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

Options

A. `-92.38 kJ`
B. `-87.42kJ`
C. `-97.34kJ`
D. `-89.9kJ`

Answer

b

Explanation

`Delta=DeltaU+Delta nRT`
[where `Deltan`= no. of molecules of products - no. of molecules of reactants]
`or . -92.38xx1000= DeltaU-2xx8.314xx298`
or , `Delta U= -87424 J= -87.421 KJ`