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. `- 87.42` kJ
C. `-97.34` kJ
D. `- 89.9` kJ

Answer

B

Explanation

`N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)`
`Delta n_(g)= 2- (1+3)= -2`
`W = Delta n_(g) RT`
`= -2 xx 8.314 xx 298 xx 10^(-3)kJ`
`= -4.955kJ`
`Delta H= -92.38kJ`
`Delta H= Delta U= W`
`Delta U= Delta H- W`
`= -92.38 - (-4.955)kJ`
`= -87.42kJ`