Question 11 of 69intermediate🔧 ApplyNumerical3 marks

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

2HI(g)⇌H2(g)+I2(g)2HI (g) \rightleftharpoons H_2 (g) + I_2 (g)
Correct Answer

Kp=4.0K_p = 4.0

Exercise: EXERCISES | Q: 6.11 | (Page 42)
For More Understanding

Explanation

The initial pressure of HI is 0.2 atm, which decreases to 0.04 atm at equilibrium. The change in pressure is 0.16 atm. Using the stoichiometry of the reaction 2HI(g)⇌H2(g)+I2(g)2\text{HI} (g) \rightleftharpoons \text{H}_2 (g) + \text{I}_2 (g), the decrease in HI pressure corresponds to half that amount for the formation of H2 and I2. Thus, pH2=0.08p_{\text{H}_2} = 0.08 atm and pI2=0.08p_{\text{I}_2} = 0.08 atm. Substituting these into the KpK_p expression yields 4.0. Since the exponents in the numerator and denominator are the same, KpK_p has no unit.

Solution Steps

  1. Initial and Equilibrium Pressures: The initial pressure of HI is 0.20.2 atm. At equilibrium, the partial pressure of HI is 0.040.04 atm. The decrease in pressure of HI = 0.2−0.04=0.160.2 - 0.04 = 0.16 atm.

  2. Calculate Partial Pressures of Products: For the equilibrium 2HI(g)⇌H2(g)+I2(g)2\text{HI} (g) \rightleftharpoons \text{H}_2 (g) + \text{I}_2 (g), 2 moles of HI decompose to form 1 mole of H2\text{H}_2 and 1 mole of I2\text{I}_2. Therefore, the partial pressure of H2\text{H}_2 formed = 0.162=0.08\frac{0.16}{2} = 0.08 atm. The partial pressure of I2\text{I}_2 formed = 0.162=0.08\frac{0.16}{2} = 0.08 atm.

  3. Calculate KpK_p: The expression for KpK_p is:

    Kp=pH2×pI2(pHI)2K_p = \frac{p_{\text{H}_2} \times p_{\text{I}_2}}{(p_{\text{HI}})^2}

    Substituting the equilibrium partial pressures:

    Kp=0.08×0.08(0.04)2=0.00640.0016=4.0K_p = \frac{0.08 \times 0.08}{(0.04)^2} = \frac{0.0064}{0.0016} = 4.0

    Since the exponents of the numerator and denominator are the same, KpK_p has no unit.

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