Enthalpies of combustion of carbon, hydrogen and methane are - 393 kJ, - 286 kJ, and -890 kJ, respectively. Calculate the enthalpy of formation of methane.

Explanation

Required equation
`C(s) + 2H_(2)(g) to CH(g) " " Delta H = ?`
Given :
`C(s) + O_(2)(g) to CO_(2)(g) " " Delta H = - 293 kJ " " (1)`
`H_(2)(s) + ((1)/(2))O_(2)(g) to H_(2) O(l) " "DeltaH = - 286 kJ " " (2)`
`CH_(4)(g) + 2O_(2)(g) to CO_(2)(g) to CO_(2)(g) + 2H_(2)O(l) " " Delta H = - 890 kJ " " (3) `
Reversing Eq.(3),
`CO_(2) (g) + 2H_(2)O(l) to CH_(4) (g) + 2O_(2)(g) " "Delta H = 890 kJ " " (4)`
`(1) + (2) xx (2) + (4) ` , we get
`C(s) + O_(2)(g) to CO_(2)(g) " " Delta H = - 393 kJ`
`2H_(2)(g) + O_(2)(g) to 2H_(2) O(l) " .. Delta H = 2 xx (-286)kJ`
`CO_(2) + 2H_(2)O(l) to CH_(4) (g) + 2O_(2)(g) " " Delta H = 890 kJ`
`C(s) + 2H_(2)(g) to CH_(4)(g) " " Delta H = - 75 kJ`
`therefore ` Enthalpy of methane is - 75 kJ.