Delta g at 298 k for this reaction is -33.3kj/mol. calculate delta g at 298k for a reaction mixture that consists of 1.9 atm n2, 2.3

To determine change in Gibbs Free Energy, we use the equation

ΔG = ΔG° + RT ln Q

where ΔG is the change in free energy at temperature T
          ΔG° is the free energy at standard temperature
            R is the universal gas constant
             T is the temperature of the system
            Q is the reaction quotient

First, we determine the reaction quotient. It is the ratio of the concentration or partial pressure of the products and the reactants. For this case, we do as follow:
P2(g) + 3 Cl2(g) -> 2 PCl3(g) 

Q = P(PCl3) ^2 / (P(P2)(P(Cl2 ^ 3)))
Q = (0.65 atm)^2 / (1.5 atm) (1.6 atm)^3
Q = 0.0688 

Substituting to the equation for free energy,

ΔG = ΔG° + RT ln Q
ΔG = -33300 J/mol + (8.314 J/mol-K)(298 K) ln 0.0688
ΔG = -39931.35 J/mol

The answer is b on plato.

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