So I have the reaction

2HBr(g) <--> H2(g)+Br2(g)

I'm supposed to find the equilibrium constant, K, using free energy (G) values. The values in my textbook for G of H2 and Br2 were zero, and the value for HBr(g) was -53.5.

So, to find delta G I did products - reactants and got (0+0)-(2(-53.5)) and got that the delta G is +107.

Next I plugged that into the equation
deltaG = -RTlnK.

107kJ/mol(1000J/1kJ) = -8.314(298K)lnK
-43.2 = lnK
K=1.75x10^-19

I thought that should be the answer, but my homework says it's wrong. What am I doing wrong here? I've done several others of the same type of problem and gotten them right, so I don't know what the difference is.

Nevermind . . . the G value for Br2 isn't 0, which was throwing off my calculations. The answer ended up being 4.96x10^-20

To find the equilibrium constant (K) using free energy (ΔG) values, you have taken the correct steps so far. However, there is a small error in your calculation. Let's go through the process again:

1. Start by finding the ΔG for the reaction. You correctly subtract the ΔG of the reactants from the ΔG of the products:

ΔG = (0 + 0) - 2(-53.5) = +107 kJ/mol

2. Convert the units of ΔG from kJ/mol to J/mol:

ΔG = 107 kJ/mol × (1000 J/1 kJ) = 107,000 J/mol

3. Use the equation ΔG = -RT ln K to solve for ln K. Rearrange the equation and plug in the values:

ln K = -ΔG / (RT)

ln K = -107,000 J/mol / [(8.314 J/mol·K) × (298 K)]

ln K ≈ -43.19

4. Finally, solve for K by taking the exponential of both sides:

K ≈ exp(-43.19)

K ≈ 1.45 × 10^(-19)

Your calculation for the equilibrium constant (K) is not entirely correct. It appears you made a mistake in your calculation of ln K, resulting in a slightly different value. Thus, the correct equilibrium constant for the given reaction is approximately 1.45 × 10^(-19), not 1.75 × 10^(-19).