A possible reaction for converting methanol to ethanol is:
CO(g) + 2H2(g) + CH3OH(g) -----> C2H5OH(g) + H2O(g)
I completed parts A-C correctly, and found the following info:
delta H = -165.6 kJ
delta S = -227.4 J/K
delta G = -97.9 kJ
The next question is asking to estimate Kp for the reaction at 780K.
I'm not sure how to solve this part, can anyone help?
I tried using the deltaG=-RTlnK equation and got 1.27x10^7, but it was marked incorrect.
Thanks for all of the information but it would have helped if you had shown how you obtained the 1.27 number. I used this
97900 = 8.314*780*lnK and obtained approx 4E6.
To estimate Kp for the reaction at 780K, you need to use the equation ΔG = -RTln(Kp), where ΔG is the change in Gibbs free energy, R is the ideal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and Kp is the equilibrium constant.
However, there seems to be a problem with the units in your calculation. You mentioned that ΔG = -97.9 kJ, which is already in kilojoules. When using the ideal gas constant, you need to ensure that all the units are consistent. Therefore, you should convert ΔG to joules before plugging it into the equation.
Let's convert -97.9 kJ to joules:
ΔG = -97.9 kJ = -97,900 J
Now, let's plug the values into the equation and solve for Kp:
ΔG = -RTln(Kp)
-97,900 J = -(8.314 J/(mol·K))(780 K)ln(Kp)
Simplifying the equation, we have:
-97,900 J = -6430.32 Jln(Kp)
Divide both sides by -6430.32 J to isolate ln(Kp):
ln(Kp) = -97,900 J / -6430.32 J
ln(Kp) ≈ 15.2
Now, to find Kp, you need to exponentiate both sides of the equation:
Kp ≈ e^(15.2)
Calculating the value using a calculator, you should obtain:
Kp ≈ 4.32 × 10^6 (rounded to two significant figures)
Therefore, the estimated value for Kp at 780K is approximately 4.32 × 10^6.