InterviewSolution
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InterviewSolution
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The ethalpy of formation of methane at constant pressure and `300 K` is `-75.83 kJ`. What will be the heat of formation at constant volume? `[R = 8.3 J K^(-1) mol^(-1)]` |
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Answer» The equation for the formation of methane is `{:(C(s)+,2H_(2)(g)rarr,CH_(4)(g),DeltaH^(Theta) =- 67.83kJ,,),(,2mol,1mol,,):}` `Deltan = (1-2) =- 1` Given `DeltaH^(Theta) =- 75.83 kJ R = 8.3 xx 10^(-3) kJ K^(-1) mol^(-1), T = 300 K` Applying `DeltaH^(Theta) = DeltaU^(Theta) + DeltanRT` `- 75.83 = DeltaU^(Theta) +(-1) (8.3 xx 10^(-3)) (300)` So, `DeltaU^(Theta) =- 75.83 + 2.49 =- 73.34 kJ` |
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