Answer:
[tex]p_{CO}=1.41atm\\p_{Cl_2}=1.41atm\\p_{COCl2}=0.59atm[/tex]
Explanation:
Hello,
In this case, since the undergoing chemical reaction is:
[tex]Kp=\frac{p_{CO}^{eq}p_{Cl_2}^{eq}}{p_{COCl_2}^{eq}}[/tex]
Thus, if we introduce the changing pressure, [tex]x[/tex], the law of mass action becomes:
[tex]Kp=\frac{(x)(x)}{(2.0atm-x)}[/tex]
Now, by taking the given value of Kp we solve for x as follows:
[tex](2.00atm-x)Kp=x^2\\x^2+Kp*x-2.00atm=0\\x^2+6.8x10^{-9}-2=0\\x_1=1.41atm\\x_2=-1.41atm[/tex]
therefore, the result is 1.41 atm so the equilibrium pressures turn out:
[tex]p_{CO}=1.41atm\\p_{Cl_2}=1.41atm\\p_{COCl2}=2.00-1.41=0.59atm[/tex]
Best regards.