15 moles of $$\text{H}_2$$ and 5.2 moles of $$\text{I}_2$$ are mixed and allowed to attain equilibrium. Concentration of HI is found to be 10 moles the equilibrium constant for the formation of HI is ____________.

Correct Answer: 50

Solution

$$\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$$. Forming 10 mol of HI uses 5 mol each of $$\text{H}_2$$ and $$\text{I}_2$$, leaving 10 mol of $$\text{H}_2$$ and 0.2 mol of $$\text{I}_2$$. Since the volume cancels, $$K = \frac{10^2}{10 \times 0.2} = 50$$.

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