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.

Explanation

`"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