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

Answer

A

Explanation

According to van't Hoff equation ,
` log = (K'_(p))/(K_(p))= -(DeltaH)/(2.303 R) (1/T_(2)-1/T_(1))`
For exothermic reaction, `Delta H=-ve`
Also ,as `T_(2) gt T_(1), (1/(T_(2)) - 1/T_(1)) = - ve `
` :. log = (K'_(p))/(K_(p)) = -ve or log K'_(p) - - log K_(p) = -ve`
i.e., ` log .K'_(p) lt log K_(p) or K'_(p) lt K_(p) or K_(p) gt K'_(p)`