Answer:
\[{{Q}_{c}}=\frac{[{{H}_{2}}][{{I}_{2}}]}{{{[HI]}^{2}}}=\frac{1\times
{{10}^{-5}}\times 1\times {{10}^{-5}}}{{{(2\times {{10}^{-5}})}^{2}}}=0.25\]
\[{{Q}_{c}}>{{K}_{c}}\]
Thus, reaction will be fast in
backward direction, i.e., it will proceed in backward direction to decrease the
value of\[{{Q}_{c}}\].
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