Please explain the answers thoroughly as well as allthe necessary steps arriving

Question

Please explain the answers thoroughly as well as allthe necessary steps arriving at the answers. Thanks. Calculate the equilibrium constant for the given reactioncarried out at 45 oC. Reaction: The formation ofcarbon monoxide as from the combustion (using pure oxygen gas) ofcarbon graphite. Please explain the answers thoroughly as well as allthe necessary steps arriving at the answers. Thanks. Please explain the answers thoroughly as well as allthe necessary steps arriving at the answers. Thanks. Calculate the equilibrium constant for the given reactioncarried out at 45 oC. Reaction: The formation ofcarbon monoxide as from the combustion (using pure oxygen gas) ofcarbon graphite.

Explanation / Answer

Formation of CO is C(graphite) + (1/2)O2 (g)..............>CO (g) rG0 = 1mol *fG0 (CO (g)) -{1mol*fG0 (C (graphite)) +(1/2)mol * fG0 (O2(g)) }          = 1mol*-137.17kJ/mol - {1mol * 0kJ/mol + 1mol * 0kJ/mol}          =-137.17kJ We know G = G0 + RTlnQ When reaction is at equilibria then G = 0 hence G0 =-RTlnK                    (Q = K at equilibrium) K = e (-G0   /RT)         = e(-(-137170J) / 8.314 * (45+273.2))         = e51.86        = 3.33x10^22        = 3.33x10^22

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