Question

If Kc = 0.405 at 40.°C and Kc = 0.575 at 90.°C, what is ΔH° for the reaction?. RXN is simply x yields y and reverse

Answer 1

We apply arrhenius equation to solve this problem which relates Kc to the change in temperature and that ΔH° is the constant. The arrhenius eqn is :

ln (Kc2)/ ln (Kc1) = ΔH°/R * [(1/T1)- (1/T2)] where T is in kelvin.
Thus,
ln (0.575)/ ln (0.405) = ΔH°/8.314 * [(1/(40+273)- (1/(90+273)]
 ΔH° = 11566.81 J/mol