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The relationship between the equilibrium...

The relationship between the equilibrium constant of a reaction and temperature is
(`T_(2) gt T_(1)` in all the options)

A

`log(K_(1)/K_(2))=(DeltaH)/R(1/T_(1)-1/T_(2))`

B

`log(K_(1)/K_(2))=(Ea)/R(1/T_(1)-1/T_(2))`

C

`log(K_(2)/K_(1))=(Ea)/(2.303R)(1/T_(1)-1/T_(2))`

D

`log(K_(2)/K_(1))=(DeltaH)/(2.303R)(1/T_(1)-1/T_(2))`

Text Solution

Verified by Experts

The correct Answer is:
D
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Assertion : The equilibrium constant of an exothermic reaction decreases as temperature increases. Reason : log(K_(2)/K_(1))=(DeltaH)/(2.303R)(1/T_(1)-1/T_(2)) where T_(2) gt T_(1) . Since for an exothermic reaction DeltaH is -ve, it follows that K_(2)/K_(1) lt 1 or K_(2) lt K_(1) .