Quiz 6.1 and 6.2 Chemical Equilibrium

Which is the correct statement for a chemical equilibrium?

• The rate of forward reaction is the same as that of reverse reaction.

• The total concentration of product equals of that reactants.

• The rate constant of forward reaction equals that of the reverse reaction.

• The product of equilibrium concentration of products and reactants equals Kc value.

A mixture of gases is allowed to reach equilibrium at 700°C in a 12.0 L flask. At equilibrium, the mixture contains 0.208 M SO₂, 1.12 x 10^-6 M O₂ and 0.725 M SO₃.

2SO₂ (g) + O₂ (g) ≶ 2SO₃(g)

What is the Kc?

• 9.22 x 10^-8

• 3.11 x 10⁶

• 1.08 x 10⁷

• 4.56 x 10⁸

Which of the following statement is correct about the reaction quotient, Q_c and the equilibrium constant, K_c.

• For the same reaction, K_c is larger than Q_c

• For the same reaction, Q_c and K_c have the same value.

• The reaction mixture is richer in product if Q_c < K_c

• The value of Q_c changes until it reaches the value of K_c at equlibrium.

Answer explanation

When Q_c < K_c, the equilibrium shifts to the right. Hence, more product is formed.

Of the following reactions, which one gives the same value of K_p and Kc?

• H₂(g) + I₂(s) ↔ 2HI (g)

• H₂(g) + Cl₂(g) ↔ 2HCl (g)

• COCl₂(g) ↔ CO₂ (g) + Cl₂ (g)

• 2NOCl(g) ↔ 2NO(g) + Cl₂ (g)

Answer explanation

K_p = K_c(RT)^Δn

Find equation with Δn = 0.

Hence

K_p = K_c(RT)⁰

∴ K_p = K_c

Which of the following reactions gives the same value of K_p and K_c?

• 2A (g) ↔ B (g)

• 2A (g) ↔ B (g) + C (g)

• A (g) + B (g) ↔ C(g)

• A (g) + B (s) ↔ 2C(g)

Answer explanation

K_p = K_c(RT)^Δn

Find equation with Δn = 0.

Hence

K_p = K_c(RT)⁰

∴ K_p = K_c

Calculate K_c for the reaction at 827K

2CO₂(g) ↔ 2CO (g) + O₂(g)

[K_p= 1.78 x 10³]

• 2.59 x 10^-1

• 2.62 x 10¹

• 1.21 x 10⁵

• 1.22 x 10⁷

Answer explanation

K_p = K_c(RT)^Δn

Δn = 3 - 2 = 1

The reaction: 4ClO₃^- ↔ Cl^- + 3ClO₄^- has an equilibrium constant of 1.0 x 10²⁹. This value indicates that the reaction occurs

• very fast

• spontaneously

• to form more ClO₄^-

• to form a little ClO₄^-

Answer explanation

The bigger the value of K, it will favour forward reaction. Hence more product can be formed.

However, K does not affect the rate of reaction or making the reaction faster.

The expression for the equilibrium constant, K_c for the reaction 2SO₂(g) + O₂(g) ↔ 2SO₃(g) is

• Kc = [SO₃] / [SO₂][O₂]

• Kc = [SO₂]²[O₂] / [SO₃]

• Kc = [SO₃]² / [SO₂]² [O₂]

• Kc = [SO₂][O₂] / [SO₃]

At 250^oC, the equilibrium constant, K_c for the reaction

PCl₃ (g) + Cl₂(g) ↔ PCl₅(g)

is 26. Pick the correct expression for the value of equilibrium constant K_p.

• 26

• 26RT

• 26/RT

• 26√RT

Answer explanation

K_p = K_c(RT)^Δn

Δn = 1- 2 = -1

K_p = 26(RT)^-1

The equilibrium constant, K_c for the reaction

N₂O₄(g) ↔ 2NO₂(g)

is 0.212 at 100°C. What is the value of K_p?

• 1.74

• 6.49

• 14.28

• 198.62

Answer explanation

K_p = K_c(RT)^Δn

Δn = 2-1 = 1

K_p = K_c(RT)