Calculate the enthalpy of formation of methane given that the enthalpies of combustion of methane, graphite and hydrogen are 890.2 KJ, 393.4 KJ and 285.7 KJ `mol^(-1)` respectively

Explanation

We are given `: (i) CH_(4) + 2O_(2) rarr CO_(2)+ 2H_(2)O,DeltaH== -890.2 kJ mol^(-1)`
(ii) `C+ O_(2) rarr CO_(2), Delta H = - 393.4 kJ mol^(-1) ` `(iii) H_(2)+(1)/(2)O_(2) rarr H_(2)O , Delta H = - 285.7 kJ mol^(-1)`
We aim at `: C+ 2 H_(2) rarr CH_(4), Delta H = ?`
In order to get this thermochemical equation, multiply eqn. (iii) by 2 and it to eqn. (ii) and then subtract eqn. (i) from their sum. We get `:`
`C+2H_(2) rarr CH_(4), Delta H = -393.4+2(-285.7) - (-890.2) kJ mol^(-1) = -74.6 kJ mol^(-1)`
Hence, the heat of formation of methane is ` : Delta _(f) H= -74.6 kJ mol^(-1)`