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$$\ce{NH2COONH4(s) <=> 2 NH3(g) + CO2(g)} \qquad K = \left(\frac{p(\ce{NH3})}{p^\circ}\right)^2\times\frac{p(\ce{CO2})}{p^\circ}$$

Assuming that you start this experiment with pure solid $\ce{NH2COONH4}$ in an evacuated container and allow it to dissociate until equilibrium is reached, show that the equilibrium constant can be rewritten as

$$K = \frac{4}{27}\left(\frac{p_\mathrm{tot}}{p^\circ}\right)^3.$$

Let $n_0$ be the initial amount of ammonium carbamate and $\alpha$ be the factor describing the extent of dissociation:

$$ \begin{array}{cccccccc} \ce{&NH2COONH4(s) &<=> &2 NH3(g) &+ &CO2(g)} &\quad &\text{Total}\\ & n_0 && 0 && 0 && n_0\\ & (1 - \alpha)n_0 && 2\alpha n_0 && \alpha n_0 && (1 + 2\alpha)n_0\\ \end{array} $$

We can express mole fraction $x_i$ and partial pressures $p_i$ for $i$th component as follows:

$$x(\ce{NH2COONH4}) = \frac{n(\ce{NH2COONH4})}{n_\mathrm{tot}} = \frac{(1 - \alpha)n_0}{(1 + 2\alpha)n_0} = \frac{1 - \alpha}{1 + 2\alpha}\tag{1}$$

$$x(\ce{NH3}) = \frac{n(\ce{NH3})}{n_\mathrm{tot}} = \frac{2\alpha}{1 + 2\alpha}\tag{2}$$

$$x(\ce{NH3}) = \frac{n(\ce{CO2})}{n_\mathrm{tot}} = \frac{\alpha}{1 + 2\alpha}\tag{3}$$

$$p(\ce{NH3}) = x(\ce{NH3})\cdot p_\mathrm{tot} = \frac{2\alpha p_\mathrm{tot}}{1 + 2\alpha}\tag{4}$$

$$p(\ce{CO2}) = x(\ce{CO2})\cdot p_\mathrm{tot} = \frac{\alpha p_\mathrm{tot}}{1 + 2\alpha}\tag{5}$$

Finally, plugging in the partial pressures into the given expression for the equilibrium constant:

$$ \begin{align} K &= \left(\frac{p(\ce{NH3})}{p^\circ}\right)^2 \times \frac{p(\ce{CO2})}{p^\circ} \\ &= \left(\frac{2\alpha p_\mathrm{tot}}{(1 + 2\alpha)p^\circ}\right)^2 \times \frac{\alpha p_\mathrm{tot}}{(1 + 2\alpha)p^\circ} \\ &= \frac{4\alpha^3p_\mathrm{tot}^3}{(1 + 2\alpha)^3(p^\circ)^3} \tag{6} \end{align} $$

This leads me to the right answer if I make an assumption that all of the ammonium carbamate dissociates. The question doesn’t say I could do this, hence why I’m not sure how to proceed. Any suggestions are greatly appreciated for deriving this.

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    $\begingroup$ Twice as many moles of $\ce{NH3}$ are formed as $\ce{CO2},$ so the partial pressure of $\ce{NH3}$ is $\frac{2}{3}p_\mathrm{tot}$ and the partial pressure of $\ce{CO2}$ is $\frac{1}{3}p_\mathrm{tot}$, where $p_\mathrm{tot}$ is the total pressure of the gases formed. Also, the carbamate is a solid species, so it is not involved in the equilibrium constant. $\endgroup$
    – Chet Miller
    Commented Feb 1, 2021 at 15:22

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Let:

$C$ represent $\ce{NH3}$

$D$ represent $\ce{CO2}$

The expression of $K$ given can be set in terms of $K_P$:

$$K=P_C^2\;P_D\;\left(\frac{1}{P^o}\right)^{\Delta n}$$

$$K=K_P\;\left(\frac{1}{P^o}\right)^{\Delta n}$$

$K_P$ can be set in terms of $K_X$:

$$K_P=K_X\;\left(P\right)^{\Delta n}$$

Substituting above:

$$K=K_X\;\left(\frac{P}{P^o}\right)^{\Delta n}$$

For this reaction, $\Delta n=3$, so we have:

$$K=K_X\;\left(\frac{P}{P^o}\right)^3$$

Now, all we need to do is evaluate $K_X$.

Since total pressure $P$ only depends on products $C$ and $D$, and considering the fact that $C$ and $D$ are in a $2:1$ molar ratio:

$$X_C=\frac{c}{\Delta n}=\frac{2}{3}$$

$$X_D=\frac{d}{\Delta n}=\frac{1}{3}$$

We can now evaluate $K_X$:

$$K_X=X_C^2\;X_D=\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)=\frac{4}{27}$$

Finally, substituting this value for $K_X$ above, we get:

$$K=\frac{4}{27}\;\left(\frac{P}{P^o}\right)^3$$

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