Calculate free energy when..?

Question

Calculate the reaction free energy of;

2H2(g) + O2(g) 2H2O(g)

when the concentrations are 0.026 mol L-1 (H2), 0.33 mol L-1 (O2), and 1.84 mol L-1 (H2O), and the temperature is 700K. For this reaction Kc = 45 at 700 K.

Answer 1

@ equilibrium

dGo = -RTlnK

dGo = - 8.314 Joules(700K) (ln45)

dGo = - 8.314 Joules(700K) 3.807

dGo = -22,154 Joules

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@ 0.026 mol L-1 (H2), 0.33 mol L-1 (O2), and 1.84 mol L-1 (H2O),

dG = dGo + RTlnQ

dG = -22,154 Joules + (8.314 Joules etc.)(700K) ln prod/reactants

dG = -22,154 J + (5,820.) ln [H2O]^2 / [H2]^2 [O2]

dG = -22,154 J + (5,820.) ln [1.84]^2 / [0.026]^2 [0.33]

dG = -22,154 J + (5,820.) ln 15,177

dG = -22,154 J + (5,820.) 9.628

dG = -22,154 J + 56,032

dG =+ 33,878

your answer is (3 sig figs): +33.9 kJ

but it is presented as a 2 sig fig problem, you should round off to 34Kj