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.
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!
2 answers
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