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The reaction $$\ce{A.3H2O(s) <=> A.H2O(s) + 2H2O (g)}$$ has a $K_p = \pu{400 mmHg2}$

One millimole each of $\ce{A.3H2O}$ and $\ce{A.H2O}$ are taken in a $\pu{1 liter}$ container at $\pu{300 K}$

Relative humidity is $80 \%$ and saturation water vapor pressure is $\pu{85 mm Hg}$ at $\pu{300 K}$. Find the final composition of the container


My attempt

\begin{align} K_p &=(P_\ce{H2O})^2 = 400 \\ \therefore P_\ce{H2O} &= \pu{20 mm Hg} \end{align}

Now,

\begin{array}{lc} \hline \ce{&A.3H2O &<=> &A.H2O &+ &2H2O(g)} \\ \hline i &1\times10^{-3} && 1 \times10^{-3} && \pu{68 mm Hg} \\ c &+x && -x && -P \\ e &2\times10^{-3} && 0 && \pu{30 mm Hg} \\ \hline \end{array}

$\pu{38 mm Hg}$ drops because $\pu{2 millimoles}$ of water vapor was consumed

$$ P=\frac{1}{12}\frac{300\times 2\times 10^{-3}}{1} = \pu{38 mm Hg}$$

The answer should have been $\pu{20 mm Hg}$ pressure of water vapor but I am afraid there isn't enough reactant to make that much vapor pressure. I asked a friend and he suggested that we can take $\pu{2 millimoles}$ each of the reactants as they are solid and don't influence the $K_p$ expressions while also not influencing the ratio of the added moles.

I do not get why this can be done but this allows for the vapor pressure at equilibrium to become $\pu{20 mm Hg}$ as required

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  • $\begingroup$ For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details $\endgroup$ – Safdar Faisal Sep 10 '20 at 11:20
  • $\begingroup$ Nice! I will try to do so when free $\endgroup$ – TheOverlord Sep 10 '20 at 12:45
  • $\begingroup$ I assume that the reaction would end at that point.. Don't listen to your friend.. Equilibrium matters only when you have reactants left at point of equilibrium.. $\endgroup$ – Safdar Faisal Sep 10 '20 at 12:58
  • $\begingroup$ But that does not satisfy the K_{p} of the reaction $\endgroup$ – TheOverlord Sep 10 '20 at 13:25
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This takes the assumption that the water vapour present acts as an ideal gas. Therefore $PV=nRT$ is valid.


Flaws/Misconceptions:

(1) You directly used pressure as you have in the RICE table.

Using pressure in a RICE table is not very useful since the RICE table deals with the amount of reactants and products and not pressure.

(2) The reaction doesn't reach equilibrium

The reaction reaches equilibrium as shown below. This means that the question in title doesn't need to be considered.
However, in such a case where equilibrium isn't reached, then the reaction stops due to lack of reagents to go forward. When you add more reagent it keeps moving forward until equilibrium is reached after which addition of solid reactants wouldn't cause a reaction to take place.


Now, as OP has stated, the $P_\ce{H2O}= \pu{20 mm Hg}$ at equilibrium. and it is given that relative humidity is $80\%$ for a saturation vapor pressure of $\pu{85 mm Hg}$ and so that means the intial pressure exerted by $\ce{H2O(g)}$ is:

$$P_\ce{H2O}= \frac{8}{10}\cdot 85=\pu{68 mm Hg}$$

Now using the ideal gas law, for the initial condition, we get:

\begin{align} n_i(\ce{H2O})&= \frac{68}{76}\cdot\frac{1}{0.0821 \times 300} \\ &\approx\pu{0.00036 mol} \end{align}

Similarly for the final case, we get:

\begin{align} n_f(\ce{H2O})&= \frac{20}{76}\cdot\frac{1}{0.0821 \times 300} \\ &\approx\pu{0.00011 mol} \end{align}

Therefore, $\pu{0.00025 mol}$ of $\ce{H2O}$ reacts for the condition to reach equilibrium. Now drawing the RICE table for the reaction we get:

\begin{array}{lc} \hline \ce{&A.3H2O &<=> &A.H2O &+ &2H2O(g)} \\ \hline i &1\times10^{-3} && 1 \times10^{-3} && 3.6 \times 10^{-4} \\ c &1.25\times10^{-4} && -1.25\times10^{-4} && -2.5\times10^{-4} \\ e &1.125\times10^{-3} && 8.75\times 10^{-4} && 1.1\times10^{-4} \\ \hline \end{array}

So the final composition is: \begin{array}{lc} \hline \text{Component} &\text{Amount} \\ \hline \ce{A.3H2O} & \pu{1.125 \times10^{-3} mol} \\ \ce{A.H2O} & \pu{0.875 \times10^{-3} mol} \\ \ce{H2O} & \pu{1.1 \times10^{-4} mol} \\ \hline \end{array}

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