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The equilibrium constant of a reaction a...

The equilibrium constant of a reaction at 298 K is `5 xx 10^(-3)` and at 1000 K is `2 xx 10^(-5)` What is the sign of `triangleH` for the reaction.

A

`Delta H^(@)= 0`

B

`Delta H^(@)` is negative

C

`Delta H^(@)` is positive

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta G^(@)= Delta G^(@)- T Delta S^(@)= -RT ln K`
`T_(1)= 298K, T_(2)= 1000K, K_(1)= 5 xx 10^(-3), K_(2)= 2 xx 10^(-5) (Delta H^(@)- T_(1) Delta S^(@))/(RT_(1))= - ln K_(1)` ...(i)
`(Delta H^(@) - T_(2) Delta S^(@))/(RT_(2))= -ln K_(2)` ...(ii)
Subtracting eq (ii) from eq (i)
`(Delta H^(@))/(RT_(1)) - (T_(1) Delta S^(@))/(RT_(1))- (Delta H^(@))/(RT_(2))+ (T_(2) Delta S^(@))/(RT_(2)) = ln K_(2)- ln K_(1)`
`(Delta H^(@))/(R ) ((1)/(T_(1))- (1)/(T_(2))) = ln ((K_(2))/(K_(1)))`
As `K_(2) lt K_(1) and T_(2) gt T_(1)`
So ln `((K_(2))/(K_(1))) lt 0 and ((1)/(T_(1))- (1)/(T_(2))) gt 0`
So `Delta H^(@)` must be negative
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