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 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?

Answer 1

Answer:

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