CO(g) + 2H2(g) ↔ CH3OH(g) Kc=14.5 (500K)

a. In which direction will the reaction proceed if the reaction vessel contains 0.500 atm CO, 0.500 atm H2, and 0.500 atm CH3OH? Show your calculations for full credit. Remember that Q and K both have to be calculated with either concentration or pressure so that the comparison between Q and K is valid.

I've never heard of a class called "Columbus State" -- you'd get help faster if you follow directions.

I would have answered much faster if I had seen "chemistry" posted. Instead Writeacher did you a favor and corrected the heading or I never would have looked at "columbus State".

The easy way to do this is to change Kc to Kp. Kp = Kc(RT)^delta n.
Then run a Q rxn for it.Then compare Kp with Qp. I can help further if you need it.

To determine the direction in which the reaction will proceed, we need to compare the reaction quotient (Q) with the equilibrium constant (K). Q is calculated using the same formula as K, but with the concentrations or pressures of the reactants and products at a specific point during the reaction.

In this case, we are given the initial pressures of CO, H2, and CH3OH, so we can use pressure to calculate Q.

The reaction quotient, Q, is calculated as follows:

Q = (P(CH3OH)) / ((P(CO))^1 * (P(H2))^2)

Given that P(CH3OH) = 0.500 atm, P(CO) = 0.500 atm, and P(H2) = 0.500 atm, we can substitute these values into the formula:

Q = (0.500) / ((0.500)^1 * (0.500)^2)
= 0.500 / (0.500 * 0.250)
= 0.500 / 0.125
= 4

Now, we can compare Q with K to determine the direction in which the reaction will proceed.

Since Kc = 14.5, and Q = 4, we can make the following comparison:

Q < K
4 < 14.5

Since Q is less than K, it means more reactants need to convert into products to reach equilibrium. Therefore, the forward reaction will proceed to the right to reach equilibrium.

In summary, based on the comparison of Q and K, the reaction will proceed in the forward direction (from left to right) to reach equilibrium.