Calculate the standard free energy change for the formation of methane at 298 K. C(graphite)+2H2(g)→CH4(g) Given that, ΔofCH4=−74.81KJmol−1 and ΔS∘ofCgraphite,H2(g)andCH4(g)are5.70K−1mol−1,and186.3JK−1mol−1 respectively.
A
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B
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C
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D
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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∘/KJmol−1 0 0 - 74.81 S/JK−1mol−1M 5.70 130.7 186.3 Now using the above data ΔrH∘=ΔfH∘CH4(g)−{ΔfH∘C(graphite)+2ΔfH∘H2(g)} = −74.81KJmol−1−(0+0)=−74.81KJmol−1 ΔrS∘m=S∘mCH4(g)−{S∘mC(graphite)+2S∘mH2(g)} =[1(186.3JK−1mol−1)]−[1(5.70JK−1mol−1)+2(130.7JK−1mol−1)]=−80.8JK−1mol−1SubstitutingthevalueofΔrH∘andΔrS∘inGibbsequation,weget,ΔrG∘=ΔrH∘−TΔfS∘=−74.81KJmol−1−(298K)(−80.8×103JK−1mol−1)=74.81KJmol−1+24.1KJmol−1=50.7KJmol−1