A) 28
B) 64
C) 32
D) 16
Correct Answer: B
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\[{{K}_{c}}\]).You need to login to perform this action.
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