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The equilibrium constants $K_{\mathrm{p},1}$ and $K_{\mathrm{p},2}$ for the reactions $\ce{X(g) <=> 2 Y(g)}$ and $\ce{Z(g) <=> P(g) + Q(g)}$, respectively, are in the ratio of $1:9$. If the degree of dissociation of $\ce{X}$ and $\ce{Z}$ are equal then the ratio of total pressure at these equilibria are?

My Approach: Let the degree of dissociation be $\alpha$

So $K_\mathrm{c,1}$ for the first reaction is $K_\mathrm{c,1} = \frac{(2\alpha)^2}{1+\alpha}$

Similarly, $Kc_2$ for the second reaction is $K_\mathrm{c,2} = \frac{(\alpha)(\alpha)}{1+\alpha}$

I don't know how to proceed from here and relate these to the ratio of $K_{\mathrm{p},1}$ and $K_{\mathrm{p},2}$. I need help here.

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  • $\begingroup$ What are the states of matter in those reactions? $\endgroup$ – andselisk Jan 18 '18 at 13:31
  • $\begingroup$ Nothing is given. I assumed them to be gas. $\endgroup$ – Piano Land Jan 18 '18 at 13:41
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The equilibrium expressions for these reactions are:

$$ K_{\text{p},1} = \frac{p_{\text{Y}}^2}{p_{\text{X}}}$$ $$ K_{\text{p},2} = \frac{p_{\text{P}}p_{\text{Q}}}{p_{\text{Z}}}$$

Since $p_\text{A} = x_\text{A}p_\text{T}$ for component $\text{A}$, we can rewrite them as:

$$ K_{\text{p},1} = \frac{(x_\text{Y}p_{\text{T},1})^2} {x_\text{X}p_{\text{T},1}} = \frac{x_{\text{Y}}^2p_{\text{T},1}}{x_\text{X}}$$ $$ K_{\text{p},2} = \frac{(x_\text{P}p_{\text{T},2})(x_\text{Q}p_{\text{T},2})}{x_\text{Z}p_{\text{T},2}} = \frac{x_{\text{P}}x_{\text{Q}}p_{\text{T},2}}{x_\text{Z}}$$

Taking the ratios:

$$ \frac{ K_{\text{p},1} }{K_{\text{p},2}} = \frac{x_{\text{Y}}^2x_\text{Z}}{x_\text{X}x_\text{P}x_\text{Q}}\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}\label{eqn}\tag{1}$$

We're asked the ratios of the total pressures, and we're given the ratios of the equilibrium constants, so we just need to find the molar fractions ratio in that equation. We know that the degree of dissociation, let's call it $\alpha$, is the same for both reactions.

Assume the first equilibrium starts with $m$ moles of $\text{X}$:

$$\begin{array}{cccc} &\ce{X(g) &<=> &2 Y(g)}\\ \hline \text{I} &m & &0\\ \text{C} &-\alpha m & &+2 \alpha m\\ \text{E} &m - \alpha m & & 2 \alpha m \end{array} $$

Then, the molar fractions are:

$$x_\text{X} = \frac{n_\text{X}}{n_\text{T}} = \frac{m - \alpha m}{m - \alpha m + 2 \alpha m} = \frac{1 - \alpha}{1 + \alpha}$$ $$x_\text{Y} = \frac{n_\text{Y}}{n_\text{T}} = \frac{2\alpha m}{m - \alpha m + 2 \alpha m} = \frac{2\alpha}{1 + \alpha}$$

Similarly, for the second equilibrium, assuming we start with $r$ moles of $\text{Z}$:

$$\begin{array}{cccccc} &\ce{Z(g) &<=> & P(g) &+ &Q(g)}\\ \hline \text{I} &r & &0 & & 0\\ \text{C} &-\alpha r & &+\alpha r & &+\alpha r\\ \text{E} &r - \alpha r & &\alpha r & & \alpha r \end{array} $$

And we calculate the mole fractions of each species:

$$x_\text{Z} = \frac{n_\text{Z}}{n_\text{T}} = \frac{r - \alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{1 - \alpha}{1 + \alpha}$$ $$x_\text{P} = \frac{n_\text{P}}{n_\text{T}} = \frac{\alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{\alpha}{1 + \alpha}$$ $$x_\text{Q} = \frac{n_\text{Q}}{n_\text{T}} = \frac{\alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{\alpha}{1 + \alpha}$$

Plugging the found molar fractions into $\ref{eqn}$, we have:

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

Which simplifies to: $$\frac{ K_{\text{p},1} }{K_{\text{p},2}} = 4\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$

Since $\displaystyle \frac{ K_{\text{p},1} }{K_{\text{p},2}} = \frac{1}{9}$, we get:

$$\frac{1}{9} = 4\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$

And finally:

$$\frac{p_{\text{T},1}}{p_{\text{T},2}} = \frac{1}{36}$$

So the total pressures are on a ratio of $1:36$.

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