Heat of combustion of methane at constant volume and 298 K has been found to be `-885.0 kJ mol^(-1).` Calculate the enthalpy of combustion under constant pressure. (`R=8.314 Jmol-K^(-1)`).

Explanation

The combustion of methane is
`CH _(4) (g) + 2 O _(2) (g) to CO _(2) (g) + 2 H _(2)O (l)`
Hence, `Delta n _(g) = 1 `mol ( gaseques product) -2 moles (gaseous reactants) `=-2` mols.
Change in enthalpy, `Delta H = Delta U + Delta n _(g) RT`
`Delta H =- 885000 J mol ^(-1) + (-2 mol) ( 8.314 JK ^(-1) mol ^(-1)) (298 K)`
`Delta H =- 88 5000 J mol ^(-1) - 4955 mol ^(-1) =- 889955 Jmol ^(-1) = - 889.955 kJ mol ^(-1)`