Equilibrium

NCERT Solution

Part 1

Question 1: A liquid is in equilibrium with its vapor in a sealed container at a fixed temperature. The volume of container is suddenly increased.

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

Answer: No change

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

Answer: Rate of evaporation increases while rate of condensation decreases.

Answer: Vapor pressure remains the same as it does not depend on volume but on temperature.

Question 2: What is K_c 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)

Answer: `K_c=([SO_3]^2)/([SO_2]^2[O_2])`

`=(1.90^2)/(0.60^2xx0.82)`

`=(3.61)/(0.36xx0.82)=12.20`

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

I₂ (g) ⇌ 2I (g)

Calculate K_p for the equilibrium.

Answer: Partial pressure of Iodine atoms = 10⁵ × 40% = 4 × 10⁴ Pa

Partial Pressure of Iodine molecules = 10⁵ × 60% = 6 × 10⁴ Pa

`K_p=((4xx10^4Pa)^2)/(6xx10^4Pa)`

= 2.6 × 10⁴ Pa

Question 4: Write the expression for the equilibrium constant K_c for each of the following reactions:

(a) 2NOCl (g) ⇌ 2NO (g) + Cl₂ (g)

Answer: `([NO]^2[Cl_2])/([NOCl]^2)`

(b) 2Cu(NO₃)₂ (s) ⇌ 2CuO (s) + 4NO₂ (g) + O₂ (g)

Answer: `([CuO]^2[NO_2]^4[O_2])/([Cu(NO)_3]^2)`

Answer: `([CH_3COOH][C_2H_5OH])/([CH_3COOC_2H_5][H_2])`

(d) Fe³⁺ (aq) + 2OH^- (aq) ⇌ Fe(OH)₃ (s)

Answer: `([Fe(OH)_3])/([Fe^(3+)][OH^-]^2)`

(e) I₂ (s) + 5F₂ ⇌ 2IF₅

Answer: `([IF_5]^2)/([I_2][F_2]^5)`

Question 5: Find out the value of K_c for each of the following equilibria from the value of K_p:

(a) 2NOCl (g) ⇌ 2NO (g) + Cl₂ (g): K_p = 1.8 × 10^-2 at 500 K

Answer: We know, K_p = K_c (RT)^Δng

K^p = 1.8 × 10^-2 atm

Δ_ng = 3 – 2 = 1

R = 0.0821 L atm K^-1 mol^-1

T= 500 K

So, `K_c=(K_p)/((RT)^(Δng))`

`=(1.8xx10^(-2))/(0.0821xx500)`=

= 4.4 × 10^-4 mol L^-1

(b) CaCO₃ (s) ⇌ CaO (s) + CO₂: K_p = 167 at 1073 K

Answer: `K_c=(K_p)/((RT)^(^Delta;ng))`

`=(167)/(0.0821xx1073)=1.9` mol L^-1

Question 6: For the following equilibrium, K_c = 6.3 × 10¹⁴ at 1000 K

NO (g) + O₃ ⇌ NO₂ (g) + O₂ (g)

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

Answer: K_c for reverse reaction will be reciprocal of K_c for forward reaction.

`K_c=1/(6.3xx10^(14))`

`=(10^(-14))/(6.3)=1.6xx10^(-13)`

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

Answer: Volumes of pure liquids and solids generally remain constant. So, they can be ignored while writing the equilibrium constant expression.

Question 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 K_c = 2.0 × 10^-37, determine the composition of equilibrium mixture.

Answer: We know that if K_c < 10^-3 then reaction rarely proceeds. Here, K_c is much less than that. So, the reaction will rarely proceed.

Let us assume that x moles of N₂ take part in reaction. According to equation, `x/2` moles of O₂will react with this to form x mole of N₂O.

	2N₂ (g)	O₂ (g)	2N₂O (g)
Initial mole L^-1	`(0.482)/(10)`	`(0.933)/(10)`	Zero
Mole L^-1 at equilibrium	`(0.482-x)/(10)`	`(0.933-x/2)/(10)`	`x/(10)`

Now, applying the Law of Equilibrium we get following equation:

`K_c=([N_2O]^2)/([N_2]^2[O_2])`

Or, `2.0xx10^(-7)=((x/(10))^2)/(((0.482)/(10))^2xx((0.933)/(10)))`

`=(0.01x^2)/(2.1676xx10^(-4))`

`x^2=43.352xx10^(-40)`

Or, `x=6.6xx10^(-20)`

Equilibrium mixture:

Molar concentration of N₂ = 0.0482 mol L^-1

Molar concentration of O₂ = 0.0933 mol L^-1

Molar concentration of N₂O = `0.1xx\x=0.1xx6.6xx10^(-20)` mol L^-1

= 6.6 × 10^-21 mol L^-1

Question 9: Nitric oxide reacts 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.

Answer: According to balanced equation, 2 mole of NO reacts with 1 mole of Br₂ to give 2 mole of NOBr.

So, if 0.0518 M of NOBr is participating in reaction then same amount of NO is participating and half of it (0.0259 M) of Br₂ is participating in reaction.

Original amount of NO = 0.087

Amount at equilibrium = 0.087 – 0.0518 = 0.0352 M

Amount of Br₂ at equilibrium = 0.0437 – 0.0259 = 0.0178 N

Amount of NOBr at equilibrium = 0.0518 (given)

Question 10: At 450K, K_p = 2.0 × 10¹⁰ bar for the given reaction at equilibrium. What is K_c at this temperature?

2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)

Answer: We know, K^p = K_c(RT)^Δng

R = 0.083 L bar K^-1 mol^-1, T = 450 K and Δ_ng = 2 – 3 = - 1

So, K_c `=(2.0xx10^(10))/((0.083xx450)^(-1))`

`=2.0xx10^(10)xx0.083xx450=7.47xx10^(11)` M^-1

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

2HI (g) ⇌ H₂ (g) + I₂ (g)

Answer: Initial pressure of HI = 0.2 atm

Pressure of HI at equilibrium = 0.04

So, reduction in pressure = 0.2 – 0.04 = 0.16

This will be the total pressure of mixture of H₂ and I₂

So, pressure of H₂ = `(0.16)/2=0.08`

Pressure of I₂ will be same, i.e. 0.08

Now K^p can be calculated as follows:

`K_p=(0.08xx0.08)/(0.04xx0.04)=4.0` atm

Question 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 vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction N₂ (g) + 3H₂ (g) ⇌ 2NH₃ is 1.7 × 10². Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer: Concentration quotient of equation can be calculated as follows:

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

`=((8.13)/(20))/((1.57)/(20)((1.92)/(20))^3)`

`=(8.13xx20xx20xx20xx20)/(20xx1.57xx1.92xx1.92xx1.92)`

`=(8.13xx8000)/(11.11)=6xx10^3`

Here, Q_c ≠ K_c

So, the reaction mixture is not at equilibrium.

Since Q_c > K_c, the reaction will shift to right.

Question 13: The equilibrium constant expression for a gas reaction is

`K_c=([NH_3]^4[O_2]^5)/([NO]^4[H_2O]^6)`

Write the balanced chemical equation corresponding to this expression.

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

Question 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.

Answer: As per balanced equation, 1 M of H₂ reacts with 1 M of CO to give 1 M of each product. (Per Liter Reaction Mixture)

So, 40% of 1 M = 0.4 M of H₂ will react with 0.4 M of CO to give 0.4 M of each product.

This means, 0.6 M of each reactant is left at equilibrium.

Hence, equilibrium constant can be calculated as follows:

`K_c=([H_2][CO_2])/([H_2O][CO])`

`=(0.4xx0.4)/(0.6xx0.6)=0.44`

Question 15: At 700 K, equilibrium constant for the following reaction is 54.8.

H₂ (g) + I₂ (g) ⇌ 2HI (g)

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

Answer: `K_c=([HI]^2)/([H_2][I_2])`

Or, `54.8=(0.5^2)/(x^2)` If x is the concentration of each reactant.

Or, `x^2=(0.25)/(54.8)=0.00456`

Or, `x=sqrt(0.00456)`