# What happens when the amount given doesn't allow for equilbrium to be achieved in a reaction?

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 $$\pu{1 l}$$ container at $$\pu{300 K}$$

Relative humidity is $$80\ \%$$ and saturation water vapor pressure is $$\pu{85 mmHg}$$ 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 mmHg} \end{align}

Now,

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

$$\pu{38 mmHg}$$ drops because $$\pu{2 mmol}$$ of water vapor was consumed

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

The answer should have been $$\pu{20 mmHg}$$ 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 $$\pu{2 mmol}$$ 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 $$\pu{20 mmHg}$$ as required

• 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.. Sep 10 '20 at 12:58
• But that does not satisfy the K_{p} of the reaction Sep 10 '20 at 13:25

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

If the relative humidity is $$80$$% and the saturation water vapor pressure is $$\pu{85 mmHg}$$ at $$\pu{300 K}$$, it means that the vapor pressure in the flask is $$\pu{0.8 · 85 = 68 mmHg}$$. We will now show that this value is much higher than the value coming from equilibrium constant of the reaction. As this constant is : $$\pu{K_p = 400 mmHg^2}$$, the value of the water pressure when both hydrates are present simultaneously is, at equilibrium, $$\pu{√K_p = 20 mmHg}$$. So there is too much water in the gas phase, : the system is not at equilibrium. To be at equilibrium, the water pressure should be $$\pu{20 mm Hg}$$. It must be reduced by $$\pu{\Delta p_w = 68 mm Hg - 20 mm Hg = 48 mm Hg}$$

As a consequence, the gas phase must loose a large part of its water by reacting with $$\ce{A·H2O}$$, so as to produce more $$\ce{A·3H2O}$$ and attain at the end the final vapor pressure $$\pu{20 mmHg}$$. The number of mole $$\ce{H2O}$$ to be adsorbed by $$\ce{A·H2O}$$ is $$\pu{\Delta n_w}$$ given by $$\pu{\Delta n_w = \frac{\Delta p_w V}{RT} = \frac{48~ mmHg}{760~ mm Hg/ atm} · 101325 Pa/atm · \frac{0.001~ m^3}{8.314~J K^{-1} mol^{-1} ~300~ K} = 2.565 10^{-3} mol}$$ The published equation states that $$1$$ millimole $$\ce{A·H2O}$$ can adsorb $$2$$ millimole $$\ce{H2O}$$ And here the vapor contains more than $$2$$ millimoles water. It means that all $$\ce{A·H2O}$$ ($$1$$ millimole) will be transformed into $$\ce{A·3H2O}$$, which is to be added to the original amount of this trihydrate (1 millimole)

The final composition of the mixture will be :

• $$0$$ millimole $$\ce{A·H2O}$$

• $$2$$ millimole $$\ce{A·3H2O}$$

• $$2.565$$ millimole - $$2$$ millimole = $$0.565$$ millimole water in the vapor phase.