Equilibrium Kp Exam Practice
Exam-style practice questions with detailed mark schemes and examiner tips.
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 NO2 = 0.60 / 1.20 = 0.50 [1]
- Mole fraction of NO = 0.40 / 1.20 = 0.333 (or 1/3) AND Mole fraction of O2 = 0.20 / 1.20 = 0.167 (or 1/6) [1]
(b)
- pNO2 = 0.50 × 180 = 90 kPa [1]
- pNO = 0.333 × 180 = 60 kPa [1]
- pO2 = 0.167 × 180 = 30 kPa [1]
(c)
- Sum of partial pressures = pNO2 + pNO + pO2 = 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 Kp 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 Kp at this temperature, and state its units, if any. [3]
Show Mark Scheme
(a)
- Kp = (pHI)^2 / (pH2 × pI2) [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: pH2 = 0.125 × 120 = 15 kPa; pI2 = 15 kPa; pHI = 0.75 × 120 = 90 kPa [1]
(d)
- Kp = 90^2 / (15 × 15) = 8100 / 225 [1]
- Kp = 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 Kp expression for this reaction. [1]
(b) Explain the effect, if any, of increasing the temperature on the value of Kp. [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 Kp. [2]
Show Mark Scheme
(a)
- Kp = (pSO3)^2 / ((pSO2)^2 × pO2) [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 Kp [1]
(c)
- Kp 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 Kp expression increases more than the numerator. To restore the ratio to the constant value of Kp, the equilibrium must shift to the right (increasing product and reducing reactants) [1]