Question

Enthalpy of sublimation of graphite is 761 kJ `mol^(-1)`

Calculate standard enthalpy of formation of `CH_(4)`

Answer 1

Correct Answer - Standard enthalpy of formation of `CH_(4)= Delta_(f) H_(CH_(4(g)))^(o)`
`=- 67.2 kJ mol^(-1)`
Given : `Delta _(sub) H_("graphite")^(o) = 716 kj mol^(-1)`

Thermochemical equation for the formation of `CH_(4)`
`C_((s)) + 2H -H_((g)) to [H- overset(H)overset(|)underset (H)underset(|)(C ) - H_((g))]`
`Delta_(f)H^(o) = [ underset("of graphite " + 2Delta H_(H-H)^(o))"Enthalpy of sublimation "] = [4 Delta H_(C-H)^(o)]`
`= [Delta _(sub)H_("graphite")^(o) + 2Delta H_(H-H)^(o) ]- [4Delta H_(C-H)^(o)]`
`= [ 716 + 2xx 436.4] - [4 xx 414]`
`= [716 + 872.8] - [ 1656]`
`= 1588.8 - 1656`
`= - 67.2 kJ mol^(-1)`