In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.11 L flask with 0.822 mol of HI gas and allows the reaction to proceed at 428°C: 2HI (g) ⇋ H2 (g) + I2 (g)

At equilibrium, the concentration of HI = 0.055 M. Calculate Kc. Enter to 4 decimal places.

HINT: Look at sample problem 17.6 in the 8th ed Silberberg book. Write a Kc expression. Find the initial concentration. Fill in the ICE chart. Put the E (equilibrium) values into the Kc expression.

Question

Chemistry

Kody

9 December, 13:10

In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.11 L flask with 0.822 mol of HI gas and allows the reaction to proceed at 428°C: 2HI (g) ⇋ H2 (g) + I2 (g)

At equilibrium, the concentration of HI = 0.055 M. Calculate Kc. Enter to 4 decimal places.

HINT: Look at sample problem 17.6 in the 8th ed Silberberg book. Write a Kc expression. Find the initial concentration. Fill in the ICE chart. Put the E (equilibrium) values into the Kc expression.

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Answers (1)

Know the Answer?

Kelsey Manning · Answer 1

Kc = 168.0749

Explanation:

2HI (g) ↔ H2 (g) + I2 (g)

initial mol: 0.822 0 0

equil. mol: 2 (0.822 - x) x x

∴ [ HI ]eq = 0.055 mol/L = 2 (0.822 - x) / (1.11 L)

⇒ 1.644 - 2x = 0.055 * 1.11

⇒ 1.644 = 2x + 0.06105

⇒ 2x = 1.583

⇒ x = 0.7915 mol equilibrium

⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq

⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²

⇒ Kc = (0.7130²) / (0.055²)

⇒ Kc = 168.0749