The enthaply change `(Delta H)` for reaction, `N_(2) (g) + 3H_(2)(g) to 2NH_(3) (g)` is 92.38 kj at 298 K. What is `Delta U` at 298 K ?

Explanation

We are given that
`Delta H = - 92.38 kj/mol = - 92380 j/mol, T = 298 K`
`R = 8.314 j K^(-1) mol^(-1)`
We know that,
`Delta n_(g)` = Change is number of moles of gaseous products and gaseous reactants
= `n_(p) - n_(r) = 2 - 4 = - 2`
`Delta n_(g) = - 2`
We are to find out the `Delta U` at 298 K.
Using the relation.
`Delta H = Delta U + Delta n_(g) RT`
By putting all values in equation (i), we get
`-92380 j/mol = Delta U - 2 xx 8.13 jK^(-1)mol^(-1) xx 298 K = Delta U - 4955 j/mol`
`therefore Delta U = (- 92380 + 4955) j/mol = - 87.425 kj/mol`