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 reaction is constant in temperature range between `T_(1) and T_(2)` , it is readily observed that

Options

A. `K_(p) gt K'_(p)`
B. `K_(p) lt K'_(p)`
C. `K_(p) = K'_(p)`
D. `K_(p) =(1)/(k'_(p))`

Answer

A

Explanation

The equilibrium constant at two different temperatures for a thermodynamic process is given by
`"log"(K_(2))/(K_(1))=(DeltaH^(@))/(2.303 R) [(1)/(T_(1))-(1)/(T_(2))]`
here, `K_(1) and K_(2)` are replaced by `k_(p) and k'_(p)`
Therefore , log `(K'_(p))/(K_(p))=(DeltaH^(@))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
For exothermic reaction
`T_(2) gt T_(1) and H=-ve`
`therefore K_(p) gt K'_(p)`