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

Question

Guest

Mathematics

9 December, 21:48

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

+2

Answers (1)

Know the Answer?

Alvin Murphy · 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