The equilibrium constant for the reaction is 37 at a certain temperature.

2H2(g) + CO(g)<==> CH3OH(g)

If there are .0293 moles of H2 and .00353 moles of CH3OH at equilibrium in a 4.53-L flask, what is the concentration of CO?

.....So far I have found:
H2= .0293/4.53= .00647 M
CH3OH= .00353/4.53= 7.79x10^-6

After that I am stuck...I can't figure out the change in concentration..please help!

I think you read your calculator wrong for CH3OH. I have 7.79E-4. Also I assume K is Kc and not Kp.

.................2H2 + CO ==> CH3OH

You don't need the "change in concn".
Kc = (CH3OH)/(H2)^2(CO)
You have Kc, C3OH and H2, solve for CO.

Calculate the equilibrium constant K(eq) for the following reaction at that temperature:

2SO2 (g) + O2 (g) = 2SO3 (g)

To find the concentration of CO at equilibrium, we need to use the balanced equation for the reaction and apply the concept of equilibrium constants.

The balanced equation is:

2H2(g) + CO(g) <==> CH3OH(g)

The equilibrium constant (K) expression for this reaction is:

K = [CH3OH] / ([H2]^2 * [CO])

Given that the equilibrium constant (K) is 37, and we know the concentrations of H2 and CH3OH at equilibrium, we can rearrange the equation to solve for the concentration of CO at equilibrium:

[CO] = [CH3OH] / ([H2]^2 * K)

Substituting the values you calculated for the concentrations of H2 and CH3OH:

[H2] = 0.00647 M
[CH3OH] = 7.79x10^-6 M
K = 37

[CO] = (7.79x10^-6) / ((0.00647)^2 * 37)

Simplifying this expression will give you the concentration of CO at equilibrium.

Note: Make sure to be consistent with the units used throughout the calculations.