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1 mol of PCl5 is placed in a closed 10 L container at 300 K and allowed to reach equilibrium: PCl5(g) <=> PCl3(g) + Cl2(g), with Kc = 10⁻² at 300 K. At equilibrium the molar concentration ratio [PCl3]: [PCl5] equals x: 1.00. Find x. (Given sqrt(41) = 6.4)

1. 0.37
2. 0.27
3. 0.64
4. 0.10

Correct answer: 0.37

Solution

With C0 = 0.1 M and Kc = a²/(0.1 - a) = 0.01, solving the quadratic gives a = 0.027 M. Then x = [PCl3]/[PCl5] = 0.027/(0.1 - 0.027) ~= 0.37.

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