Question

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!

Answer 1

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.

Answer 2

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

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