The degree of dissociation of $\ce{PCl5}$ is 0.4 at 2 atm pressure and T kelvin. At what pressure will degree of dissociation of $\ce{PCl5}$ be 0.6?

Taking the degree of dissociation as 0.4, I got the partial pressure of $\ce{PCl5}$ to be $\frac{2}{2.33}$. I then thought of taking the degree of dissociation as 0.6 and multiplying the mole fraction of $\ce{PCl5}$ with the total unknown pressure and equating it to the partial pressure of $\ce{PCl5}$. But I then realized that wouldn't the partial pressure change with change in total pressure? I'm not able to figure this question out. Any help would be greatly appreciated!


1 Answer 1


The dissociation of $\ce{PCl5}$ is given by the equation:

$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$

with $\displaystyle K_\text{p} = \frac{p_{\ce{Cl2}}p_{\ce{PCl3}}}{p_{\ce{PCl5}}}$. Putting this in terms of total pressure, $\displaystyle K_\text{p} = \frac{x_{\ce{Cl2}}x_{\ce{PCl3}}}{x_{\ce{PCl5}}}p_\text{T}$.

In terms of the degree of dissociation, $\alpha$, assuming an initial amount $y$ of moles of $\ce{PCl5}$, at equilibrium, we have $n_{\ce{PCl5}} = y - \alpha y\,\text{mol}$, $n_{\ce{PCl3}} = \alpha y\,\text{mol}$, and $n_{\ce{Cl2}} = \alpha y\,\text{mol}$. Thus, their molar fractions are $x_{\ce{PCl5}} = \frac{1 - \alpha}{1 + \alpha}$, $x_{\ce{PCl3}} = \frac{\alpha}{1 + \alpha}$, and $x_{\ce{Cl2}} = \frac{\alpha}{1 + \alpha}$, so:

$$K_\text{p} = \frac{\left(\frac{\alpha}{1 + \alpha}\right)^2}{\left(\frac{1-\alpha}{1+\alpha}\right)}p_\text{T} = \frac{\alpha^2}{1-\alpha^2}p_\text{T}$$

Since at $p_\text{T} = \pu{2 atm}$, $\alpha = 0.4$, we get $K_\text{p} = \pu{0.381 atm}$. Hence, for $\alpha = 0.6$, since $K_\text{p}$ doesn't change (i.e. assuming it's an isothermic system), we get $p_\text{T} = \pu{0.677 atm}$. This is in agreement with Le Chatelier's principle: reducing the total pressure will push the equilibrium towards the side with more gaseous moles, so the dissociation constant in this case will be higher.

  • $\begingroup$ Thank you so much! I was confused why Kp would remain same but then realised since Kp=Kc(RT) ^deltaN, since Kc and T are constant and R and delta N are fixed values, Kp must remain the same. Thanks for the help! $\endgroup$ Mar 21, 2018 at 17:28

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