Question

Calculate `DeltaG_f^°` of `CH_4` from the tabulated values?

What is `DeltaG_f^°` of `CH_4(g)` at `298K` ?

Answer 1

Seemingly simple, this takes a bit of toying around (took me two hours!): we need to construct an equation from the values, as such,

`CO_2(g) + 4H_2(g) to CH_4(g) + 2H_2O(g)`

Derive the enthalpy and entropy of the reaction,

`DeltaH° = (2mol*(-241.8kJ)/(mol) +1mol * (-74.8)/(mol)) - (1mol * (-393.5kJ)/(mol)) approx (-164.9kJ)/(mol)`
`DeltaS° =(2mol * (188.8J)/(mol*K) + 1mol * (186.3J)/(mol*K)) - (4mol * (130.7J)/(mol*K) + 1mol * (213.7J)/(mol*K)) approx (-172.6J)/K`

Calculate the standard free energy of reaction from here,

`DeltaG° = (-164.9*10^3J)/(mol) + 298K * (172.6J)/K approx (-113.5kJ)/(mol)`

and finally, solve for the free energy of formation of methane,

`(-113.5kJ)/(mol) = (2mol*(-228.6kJ)/(mol)+DeltaG_(CH_4)^°) - (1mol * (-394.4kJ)/(mol))`
`therefore DeltaG_(CH_4)^° approx (-50.7kJ)/(mol)`