Question

Calculate ΔGo at 25oC for the reaction H2 + I2 D 2HI.
Starting with 1.0 atm H2 and 1.0 atm I2, What is ΔG when: 0.1%, 1%, 10%, 50%, 90%, 99%, and 99.9% has reacted.

Answer 1

For 0.1 % reaction (fraction = 0.1 / 100 = 0.001)

H2    +    I2 --------------> 2HI

(1-0.001), (1-0.001).......0.002

Kp = (0.002)^2 / (0.999) (0.999) = 4 x 10^-6

delta G = - RT ln K = - 8.314 x 298 x ln (4 x 10^-6) = 30789.32 J

For 1 % reaction (fraction = 1 / 100 = 0.01)

H2    +    I2 --------------> 2HI

(1-0.01), (1-0.01)..........0.02

Kp = (0.02)^2 / (0.99) (0.99) = 4.08 x 10^-4

delta G = - RT ln K = - 8.314 x 298 x ln (4.08 x 10^-4) = 19334.84 J

For 10 % reaction (fraction = 10 / 100 = 0.1)

H2    +    I2 --------------> 2HI

(1-0.1), (1-0.1).............0.2

Kp = (0.2)^2 / (0.9) (0.9) = 0.0494

delta G = - RT ln K = - 8.314 x 298 x ln (0.0494) = 7452.92 J

For 50 % reaction (fraction = 50 / 100 = 0.5)

H2    +    I2 -----------> 2HI

(1-0.5), (1-0.5)..........1

Kp = (1)^2 / (0.5) (0.5) = 4

delta G = - RT ln K = - 8.314 x 298 x ln (4) = -3434.64 J

For 90 % reaction (fraction = 90 / 100 = 0.9)

H2    +    I2 --------------> 2HI

(1-0.9), (1-0.9).............1.8

Kp = (1.8)^2 / (0.1) (0.1) = 324

delta G = - RT ln K = - 8.314 x 298 x ln (324) = -14322.21 J

For 99 % reaction (fraction = 99 / 100 = 0.99)

H2    +    I2 --------------> 2HI

(1-0.99), (1-0.99).............1.98

Kp = (1.98)^2 / (0.01) (0.01) = 39204

delta G = - RT ln K = - 8.314 x 298 x ln (39204) = -26204.12 J

For 99.9 % reaction (fraction = 99.9 / 100 = 0.999)

H2    +    I2 --------------> 2HI

(1-0.999), (1-0.999).............1.998

Kp = (1.998)^2 / (0.001) (0.001) = 3992004

delta G = - RT ln K = - 8.314 x 298 x ln (3992004) = -37658.61 J