Consider the reaction: `N_(2) + 3H_(2) hArr 2NH_(3)` carried out at constant pressure and temperature. If `DeltaH` and `DeltaU` are change in enthalpy and change in internal energy respectively, then:

Options

A. `Delta=0`
B. `Delta=DeltaU`
C. `DeltaH lt DeltaU`
D. `DeltaH gt DeltaU`

Answer

C

Explanation

`Deltan_(g)` for this reaction is `-ve`
(`Deltan_(g)=-4` if `NH_(3)` is considered as liquid and `Deltan_(g)=-2` if `NH_(3)` is considered as gas)
Now `DeltaH=DeltaU+Deltan_(g)RT`
`:. DeltaH lt DeltaU` for this reaction.