For the reaction `Br_(2) hArr 2Br`, the equilibrium constants at `327^(@)C` and `527^(@)C` are, respectively, `6.1xx10^(-12)` and `1.0xx10^(-7)`. What is the nature of the reaction?

For the reaction `Br_(2) hArr 2Br`, the equilibrium constants at `327^

1.	For the reaction `Br_(2) hArr 2Br`, the equilibrium constants at `327^(@)C` and `527^(@)C` are, respectively, `6.1xx10^(-12)` and `1.0xx10^(-7)`. What is the nature of the reaction?
Answer» We have `"log" K_(p2)-log K_(p1)=(DeltaH)/(2.303 R)xx(T_(2)-T_(1))/(T_(2)T_(1))` As we know from the above equation that if on increasing temperature, `K_(p)` increases, `DeltaH` becomes positive, i.e., the reaction is endothermic. Thus, from the given data, we see that the reaction is endothermic.

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For the reaction `Br_(2) hArr 2Br`, the equilibrium constants at `327^(@)C` and `527^(@)C` are, respectively, `6.1xx10^(-12)` and `1.0xx10^(-7)`. What is the nature of the reaction?

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