Question

For the reaction CO2(g) + H2(9)—-CO(g) +H30(9) AH° = 41.2 kJ and A Sº = 42.1 JK The equilibrium constant for this reaction at 252.0 K is Assume that AH and AS are independent of temperature.

Answer 1

1)

ΔH = 41.2 KJ

ΔS = 42.1 J/K

= 0.0421 KJ/K

T = 252 K

use:

ΔG = ΔH - T*ΔS

ΔG = 41.2 - 252.0 * 0.0421

ΔG = 30.5908 KJ

We have:

T = 252 K

ΔGo = 30.5908 KJ

ΔGo = 30590.8 J

use:

ΔGo = -R*T*ln Kc

30590.8 = - 8.314*252.0* ln(Kc)

ln Kc = -14.6009

Kc = 4.559*10^-7

Answer: 4.56*10^-7

2)

ΔH = -92.2 KJ

ΔS = -198.7 J/K

= -0.1987 KJ/K

T = 347 K

use:

ΔG = ΔH - T*ΔS

ΔG = -92.2 - 347.0 * -0.1987

ΔG = -23.2511 KJ

So, we have:

T = 347 K

ΔGo = -23.2511 KJ

ΔGo = -23251.1 J

use:

ΔGo = -R*T*ln Kc

-23251.1 = - 8.314*347.0* ln(Kc)

ln Kc = 8.0594

Kc = 3.163*10^3

Answer: 3.16*10^3