[lgn98868]

Struggling with this question asking to calculate Kp at 1000 degrees celsium for the reaction:

                              C_graphite                  +                CO_{2 (g)}                    2CO_g

Standard entropy°               0                          0.2137 (kj)                          0.1975 (kj)          ΔS= 0.1813 kj
Enthalphy of Form.°             0                          -393.5 (kj)                          -110.5 (kj)          ΔH= 172.5 kj

We could calculate free energy= ΔH° - T ΔS°
                                               172.5 - (1273 K) (0.1813) = so at 1273 K  ΔG= -58.2949

Use this free energy in the equation ΔG= -(RT)lnKp                        -58.2949= -(0.008314 x 1273 K) lnK
                                                                                           Solving this gives Kp= 246.65



The book says the answer is 120.  Where am I going wrong?

[lgn98868]

After thinking on it, is the Kp that I calculated per mole? Dividing that by two would give roughly 120 without worrying about sig figs.

[mjc123]

Your calculations look correct to me. If you halve the quantities, K is not halved but squarerooted, so that doesn't solve the discrepancy. Maybe the book is just wrong. It happens.

[Big-Daddy]

It doesn't help that you appear to have given entropy units of kJ. And is the standard molar entropy of carbon as graphite really 0 Jmol^-1K^-1?

[Corribus]

And is the standard molar entropy of carbon as graphite really 0 Jmol^-1K^-1?

No, it's about 5.7 J K^-1 mol^-1. The problem assumes these values are the same at 1273 K as they are are at 273 K, which is probably a poor assumption.

[Big-Daddy]

And is the standard molar entropy of carbon as graphite really 0 Jmol^-1K^-1?

No, it's about 5.7 J K^-1 mol^-1. The problem assumes these values are the same at 1273 K as they are are at 273 K, which is probably a poor assumption.

Ok, but even at 273 K (or 298.15 K which is standard) I wouldn't expect anything except H⁺ (aq) to be given S_m°=0. The data seems faulty...?

[Corribus]

Right, it's not 0, it's ~5.7. Compared to the other values, though, that's pretty negligible.