Question

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species.

For the reaction

C2H6(g)+H2(g)↽−−⇀2CH4(g)

the standard change in Gibbs free energy is Δ?°=−69.0 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are ?C2H6=0.300 atm, ?H2=0.500 atm, and ?CH4=0.950 atm?

Answer 1

The equation for Gibbs Free Energy is

G = + RT lnQ

For a gas phase reaction

aA(g) + bB(g) cC(g) + dD(g)

Q_p =   

hence

For Given Reaction

where = -69.0kJ/mol ; T = 298K ; R = 8.314 J K^-1 mol^-1

PC2H6 = 0.300 atm ; PH2 = 0.500 atm ; PCH4 = 0.950 atm

= -69000 J mol^-1 + ( 8.314 J K^-1 mol^-1 )(298 K)

= -69000 J mol^-1 + 4446.086 J mol^-1

⁼ - 64553.914 J mol^-1

= - 64.55 kJ mol^-1