Question

For the reaction below at 298.15 K, the standard enthalpy of reaction is ΔrH° = 41.6 kJ mol−1 and the thermodynamic equilibrium constant is K = 1.50×10−6.
2 NO(g) + I2(s) ⇌ 2 NOI(g)

What is ΔrS° for this reaction?

Do I need to take the K value into consideration or can I just solve using n and S values?

Answer 1

I think you must use K.
dGo = -RT*lnK
Substitute and solve for dGo.
Then dGo = dHo - TdSo and solve for dSo.

Answer 2

dGo = -RT*lnK
for this would it be -8.135(298)ln(298)?

Answer 3

No, You have substituted 298 for k but that isn't k. The other substitutions look OK to me.

Answer 4

what would k be then?

Answer 5

C'mon. K is given to you in the problem.