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
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.
To calculate the equilibrium constant (K) at 298 K for the given chemical reaction, you need to use the standard thermodynamic data from Appendix L of your text.
The equation for calculating the equilibrium constant is given by:
ΔG = -RT ln K
where ΔG is the Gibbs free energy change for the reaction, R is the gas constant (8.3145 J/(mol·K)), T is the temperature in Kelvin (298 K in this case), and K is the equilibrium constant.
First, you need to find the ΔG for the given reaction using the standard thermodynamic data. The values for ΔG are given in kJ/mol.
ΔG = (ΔGf° of CO2 + ΔGf° H2) - (ΔGf° of CO + ΔGf° H2O)
= (394.359 kJ/mol + 0 kJ/mol) - (137.168 kJ/mol + 237.15 kJ/mol)
= 394.359 kJ/mol - 374.318 kJ/mol
= 20.04 kJ/mol
Next, you can substitute the values into the equation:
20.04 kJ/mol = (8.3145 J/(mol·K)) * (298 K) * ln K
Now, you can solve for ln K:
ln K = 20.04 kJ/mol / (8.3145 J/(mol·K) * 298 K)
= 0.08
Finally, you can find K by taking the exponential of ln K:
K = e^(0.08)
≈ 1.0831
Therefore, the equilibrium constant (K) at 298 K for the given chemical reaction is approximately 1.0831.