Structured Questions
Answer the questions below in the spaces provided. Check your answers with the mark schemes and examiner tips.
Question 1: Mole Fractions & Partial Pressures
8 marksNitrogen dioxide decomposes according to the following equation:
2NO₂(g) ⇌ 2NO(g) + O₂(g)
A sample of nitrogen dioxide was placed in a sealed vessel at a constant temperature. At equilibrium, the vessel was found to contain 0.60 mol of NO₂, 0.40 mol of NO, and 0.20 mol of O₂. The total pressure of the mixture was 180 kPa.
(a) Calculate the mole fraction of each gas in the equilibrium mixture. [3]
(b) Calculate the partial pressure, in kPa, of each gas at equilibrium. [3]
(c) Show that the sum of the partial pressures equals the total pressure. [2]
Show Mark Scheme
(a)
- Total moles of gas = 0.60 + 0.40 + 0.20 = 1.20 mol [1]
- Mole fraction of \(\text{NO}_2 = \frac{0.60}{1.20} =\) 0.50 [1]
- Mole fraction of \(\text{NO} = \frac{0.40}{1.20} =\) 0.333 (or 1/3) AND Mole fraction of \(\text{O}_2 = \frac{0.20}{1.20} =\) 0.167 (or 1/6) [1]
(b)
- \(p(\text{NO}_2) = 0.50 \times 180 =\) 90 kPa [1]
- \(p(\text{NO}) = 0.333 \times 180 =\) 60 kPa [1]
- \(p(\text{O}_2) = 0.167 \times 180 =\) 30 kPa [1]
(c)
- Sum of partial pressures = \(p(\text{NO}_2) + p(\text{NO}) + p(\text{O}_2) = 90 + 60 + 30\) [1]
- Sum = 180 kPa, which is equal to the total pressure of 180 kPa [1]
Question 2: Equilibrium Kp Calculation (HI Formation)
9 marksHydrogen gas and iodine vapour react to form hydrogen iodide:
H₂(g) + I₂(g) ⇌ 2HI(g)
Initially, 1.00 mol of H₂ and 1.00 mol of I₂ were mixed and allowed to reach equilibrium in a sealed container at 700 K. At equilibrium, 1.50 mol of HI had been formed. The total pressure of the mixture was 120 kPa.
(a) Write the \(K_p\) expression for this reaction. [1]
(b) Calculate the number of moles of H₂ and I₂ present at equilibrium. [2]
(c) Calculate the mole fraction and partial pressure of each gas in the equilibrium mixture. [3]
(d) Calculate the value of \(K_p\) at this temperature, and state its units, if any. [3]
Show Mark Scheme
(a)
- \(K_p = \frac{p(\text{HI})^2}{p(\text{H}_2) \times p(\text{I}_2)}\) [1]
(b)
- According to the equation, 1 mole of H₂ (or I₂) produces 2 moles of HI. Moles of H₂ (or I₂) reacted = 1.50/2 = 0.75 mol [1]
- Moles of H₂ at equilibrium = 1.00 − 0.75 = 0.25 mol; Moles of I₂ at equilibrium = 1.00 − 0.75 = 0.25 mol [1]
(c)
- Total moles at equilibrium = 0.25 + 0.25 + 1.50 = 2.00 mol [1]
- Mole fractions: H₂ = 0.125; I₂ = 0.125; HI = 0.75 [1]
- Partial pressures: \(p(\text{H}_2) = 0.125 \times 120 =\) 15 kPa; \(p(\text{I}_2) =\) 15 kPa; \(p(\text{HI}) = 0.75 \times 120 =\) 90 kPa [1]
(d)
- \(K_p = \frac{90^2}{15 \times 15} = \frac{8100}{225}\) [1]
- \(K_p =\) 36.0 [1]
- No units (units cancel completely) [1]
Question 3: Qualitative Shifts & Kp
6 marksConsider the equilibrium reaction below:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (ΔH = −197 kJ mol⁻¹)
(a) Write the \(K_p\) expression for this reaction. [1]
(b) Explain the effect, if any, of increasing the temperature on the value of \(K_p\). [3]
(c) Explain why increasing the total pressure of the system shifts the position of equilibrium to the right, but has no effect on the value of \(K_p\). [2]
Show Mark Scheme
(a)
- \(K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \times p(\text{O}_2)}\) [1]
(b)
- The forward reaction is exothermic [1]
- According to Le Chatelier's principle, increasing the temperature shifts the equilibrium in the endothermic (reverse) direction to absorb heat [1]
- The partial pressure of SO₃ decreases and the partial pressures of SO₂ and O₂ increase, which decreases the value of \(K_p\) [1]
(c)
- \(K_p\) is only affected by temperature, so it remains constant [1]
- Increasing pressure increases all partial pressures. Since there are more moles of gas on the left (3 moles) than on the right (2 moles), the denominator of the \(K_p\) expression increases more than the numerator. To restore the ratio to the constant value of \(K_p\), the equilibrium must shift to the right (increasing product and reducing reactants) [1]