Calculate the equilibrium constants at 25C for each of the following reactions

Calculate the equilibrium constants at 25C for each of the following reaction:

2H2S(g) <==> 2H2(g) + S2(g)

I know the equation for this one is Delta G=-RTlnK, but I keep on getting really weird numbers.

Looking up the gibbs free energy in the back of my book I found that:
H2S: -33.4 kJ/mol
H2: 0 kJ/mol
S2: 79.7 kJ/mol

So by looking at the reaction I multiplied H2S by two and got -66.8, and because delta G is products minus reactants I got -146.5. I then converted that from kJ to J, which was 146500.

Putting it all together I have: -146500=(-8.314)(298)(ln(k))

And using wolframalpha to solve for k I keep on getting a ridiculous number (e^36625000/619393)

What am I doing wrong here?

You're right that Delta G is products minus reactants.

But here you did reactants minutes products.

I would recommend using kJ. The numbers are smaller and easier to keep track of. So that means change 8.314 to the kJ version=8.314x10^-3

Edit:

Products are on the right side. Products are what you get after a reaction. This is your H2 and S2 then minus the H2S

2H₂S ↔ 2H₂ + S₂

ΔG° = (1000 J / kJ)[2(0 kJ) + 79.7 kJ] – [2(–33.56 kJ)] = 146820 J

ΔG° = –RTlnK
146820 J = –(8.314 J /(mol·K))(25 + 273.15 K)ln(K)
K = 1/e^(1468200000/24788191) = 1.89 × 10⁻²⁶

Basically you got the sign of ΔG° wrong. ΔG° for H₂S is negative so the negatives cancel in products – reactants.