17.1 The Equilibrium Law

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Question 1

Marks: 1

Which is the correct K_c expression for the reaction between hydrogen and fluorine?

H₂ (g) + F₂ (g) ⇌ 2HF (g)

$\frac{{[HF]}^{2}}{[H_{2}] [F_{2}]}$
$\frac{[HF]}{[H_{2}] [F_{2}]}$
$\frac{2 [HF]}{[H_{2}] [F_{2}]}$
$\frac{[H_{2}] [F_{2}]}{{[HF]}^{2}}$

Key Concepts

Explaining Le Châtelier’s Principle

Question 2

Marks: 1

The K_c expression for the following reaction between hydrogen and iodine is shown

H₂ (g) + I₂ (g) ⇌ 2HI (g)

$K_{C} = \frac{{[HI]}^{2}}{[H_{2}] [I_{2}]}$

At equilibrium there were 0.234 moles of HI, 0.150 moles of H₂ and 0.025 moles of I₂.

Which is the correct K_c expression for the reaction between hydrogen and fluorine?

$\frac{{[0.234]}^{2}}{[0.15] [0.025]}$
$\frac{[0.234]}{[0.15] [0.025]}$
$\frac{[0.15] [0.025]}{{[0.234]}^{2}}$
$\frac{[0.15] [0.025]}{[0.234]}$

Key Concepts

Equilibrium Problems

Question 3

Marks: 1

Nitrosyl chloride decomposes into nitrogen monoxide and chlorine according to the following equation. The forward reaction is endothermic

2NOCl (g) ⇌ 2NO (g) + Cl₂ (g)

Which change in condition would change the value for K_c?

Decreasing the pressure
Adding a catalyst
Increasing the temperature
Increasing the pressure

Key Concepts

Explaining Le Châtelier’s Principle

Question 4

Marks: 1

Which of the following rows correctly describes K_c and ΔG for a reaction where the products are favoured?

	K_c	ΔG
A.	> 1	< 1
B.	> 1	< 0
C.	> 0	> 0
D.	> 0	> 1

Key Concepts

Gibbs Free Energy & the Equilibrium Constant

Question 5

Marks: 1

At 300 K, iron oxidises according to the following equation:

2Fe (s) + $\frac{3}{2}$ O₂ (g) ⇌ Fe₂O₃ (s)

The standard Gibbs free energy change for this reaction is -743.05 kJ mol^-1.

The quantitative relationship between the standard Gibbs free energy change, temperature and the equilibrium constant is:

ΔG = –RT ln K

Which expression is a correct step towards calculating the value of the equilibrium constant? (R = 8.31 J K^–1 mol^–1)

$\ln K = \frac{8.31 \times 573}{- 743.05 \times 10^{3}}$
$\ln K = \frac{8.31 \times 300}{- 743.05 \times 10^{3}}$
$\ln K = \frac{- 743.05}{8.31 \times 300}$
$\ln K = \frac{- 743.05 \times 10^{3}}{8.31 \times 300}$

Key Concepts

Free Energy & Equilibrium Calculations

DP IB Chemistry: HL

Topic Questions

17.1 The Equilibrium Law

Question 1

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Question 2

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Question 3

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Question 4

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Question 5

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