`DeltaH` for the reaction,
`2C(s) + 3H_2(g) to C_2 H_6(g) " is " - 84.4 kJ " at " 25^@C`. Calculate `DeltaU` for the reaction at `25^@C. (R = 8.314 J K^(-1) mol^(-1))`

Explanation

`DeltaH = DeltaU + DeltaN_(g) RT`
`Deltan_(g) `= (moles of product gases) - (moles of reactant gases)
`Deltan_g = 1 - 3 = -2 mol`
`DeltaH = -84.4 kJ, R = 8.314 JK^(-1) mol^(-1)`
` = 8.314 xx 10^(-3) kJ K^(-1) mol^(-1)`
Substitution of these in above
` - 84.4 kJ = DeltaU + 8.314 xx 10^(-3) kJ K^(-1) mol^(-1) xx 298 K xx (-2 mol)`
`DeltaU - 4.96 kJ`
Hence, `Delta U = -84.4 kJ + 4.96 kJ = -79.44 kJ`