In phosgene gas reaction at $\pu{400 ^\circ C}$, the initial pressure $p_{\ce{CO}}= \pu{342 mmHg}$ and $p_{\ce{Cl2}}= \pu{352 mmHg}$ and the the total pressure at equlibirum is $\pu{440 mmHg}$.
$$\ce{CO + Cl_2 = COCl_2}$$ Calculate the percentage dissociation of phosgene at $\pu{400^\circ C}$ at $\pu{1 atm}$.

Let $x$ be the decrease in pressure of both $\ce{CO}$ and $\ce{Cl_2}$ \begin{array}{lcccc} & \ce{CO &+& Cl_2 &<=> &COCl_2}\\\hline \text{Initial pressure of substance} & 342 && 352 && 0\\ \text{Final pressure of substance} & 342-x && 352-x && x\\\hline \end{array}

$$342 - x + 352 - x + x = 440 \implies x = 254$$

Using the formula for $K_p$ I found it to be $\approx 13$. For the dissociation at $\pu{760 mmHg}$, $$\ce{COCl2 <=> CO + Cl2} \implies K_p= \frac{1}{13}$$

Let $y$ be the decrease in pressure of phosgene which is proportional to the amount of substance of it lost or it's degree of dissociation. $$\frac{1}{13} = \frac{y^2}{760-y} \implies y \approx 7.6$$

The value of $y$, that I have obtained, must be proportional to percentage dissociation, so percentage dissociation should be $7.6 \%$. But the answer given is $20.6 \%$.

I am not looking for the entire solution but would like to know where have I misapplied the concepts. I am doubtful about my second step, but can't figure out how it's wrong.

  • $\begingroup$ I don't understand how you computed $K_{\mathrm{p}} = 13$. I got $34.0$ Double check your work. $\endgroup$
    – Zhe
    Commented Oct 8, 2017 at 23:43

1 Answer 1


Here goes a possible answer, I tried to describe it step by step so you can see your (and my) possible errors. The first part of your calculation was conceptually good, but you make the calculation wrong.

As one of the comments say, $$K_p=\frac{(342-254)*(352-254)}{254}=33.95$$

So you have $K_p$. Then

\begin{array}{lcccc} & \ce{CO &+& Cl_2 &<=> &COCl_2}\\ \text{Initial pressure of substance} &0 && 0 && A\\ \text{Final pressure of substance} & x && x && A-x\\~\\ \end{array}

With the condition that $x + x + A-x=760$, we have $A+x=760$.

Using our first calculation:

$$K_p=\frac{(x)*(x)}{A-x}=33.95$$ (Kp is the same at same $T$)

$$K_p=\frac{x^2}{760-2x}=33.95$$ here we get x=130.229, then: dissociation of $A= 1-\frac{(A-x)}{A}$ we get dissociation of A=1-0793=0.207

or 20.7% (0.01% flew away)

  • 2
    $\begingroup$ @Abcd Neither 13 or 1/13 are correct... Also, I think you'd need to assume $K_{\mathrm{p}}$ was relatively constant over a range of pressures to solve this problem. $\endgroup$
    – Zhe
    Commented Oct 9, 2017 at 12:55
  • 1
    $\begingroup$ @Zhe Isn't $K_p$ independant of pressure? I mean is this not true generally? $\endgroup$
    – jonsno
    Commented Oct 9, 2017 at 15:27
  • $\begingroup$ We generally assume it to be independent, and it likely is fairly constant for a range of pressures, but there's no reason why it should be constant for all pressures. $\Delta G^{\varnothing}$ isn't even constant at all temperatures. $\endgroup$
    – Zhe
    Commented Oct 9, 2017 at 18:15

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