Question

Using values from Appendix C of your textbook, calculate the value of K_eq at 298 K for each of the following reactions:

(a) H₂(g) + I₂(g) 2 HI(g)

K_eq =

.

(b) 2 HBr(g) H₂(g) + Br₂(g)

K_eq =

.

(c) 3 Fe(s) + 4 CO₂(g) Fe₃O₄(s) + 4 CO(g)

K_eq =

.

Answer 1

We will use following formula

ΔG rxn = Sum of ΔG of products - Sum of ΔG reactants

a) H₂(g) + I₂(g) ---> 2 HI(g)

ΔG⁰ rxn = [2X ΔG⁰_f(HI)] - [Δ⁰G (H2) + Δ⁰G (I2)]

ΔG⁰rxn = [2 X 1.3 KJ / mole] -[ 0 + 0 ]

ΔG⁰rxn = 2.6 KJ / mole

ΔG⁰rxn = -RTlnKeq

R = 8.314 J / mole K

T = 298 K

Putting values

ΔG⁰rxn = 2.6 X 10^3J / mole = -8.314 X 298 x lnKeq

-1.049 = lnKeq

Keq = 0.3503

b) 2HBr ---> H2(g) + Br2(g)

ΔG⁰ rxn = [Δ⁰G (H2) + Δ⁰G (Br2)] - [2X ΔG⁰_f(HBr)]

ΔG⁰rxn = [ 0 + 0 ]- [2 X (-53.1)]

ΔG⁰rxn = 106.2 KJ / mole

ΔG⁰rxn = -RTlnKeq

R = 8.314 J / mole K

T = 298 K

Putting values

ΔG⁰rxn = 106.2X 10^3J / mole = -8.314 X 298 x lnKeq

-42.86 = lnKeq

Keq = 2.43 x 10^-19

c) 3Fe(s) + 4CO2(g) ---> Fe3O4(s) + 4CO(g)

ΔG⁰ rxn = [4X ΔG⁰_f(CO) + ΔG⁰_f(Fe3O4)] - [3 X Δ⁰G (Fe) + 4 X Δ⁰G (CO2)]

ΔG⁰rxn = [ 4 X (-137.2) + -1015.45] -[ 0 + 4(-394.38) ]

ΔG⁰rxn =-1564.25 +1577.52 = 13.27

ΔG⁰rxn = -RTlnKeq

R = 8.314 J / mole K

T = 298 K

Putting values

ΔG⁰rxn = 13.27 X 10^3J / mole = -8.314 X 298 x lnKeq

-5.356 = lnKeq

Keq =0.00472