When the temperature of a reversible rection is increased from 327 to 427^(@)C, the equilibrium constant K_(p) is decreased by four times. Find the enthalpy of the reaction in this temperature range.

When the temperature of a reversible rection is increased from 327 to

1.	When the temperature of a reversible rection is increased from 327 to 427^(@)C, the equilibrium constant K_(p) is decreased by four times. Find the enthalpy of the reaction in this temperature range.
Answer» Solution :`"log"((K_(p))_(2))/((K_(p))_(1))=(DELTAH^(@))/(2.303R)[(T_(2)-T_(1))/(T_(1)*T_(2))]` `"log"1/4=(DeltaH^(@))/(2.3xx8.3)[(700-600)/(600xx700)]` `-0.6201=(DeltaH^(@))/19.09xx100/420000` `thereforeDeltaH^(@)=-(0.6201xx19.09xx420000)/100` = -49734 J = -49.734 kJ