCl₅ is 8.3 × 10^-3. If decomposition is depicted as

PCl_{5 (g) ⇌ PCl3 (g) + Cl2 (g) ΔrH^Θ = 124.0 kJ mol^-1}

(a) Write an expression for K_c for the reaction.

Answer: `K_c=([PC\l_3][Cl_2])/([PC\l_5])`

(b) What is the value of K_c for the reverse reaction at the same temperature?

Answer: `K_c'=([PC\l_5])/([PC\l_3][Cl_2])`

`=1/(8/3xx10^(-3))=120.48`

(c) What would be the effect on K_c if

(i) more PCl₅ is added

Answer: No change, because there is no change in temperature

(ii) pressure is increased

Answer: Reaction shifts to LHS because of less no. of moles on LHS

(iii) temperature is increased?

Answer: As it is an endothermic reaction, so increase in temperature will shift it to RHS.

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