Question

What is the equilibrium constant for the reaction:

            SO₂ (g) + NO₂ (g) → SO₃ (g) + NO (g)

At 298 K? Use the following data: R=8.314 J/(K^.mol)

Substance        SO₂ (g)            SO₃ (g)            NO₂ (g)                       NO (g)

ΔG^o (kJ/mol) -300.2               -371                 51                                86.6

a) 6.8 ^. 10^-7

            b) 1.5 ^. 10⁶

            c) 1.014

            d) 0.986

            e) -35.2

Answer 1

Given:

Gof(SO2(g)) = -300.2 KJ/mol

Gof(NO2(g)) = 51.0 KJ/mol

Gof(SO3(g)) = -371.0 KJ/mol

Gof(NO(g)) = 86.6 KJ/mol

Balanced chemical equation is:

SO2(g) + NO2(g) ---> SO3(g) + NO(g)

ΔGo rxn = 1*Gof(SO3(g)) + 1*Gof(NO(g)) - 1*Gof( SO2(g)) - 1*Gof(NO2(g))

ΔGo rxn = 1*(-371.0) + 1*(86.6) - 1*(-300.2) - 1*(51.0)

ΔGo rxn = -35.2 KJ

We have:

T = 298 K

ΔGo = -35.2 KJ/mol

ΔGo = -35200 J/mol

use:

ΔGo = -R*T*ln Kc

-35200 = - 8.314*298.0* ln(Kc)

ln Kc = 14.2075

Kc = 1.5*10^6

Answer: b