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For a given exothermic reaction, Kp and ...

For a given exothermic reaction, `K_p and K_(p)^'` are the equilibrium constants at temperatures `T_1 and T_2` respectively. Assuming that heat of reaction is constant in temperature range between `T_1 and T_2,` it is readily water that

A

`K_(p) gt K_(p)'`

B

`K_(p) lt K_(p)'`

C

`K_(p)=K_(p)'`

D

`K_(p)=1/(K_(p)')`

Text Solution

Verified by Experts

The correct Answer is:
A
`log "" (K_p.)/(K_(p))=- (triangleH)/(2.303 R) [1/T_2-1/T_1]`
For exothermic reaction `triangleH=-ve`, i.e, heat is evolved. The temperature `T_2` is higher than `T_1`.
Thus, `(1/T_2-1/T_1)` is negative.
so, `log K_(p).-log K_(p)=-"ve or log "K_(p) gt log K_(p). or K_p gt K_(p).`
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