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Question

Calculate the standard free energy change for the formation of methane at 298 K. C(graphite)+2H2(g)CH4(g)
Given that, Δ of CH4=74.81KJmol1 and ΔS of Cgraphite,H2(g) and CH4(g) are 5.70K1mol1, and 186.3JK1mol1 respectively.

A
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Solution

The correct option is B
Let us consider values of ΔfH and S
C(graphite) H2(g) CH4(g)
ΔrH/KJmol1 0 0 - 74.81
S/JK1mol1M 5.70 130.7 186.3
Now using the above data
ΔrH=ΔfHCH4(g){ΔfHC(graphite)+2ΔfHH2(g)}
= 74.81KJmol1(0+0)=74.81KJmol1
ΔrSm=SmCH4(g){SmC(graphite)+2SmH2(g)}
=[1(186.3JK1mol1)][1(5.70JK1mol1)+2(130.7JK1mol1)]=80.8JK1mol1Substituting the value of ΔrH and ΔrS in Gibbs equation, we get,ΔrG=ΔrHTΔfS=74.81KJmol1(298K)(80.8×103JK1mol1)=74.81KJmol1+24.1KJmol1=50.7KJmol1

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