Calculate ΔG° for the following reaction at 298 K.?

Question

A+B --> 2D ΔH° = 775.0 kJ ΔS°=296.0 J/K
C --> D ΔH° = 446.0 kJ ΔS°=-218.0 J/K

A+B --> 2C

ΔG° = ? kJ

I know I am suppose to reverse the (C --> D) and add a coefficient of 2 to both the enthalpy and entropy.

I got this but it doesn't seem right.
ΔH° = 2(-446.0) - (-722.0) = -1674 kJ
ΔS° = 2(218) - (296) = .14kJ/k

ΔG° = -1674 - (.14)(298) =-1688.7 a really large number that does not seem right,
Please help and explain this to me.

Answer 1

I agree with your approach with the coefficients of 2 and reverse the C==>D. But aren't you supposed to add? You're adding the equations to get the final equation; therefore, you add dH and dS. And where did the 722 come from?
I believe you are confusing this with dH rxn = (n*dHproducts) - (n*dHreactants)

dH = 775 kJ + (2*-446) = ?
dS = 296 J + (2*218) = ?

Then dGo = dH - TdS
Don't forget to change dH to J; it's in kJ now.

Answer 2

You didn't change to the delta S into kJ.