For a given exthermic reaction , K_(p) nd K_(p)are the equilibrium constants at temperature T_(1) and T_(2) respectively . Assumbing that heat of reaction is constant in temperature range between T_(1) and T_(2)it is readily observed that

For a given exthermic reaction , K_(p) nd K_(p)are the equilibrium con

1.	For a given exthermic reaction , K_(p) nd K_(p)are the equilibrium constants at temperature T_(1) and T_(2) respectively . Assumbing that heat of reaction is constant in temperature range between T_(1) and T_(2)it is readily observed that
Answer» <P>` K_(p) gt K_(p)'` ` K_(p) lt K'_(p)` `K_(p) = K'_(p)` ` K_(p) = 1/(K'_(p)) ` Solution :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)`