Calculate K at 298 K for the following reaction…?

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NO (g) + 1/2 O2 (g) ⇋ NO2 (g)

I know that the equation to solve for this is K=e^(deltaG/-RT) and I know that the answer is 1.74*10^6 but I am unsure why this is the answer. Please show all work.

1 Answer

  • To answer this question you would need to know the delta Go of formation of the reactants and products. I’ll look them up in my textbook.

    delta Go f NO2(g) = 51.24 kJ/mole

    delta Go f NO(g) = 86.60 kJ/mole

    delta Go f O2(g) = 0

    delta Go reaction = delta Go f products – delta Go f reactants

    = (1 mole NO2)(51.24 kJ/mole) – (1 mole NO)(86.60 kJ/mole) = -35.4 kJ = -35,400 J

    K = e^(delta G/-RT) = e^((-35,400) / -(8.31)(298)) = 1.62 x 10^6

    This is very close to your answer. Perhaps whomever came up with 1.74 x 10^6 used slightly different value for the delta Go f.

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