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.

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?

DrBob222 · Answer

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

jhope · Answer

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

DrBob222 · Answer

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

jhope · Answer

what would k be then?

DrBob222 · Answer

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