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

the given reaction is

H2 + I2 ---> 2HI

we know that

dGo = dGo products - dGo reactants

so

dGo = 2 x dGo HI - dGo H2 - dGo I2

dGo = 2 x 1.3 - 0 - 0

dGo = 2.6 kJ

now

we know that

dG = dGo + RT ln Q

consider the given reaction

H2 + I2 ---> 2HI

Q = [HI]^2 / [H2 ] [I2]

1) 0.1 % reacted

[H2] = 1 - 0.001 = 0.999

[I2] = 0.999

[HI] = 2 x 0.001 = 0.002

so

Q = [0.002]^2 / [0.999] [0.999]

Q = 4 x 10-6

now

dG = (2.6 x 1000) + 8.314 x 298 x ln 4 x 10-6

dG = -28.19 kJ

2) 1 % reacted

[H2] = 1 - 0.01 = 0.99

[I2] = 0.99

[HI] = 2 x 0.01 = 0.02

so

Q = [0.02]^2 / [0.99] [0.99]

Q = 4.08 x 10-4

now

dG = (2.6 x 1000) + 8.314 x 298 x ln 4.08 x 10-4

dG = -16.73 kJ

3)   10 % reacted

[H2] = 1 - 0.1 = 0.9

[I2] = 0.9

[HI] = 2 x 0.1 = 0.2

so

Q = [0.2]^2 / [0.9] [0.9]

Q = 0.04938

now

dG = (2.6 x 1000) + 8.314 x 298 x ln 0.04938

dG = -4.85 kJ

4)

50 % reacted

[H2] = 1 - 0.5 = 0.5

[I2] = 0.5

[HI] = 2 x 0.5 = 1

so

Q = [1]^2 / [0.5] [0.5]

Q = 4

now

dG = (2.6 x 1000) + 8.314 x 298 x ln 4

dG = 6.03 kJ

5)

90 % reacted

[H2] = 1 - 0.9 = 0.1

[I2] = 0.1

[HI] = 2 x 0.9= 1.8

so

Q = [1.8]^2 / [0.1] [0.1]

Q = 324

now

dG = (2.6 x 1000) + 8.314 x 298 x ln 324

dG = 16.92 kJ

6)

99 % reacted

[H2] = 1 - 0.99 = 0.01

[I2] = 0.01

[HI] = 2 x 0.99= 1.98

so

Q = [1.98]^2 / [0.01] [0.01]

Q = 39204

now

dG = (2.6 x 1000) + 8.314 x 298 x ln 39204

dG = 28.80 kJ

7)

99.9 % reacted

[H2] = 1 - 0.999 = 0.001

[I2] = 0.001

[HI] = 2 x 0.999 = 1.998

so

Q = [1.998]^2 / [0.001] [0.001]

Q = 4 x 10^6

now

dG = (2.6 x 1000) + 8.314 x 298 x ln 4 x 10^6

dG = 40.26 kJ