Question:
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
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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For reactions with equal moles of gas on both sides, volume cancels out.
Updated On: Apr 20, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
For H\(_2\) + I\(_2\) \(\rightleftharpoons\) 2HI, \(K_C = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\).
Step 2: Detailed 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: Final Answer:
64
For H\(_2\) + I\(_2\) \(\rightleftharpoons\) 2HI, \(K_C = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\).
Step 2: Detailed 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: Final Answer:
64
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