May I get help on this?

Question

At 25 ∘C , the equilibrium partial pressures for the reaction 3A(g)+4B(g)↽−−⇀2C(g)+3D(g) were found to be 𝑃A=4.71 atm, 𝑃B=4.26 atm, 𝑃C=5.75 atm, and 𝑃D=4.58 atm. What is the standard change in Gibbs free energy of this reaction at 25 ∘C ? Δ𝐺∘rxn= KJ/mol

Answer 1

Here you're given stuff about equilibrium partial pressures and asked to find Gibbs free energy. Your first thought should be "How do I get from those to ∆G?" That's what's going to guide us through solving the problem. Remember that the equilibrium constant K is related to ∆G through this equation: ∆G = -RT ln(K). So if we find K, because we're given the temperature we should be able to find free energy. All K is is to set up the equilibrium expression for the gases you've got going on here:

K = P_c² * P_D³ / (P_A³ * P_B⁴)

You solve this equation and plug it into -RT ln(K) in order to find the free energy for this reaction:

K = .0923 ish

T = 298 because we always want it in Kelvin, not Celsius

∆G = -RT ln(K) = -(8.314)(298) ln(.0923) = 5906 J

This makes sense because if the K value is very low, this means that there are a lot more reactants (denominator) than products (numerator). Therefore, our free energy is a big positive number, meaning that the reaction is not spontaneous.