Calculate the `Delta H` at 298 K for the reaction `(1)/(2)N_(2)(g) + (3)/(2)H_(2)(g) rarr NH_(3)(g)` given that `Delta H` for the formation of `NH_(3)` has a valve of `-46.0 kJ mol^(-1)`
`(R = 8.314 JK^(-1) mol^(-1))`.

Explanation

Using the eqn. `Delta H = Delta U + Delta nRT`
`Delta H = -46.0 kJ mol^(-1) = -46000 J mol^(-1), Delta n = 1 - 2 = -1, R = 8.31 JK^(-1) mol^(-1)`, `T = 298 K " "Delta H = ?`
`Delta U = Delta H - Delta nRT`
`= -46000 J mol^(-1) -[(-1) xx 8.314 JK^(-1) mol^(-1) xx 298 K]`
`= 43523 J mol^(-1) = -43.52 kJ mol^(-1)`
Change in the internal energy for this reaction `= -43.52 kJ mol^(-1)`.