Question

Calculate Grxn at 298 K under the conditions shown below for the following reaction.

3 O2(g) ? 2 O3(g); ?G� = +326 kJ

P(O2) = 5.2 atm, P(O3) = .14 atm

Is this reaction spontaneous? Explain and show all work.

Answer 1

3 O₂ (g) ---------> 2 O₃(g)

Given

ΔG⁰ = +326 kJ =326000J

P(O₂) = 5.2 atm

P(O₃) = .14 atm

Use following formula to calculate Δ Grxn

ΔGrxn = ΔG° + RT ln(Q)

Here

R =8.314 Joules/moleK

Q = (PO₃)^2/pressure (PO₂)^3

T = 298 K

ΔG = +326,000 Joules + 8.314 Joules/mole K (298K) ln (5.2)^2/(0.14)^3

ΔG =+326,000 Joules + 2477 ln(9854)

ΔG =+326,000 Joules + 2477(9.196)

ΔG = 326,000 J +22778.49 J

ΔG = +348778.49 J = +348.78KJ

ΔG rxn is positive, so the reaction is not spontaneous.