The enthalpy change `(DeltaH)` for the reaction
`N_2(g) + 3H_2 (g) to 2NH_3(g)`
is -92.38 kJ at 298 K. What is `DeltaU` at 298 K ? `(R = 8.314 j K^(-1) mol^(-1))`

Options

A. `-0.3024`
B. `-0.6048`
C. `-0.12`
D. None

Answer

A

Explanation

`{:(C_(2)H_(4),+,H_(2),to,C_(2)H_(6)),(50ml,,50m,,0),(x,,x,,50ml):}`
`DeltaH = DeltaU + P(DeltaV)`
`-0.31 = DeltaU + 1.5 xx 1.01 xx 10^(5) (-50 xx 10^(-6)) xx 10^(-3)`
`DeltaU = -0.3024` kJ