Question

CO(g) + Cl2 (g) ⇌ COCl2 (g) Kp = 22.5 at 395 °C . A sample of COCl2 (g) is introduced into a constantvolume vessel at 395 °C and observed to exert an initial pressure of 0.351 atm. When equilibrium is established, what will be the total pressure within the vessel?

Answer 1

Answer:

$P_T=0.456atm$

Explanation:

Hello,

In this case, it is convenient to rewrite the chemical reaction considering the COCl₂ as the reactant, so we must invert the equilibrium constant as shown below:

$Kp'=\frac{1}{Kp} =\frac{p_{CO}^{eq}p_{Cl_2}^{eq}}{p_{COCl_2}^{eq}}=0.0444$

Thus, by introducing the change $x$ due to reaction's progress, one obtains:

$0.0444=\frac{x^2}{0.351-x}$

Solving for $x$ via quadratic equation or solver, one obtains:

$x_1=-0.149atm\\x_2=0.105atm$

Clearly, the solution is $x=0.105atm$ for which the total pressure at equilibrium is:

$P_T=p_{CO}^{eq}+p_{Cl_2}^{eq}+p_{COCl_2}^{eq}=0.105atm+0.105atm+(0.351atm-0.105atm)\\P_T=0.456atm$

Best regards.