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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

`Delta H = Delta U + Delta n_(g)RT`
Here, `Delta H=-92.38 kJ =-92380 J, Delta n_(g)=2-4=-2` mol
`therefore -92380 J = Delta U+(-2 mol)xx(8.314 JK^(-1)mol^(-1))(298 K)`
`-92380 J = Delta U - 4955.144 J`
`Delta U = -92380+4955.144 J`
`Delta U=-87424.856 J` or `Delta U = -87.42 J`
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