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 observed that

Options

A. `K_(p)=1/(K_(P)^('))`
B. `K_(p) gt K_(p)^(')`
C. `K_(p) lt K_(p)^(')`
D. `K_(p)=K_(p)^(')`

Explanation

As per `log"" (K_(P)^(.))/(K_(P))=(triangle H)/(2.303R) (1/T_(1)-1/T_(2))`
If `triangleH` is zero as heat of reaction is constant then log `K_(P)^(.)=log K_(P)`
`therefore K_(P)^(.)=K_(P)`