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The enthalpy change (Delta H) for the re...

The enthalpy change `(Delta H)` for the reaction
`N_(2) (g)+3H_(2)(g) rarr 2NH_(3)(g)`
is `-92.38 kJ` at `298 K`. What is `Delta U` at `298 K`?

Text Solution

Verified by Experts

`DeltaH` and `DeltaU` are related as
`DeltaH = DeltaU + Deltan_(g)RT`
For the reaction,
`N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g)`
`Deltan_(g) = 2 - (1 + 3) = -2 mol, T = 298 K`
`DeltaH = -92.33 kJ = - 92380 J, R = 8.314 JK^(-1) mol^(-1)`
` -92380 = DeltaU + (-2 mol) xx (8.314 J mol^(-1)K^(-1))xx (298 K)`
`- 92380 = DeltaU - 4955`
`DeltaU = -92380 + 495 = -87425 J = -87.425 kJ`
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