A) \[\left[ {{I}_{2}}(g) \right]>\left[ {{I}^{-}}\left( g \right) \right]\]
B) \[\left[ {{I}_{2}}(g) \right]<\left[ {{I}^{-}}\left( g \right) \right]\]
C) \[\left[ {{I}_{2}}(g) \right]=\left[ {{I}^{-}}\left( g \right) \right]\]
D) \[\left[ {{I}_{2}}(g) \right]=\frac{1}{2}\left[ {{I}^{-}}\left( g \right) \right]\]
Correct Answer: A
Solution :
[a] \[\begin{align} & {{I}_{2}}\rightleftharpoons 2{{I}^{-}} \\ & \begin{matrix} 1-x & 2x \\ \end{matrix} \\ \end{align}\] \[{{K}_{c}}=\frac{{{(2x)}^{2}}}{(1-x)}={{10}^{-6}}\] Soln. shows that \[(1-x)>2x\] \[\therefore [{{I}_{2}}(g)]>[I{{ }^{-}}(g)]\]
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