Ask Question
30 March, 03:42

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)
  1. 30 March, 04:10
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers