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

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 –> 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.

2 Answers

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)

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.

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