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In a 7.0 L evacuated chamber, 0.50 mol H2 and 0.50 mol I2 react at 427 deg C: H2(g) + I2(g) <=> 2HI(g), with Kc = 49. How many moles of I2 remain unreacted at equilibrium?

1. 0.388
2. 0.112
3. 0.25
4. 0.125

Correct answer: 0.112

Solution

Let x mol of each react. (2x)²/((0.5-x)²) = 49 -> 2x/(0.5-x) = 7 -> x = 0.388. Remaining I2 = 0.5 - 0.388 = 0.112 mol.

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