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?

Question

Chemistry

Gisselle Moon

7 March, 20:12

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?

+5

Answers (1)

Know the Answer?

Soren Stokes · Answer 1

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