In: Chemistry
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.
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