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

## What is DeltaG_f^° of $C {H}_{4} \left(g\right)$ at $298 K$ ?

Dec 8, 2017

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

$C {O}_{2} \left(g\right) + 4 {H}_{2} \left(g\right) \to C {H}_{4} \left(g\right) + 2 {H}_{2} O \left(g\right)$

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)