question_answer 1)

  The value of \[{{K}_{c}}\] for the reaction \[2HI(g)\underset{{}}{\leftrightarrows}{{H}_{2}}(g)+{{I}_{2}}(g)\] is \[1\times {{10}^{-4}}\]. At a given time, the composition of reaction mixture is: \[[HI]=2\times {{10}^{5}}mol,[{{H}_{2}}]=1\times {{10}^{-5}}mol\] and \[[{{I}_{2}}]=1\times {{10}^{-5}}mol\] In which direction will the reaction proceed?

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\[{{Q}_{c}}\].