1. Tardigrade
  2. Question
  3. Chemistry
  4. The Kc for H2(g)+I2(g)leftharpoons 2HI(g) is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will be:

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Q. The $ {{K}_{c}} $ for $ {{H}_{2}}(g)+{{I}_{2}}(g)\rightleftharpoons 2HI(g) $ is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will be:

JIPMERJIPMER 2001Equilibrium

Solution:

$ \begin{matrix} {} & {{H}_{2}}(g)+ & {{I}_{2}}(g)\rightleftharpoons & 2HI(g) \\ \text{at}\,\,\text{equilibrium} & (x-a) & (x-a) & (2a) \\ \end{matrix} $ $ {{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}]\,[{{I}_{2}}]}=\frac{{{\left( \frac{2a}{V} \right)}^{2}}}{\frac{(x-a)}{V}\times \frac{x-a}{V}} $ $ =\frac{4{{a}^{2}}}{{{(x-a)}^{2}}} $ Thus, $ {{K}_{c}} $ is independent volume for this reaction, hence, $ {{K}_{c}} $ will remain equal to 64 (i.e., equal to initial $ {{K}_{c}} $ ).