Question

Using standard thermodynamic data from Appendix L of your text, calculate the equilibrium constant at 298 K for the following chemical reaction:
CO(g) + H2O(l) CO2(g) + H2(g).

Answer:

CO(g) + H2O(l) CO2(g) + H2(g) G = (394.359 + 0) (137.168 237.15) = 20.04 kJ/mol G = RT ln K 20.04 = (8.3145 x 103)(298) ln K
ln K = K = 3290

Answer 1

I think your work is sloppy. It's delta G and not G (yes I know what you mean) and you've omitted all of the negative signs (yes I could see what you did). I think you put the 103 in the wrong place but calculated as if you had it in the right place. At any rate I obtained 3257 based on your delta G values. The 20.04 kJ should be changed to J and I assume that is what the 103 is for but it's in the wrong place.