Which of the following conditions is always true at equilibrium? deltaG=0, Q=1, K=1, deltaG=deltaG(standard), deltaG(standard)=0, Q=0, deltaG(standard)=1

Question

I'm so confused, please help!

Answer 1

Here are the formulas that you need.

ΔG=-RT*lnK

ΔG=ΔGo’+RTlnQ

At equilibrium, Q=K, If K=1 then Q=1.

ln(1)=0, so ΔG=ΔGo’+RTln(1) becomes ΔG=ΔGo’

Now if Q=K, then K=1

ln(1)=0, so ΔG=-RT*ln(1)=0. So, ΔG=0.

The last three, deltaG(standard)=0, Q=0, deltaG(standard)=1, are incorrect, I believe.

Answer 2

I would go with dG = 0

Answer 3

It was a typo. When I did the work, I showed that it does equal 0. just didn't omit it.

Answer 4

My original post should say the last two, Q=0, deltaG(standard)=1, are incorrect.

Answer 5

Oh, my original post was correct. the last three are false. deltaG(standard)=ΔGo’=0 is false. Its either positive or negative, but it doesn't depend on K or Q.

Answer 6

The answer is dg=0 only. The other conditions may happen at equilibrium but are not always the case.

Answer 7

Court and Bob, you are correct. The way that I went about solving the problem, I believe, is only true for standard conditions; the problem did not state standard conditions.