Question

Q: A reaction vessel is pressurized with 1.600 atm of CO2 . At equilibrium, the pressure of CO2 is 1.200 atm. What is the equilibrium constant, KP? 2 CO(g) + O2 (g) −−−− 2 CO2 (g)

Any detail to this problem solution will help to better understand, thanks.

*I have posted this problem for help already but I'm not sure if the response was correct, mostly due to the division part. My understanding is that the sum which is 2 CO2 (g) goes first on top of division the, then 2 CO(g) + O2 (g) goes on bottom of divion line.Can someone solve this problem to better understand, thanks.

Answer 1

Given reaction : 2 CO(g) + O2 (g) −−−− 2 CO2 (g)

But in the begining 1.6atm of CO2 is present .then it is decomposes to 2 CO(g) + O2 (g).

So the actual reaction : 2 CO2 (g) <−−−−> 2 CO(g) + O2 (g)

                Initial         1.6 atm       0atm          0atm                    

at equilibrium 1.2 atm       0.4atm    0.2 atm             

               Kp = [CO]^2[O2]/[CO2]^2  

   = (0.4^2)*0.2 / (1.2^2)

              = 0.022 M