Consider the dissociation of methane: Given that Delta_rxn H degree = 74.85 KJ m

Question

Consider the dissociation of methane: Given that Delta_rxn H degree = 74.85 KJ mol^-1 and that Delta_rxn S degree = 80.67 J mol^-1 F^-1, calculate Delta_rxn G degree and the equilibrium constant k at 298 k. Assuming Delta_rxn H degree is independent of temperature, calculate K at 50 degree C Fill in the following table to calculate the degree of dissociation, alpha, of methane at 25 degree C and a total pressure of 0.1010 bar. Please note that it is the correct that C(s, graphite) is not included in the table below

Explanation / Answer

7

(a) we know that

Delta G = Delta H - T Delta S

Delta G = 74.85 KJ / mole - 298 X 80.67 J / mol / K

Let us write all the physical quantity in same unit

so Delta S = 80.67 J / m ol / K = 0.08067 KJ / mol K

Delta G = 74.85 - 24.04 = 50.81 KJ / mole

b) Delta G = -RTlnK = -8.314 X 348 X ln K

-50.81 / 8.314 X 348 = ln K

lnK = -0.0175

Taking antilog

K = 0.982

(c) Inital moles not given

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