I'm at a slight disadvantage since I haven't seen part A but I think all you need to do is convert grams to moles, then use
dGrxn= (n*dGf products) - (n*dGf reactants)
At 25∘C the reaction from Part A has a composition as shown in the table below.
Substance Pressure
(atm)
C2H2(g) 4.65
H2(g) 3.85
C2H6(g) 5.25×10−2
What is the free energy change, ΔG, in kilojoules for the reaction under these conditions?
2 answers
C2H2(g) + 2H2(g) = C2H6(g)
dG° = -RTLn(Kp)
and
Kp = P(C2H6) / (P(C2H2) x P(H2)^2)
Kp = 5.25 × 10−2 / (4.65 x 3.85^2)
Kp = 7,62 x 10-4
So
dG°=-8.31 x 298 x Ln(7,62x10-4)
dG° = 17.779 kJ > 0
dG° = -RTLn(Kp)
and
Kp = P(C2H6) / (P(C2H2) x P(H2)^2)
Kp = 5.25 × 10−2 / (4.65 x 3.85^2)
Kp = 7,62 x 10-4
So
dG°=-8.31 x 298 x Ln(7,62x10-4)
dG° = 17.779 kJ > 0
1 answer