18. NEED HELP FAST

Question

At  25 ∘C , the equilibrium partial pressures for the reaction

3A(g)+4B(g)↽−−⇀2C(g)+2D(g)

were found to be  𝑃A=5.87  atm,  𝑃B=4.85  atm,  𝑃C=4.79  atm, and  𝑃D=4.93  atm.

What is the standard change in Gibbs free energy of this reaction at  25 ∘C ?

Δ𝐺∘rxn=

Answer 1

Since the Reaction is at equilibrium, Δ𝐺=0, and Δ𝐺°= -RTlnQ, where Q is the reaction quotient. In your case, your reaction quotient will be the partial pressures, so Q= (P_C)^C(P_D)^D/(P_A)^A(P_B)^B. From there, just plug and solve.