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InterviewSolution
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The enthalpy change `(DeltaH)` for the reaction, `N_(2(g))+3H_(2(g)) rarr 2NH_(3g)` is `-92.38kJ` at `298K` What is `DeltaU` at `298K ?` |
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Answer» `DeltaH` and `DeltaU` are related as `DeltaH = DeltaU +Deltan_(g)RT` For the reaction, `N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g)` `Deltan_(g) = 2 -(1+3) =- 2mol,T = 298K` `DeltaH =- 90.00 kJ, R = 8.314 J K^(-1) mol^(-1)` `-90000 = DeltaU +(-2mol) xx (8.314 J mol^(-1)K^(-1)) xx (298K)` `-9000 = DeltaU - 4955` `DeltaU =- 90000 + 4955 = 85045J =- 85.045 kJ` |
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