For an equilibrium reaction K_(p) = 0.0260 at 25^(@)C and DeltaH = 32.4 kJ mol^(-1) . Calculate K_(p) at 37^(@)C.

For an equilibrium reaction K_(p) = 0.0260 at 25^(@)C and DeltaH = 32.

1.	For an equilibrium reaction K_(p) = 0.0260 at 25^(@)C and DeltaH = 32.4 kJ mol^(-1) . Calculate K_(p) at 37^(@)C.
Answer» <P> Solution :`T_(1) = 25 + 273 = 298 K , T_(2) = 37 + 273 = 310 K` , `Delta H = 32.4 KJ mol^(-1) = 32400 J mol^(-1) , R = 8.314 JK^(-1) mol^(-1) , K_(P_(1)) = 0.0260 , K_(P_(2)) = ?` `log (K_(2))/(0.0260)= (Delta H^(@))/(2.303 xx 8.314) [(T_(2) - T_(1))/(T_(2) T_(1))]` `log (K_(2))/(0.0260) = (32400)/(2.303 xx 8.314) ((310 - 298)/(310 xx 298))` `= (32400 xx 12)/(2.303 xx 8.314 xx 310 xx 298) = 0.2198` log `(K_(2))/(0.0260)` = anti log 0.2198 = 1.6588 `K_(2) = 1.6588 xx 0.026 = 0.0431`