Ask Question
19 January, 12:04

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.

+3
Answers (1)
  1. 19 January, 12:14
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers