For the reaction `I_(2)(g) hArr 2I(g)`, `K_(c) 37.6 xx 10^(-6)` at 1000K . If 1.0 mole of `I_(2)` is introduced into a 1.0 litre flask at 1000K, at equilibriumA. `[I_(2)]gt gt [I]`B. `[I_(2)] = [I]`C. `[I_(2)] lt [I]`D. Unpredictable

For the reaction `I_(2)(g) hArr 2I(g)`, `K_(c) 37.6 xx 10^(-6)` at 100

1.	For the reaction `I_(2)(g) hArr 2I(g)`, `K_(c) 37.6 xx 10^(-6)` at 1000K . If 1.0 mole of `I_(2)` is introduced into a 1.0 litre flask at 1000K, at equilibriumA. `[I_(2)]gt gt [I]`B. `[I_(2)] = [I]`C. `[I_(2)] lt [I]`D. Unpredictable
Answer» Correct Answer - A As `K_(c)` is `~~ 10^(-6)` therefore though the equilibrium shall move in forward direction yet `I_(2) gt gt gt I`

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For the reaction `I_(2)(g) hArr 2I(g)`, `K_(c) 37.6 xx 10^(-6)` at 1000K . If 1.0 mole of `I_(2)` is introduced into a 1.0 litre flask at 1000K, at equilibriumA. `[I_(2)]gt gt [I]`B. `[I_(2)] = [I]`C. `[I_(2)] lt [I]`D. Unpredictable

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