Equivalent amounts of \(\mathrm{H}_2\) and \(\mathrm{I}_2\) are heated in a closed vessel till equilibrium is obtained. If \(80%\) of the hydrogen can be converted to \(\mathrm{HI}\), the \(K_C\) at this temperature is

Question

For the reaction \( \text{N}_2(g) + 3 \, \text{H_2(g) \rightleftharpoons 2 \, \text{NH}_3(g) \), the equilibrium constant \( K_c \) at a certain temperature is \( 0.5 \). If the initial concentrations of \( \text{N}_2 \), \( \text{H}_2 \), and \( \text{NH}_3 \) are \( 1.0 \, \text{mol/L} \), \( 1.0 \, \text{mol/L} \), and \( 0 \, \text{mol/L} \) respectively, calculate the equilibrium concentrations of all species.}

Answer 1

Step 1: Underlying Concept:
For H\(_2\) + I\(_2\) \(\rightleftharpoons\) 2HI, \(K_C = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\).
Step 2: Explanation:
Let initial [H\(_2\)] = [I\(_2\)] = 1 M. At equilibrium, 80% H\(_2\) reacts, so:
[H\(_2\)] = 1 - 0.8 = 0.2 M, [I\(_2\)] = 0.2 M, [HI] = 1.6 M.
\(K_C = \frac{(1.6)^2}{(0.2)(0.2)} = \frac{2.56}{0.04} = 64\).
Step 3: Conclusion:
64

Equivalent amounts of \(\mathrm{H}_2\) and \(\mathrm{I}_2\) are heated in a closed vessel till equilibrium is obtained. If \(80%\) of the hydrogen can be converted to \(\mathrm{HI}\), the \(K_C\) at this temperature is

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The Correct Option is A

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