For a hypothetical reactionP(g) + Q(g) hArr R(g) + S(g) , " a graph between log K and " T^(-1)" is a straight line as hsown in the fig. in which "theta = tan^(-1) 0*5 and OA = 10. " Assuming "Delta H^(@) " is independent of temperature , calculate the equilibrium constant of the reaction at 298 K and 798 K respectively.

For a hypothetical reactionP(g) + Q(g) hArr R(g) + S(g) , " a graph be

1.	For a hypothetical reactionP(g) + Q(g) hArr R(g) + S(g) , " a graph between log K and " T^(-1)" is a straight line as hsown in the fig. in which "theta = tan^(-1) 0*5 and OA = 10. " Assuming "Delta H^(@) " is independent of temperature , calculate the equilibrium constant of the reaction at 298 K and 798 K respectively.
Answer» Solution :Effect of temperature on equilibrium constant K is given by van't Hoff equation, viz. ` log K = -(DeltaH^(@))/(2* 303 RT) +C`where C = constant of integration THUS , a plot of log or `1/T`, i.e., `T^(-1) ` is a straight LINE with SLOPE = `(Delta H^(@))/(2* 303 R)` and intercept = C `:. C = 10 and " slope " = tan theta= 05 = (DeltaH^(@))/(2303 xx 8314) or DeltaH^(@)= 9 574 " Jmol"^(-1)` `:. log K= 10 -(9574)/(2303 xx 8* 314 xx 298 ) =9* 9827` or K= Antlog `(9 9827)= 9 96 xx 10^(9)` As `DeltaH^(@)` is independent of temperature , K will also be independent of temperature . Hence , K wil be same at 298 K and 798 K.