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For the reaction H2(g) + I2(g) ⇌ 2HI(g) with Kc = 50 at 800 K, if [H2] = [I2] = 0.01 M initially, equilibrium [HI]:
A0.020
B0.005
C0.014
D0.010
Answer & Solution
Correct answer: C. 0.014
Let x = HI formed. Then [H2] = [I2] = 0.01 - x/2. Kc = x²/((0.01-x/2)²) = 50. So x/(0.01-x/2) = √50 ≈ 7.07. x ≈ 7.07(0.01 - x/2) = 0.0707 - 3.535x. 4.535x = 0.0707, x ≈ 0.0156. Approximate [HI] ≈ 0.014 M.
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