Ask Question
29 March, 04:16

At - 10.8 °C the concentration equilibrium constant Kc = 4.0 x 10^-5. for a certain reaction Here are some facts about the reaction:

If the reaction is run at constant pressure, the volume increases by 14.9%.

The constant pressure molar heat capacity Cp = 2.63 J-mol^-1. K^-1.

If the reaction is run at constant pressure, 120. kJ/mol of heat are released.

Using these facts, can you calculate Kc at - 16.°C?

+2
Answers (1)
  1. 29 March, 06:25
    0
    K2 = 9.701 x 10^-10

    Explanation:

    K1 = 4.0 x 10^-5

    K2 = ?

    T1 = - 10.8 °C + 273 = 262.2K

    T2 = - 16 °C + 273 = 257K

    ΔHrxn = 120. kJ/mol = 120000 J/mol

    The formular relating all these parameters is given as;

    ln (K2 / K1) = - ΔHrxn / R * (1 / T2 - 1 / T1)

    ln (K2 / 4.0 x 10^-5) = - 120000 / 8.314 (1 / 257 - 1 / 262.2)

    ln (K2 / 4.0 x 10^-5) = 1.1138

    ln K2 - ln4.0 x 10^-5 = 1.1138

    ln K2 = 1.1138 + ln4.0 x 10^-5

    ln K2 = 1.1138 - 10.1266

    ln K2 = - 9.0128

    K2 = 9.701 x 10^-10
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “At - 10.8 °C the concentration equilibrium constant Kc = 4.0 x 10^-5. for a certain reaction Here are some facts about the reaction: If the ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers