For a reaction whose standrad enthalpy change is -100 kJ, what final temperature is needed to double the equilibrium constant from its value at 298 K?

For a reaction whose standrad enthalpy change is -100 kJ, what final t

1.	For a reaction whose standrad enthalpy change is -100 kJ, what final temperature is needed to double the equilibrium constant from its value at 298 K?
Answer» Solution :`"log"K_(2)/K_(1)=(DeltaH^(@))/(2.303R)[(T_(2)-T_(1))/(T_(1)*T_(2))]` `LOG2=(-100000)/(2.3xx8.3)[(T_(2)-298)/(298T_(2))]` `0.301=(-100000)/(19.09)((T_(2)-298))/((298T_(2)))` Solving for `T_(2)`, we get `T_(2)~~293K`