Animated Solution for Chemistry - Chemical Equilibrium: When $5.1\text{ g}$ of solid ${\text{NH}}_4\text{HS}$ is introduced into a two litre evacuated flask at $27^{\circ }\text{C}$, $20\%$ of the solid decomposes into gaseous ammonia and hydrogen sulphide. The $K_p$ for the reaction at $27^{\circ }\text{C}$ is $x\times 10^{-2}$. The value of $x$ is ........... . [Given, $R=0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}$]

The journey to mastering chemical equilibrium often involves navigating the subtle nuances of heterogeneous reactions. In this problem, we are presented with a classic scenario: the thermal decomposition of a solid into gaseous products. This is not just a mathematical exercise; it's a window into how matter behaves under closed conditions and how we can quantify that behavior using equilibrium constants.

Imagine a sturdy, evacuated flask with a precise volume of $2\text{ L}$. The term "evacuated" is crucial here—it means the flask is completely empty, devoid of any air or other gases that could interfere with our reaction or contribute to the total pressure. The temperature is maintained at a constant $27^{\circ }\text{C}$, which we immediately convert to the absolute Kelvin scale as $300\text{ K}$.

Into this pristine environment, we introduce $5.1\text{ g}$ of solid ammonium bisulfide, ${\text{NH}}_4\text{HS}$. Before we can even think about equilibrium, we must translate this mass into the language of chemistry: moles.

The molar mass of ${\text{NH}}_4\text{HS}$ is calculated by summing the atomic masses of its constituent elements: Nitrogen ($14$), four Hydrogens ($4\times 1$), Sulfur ($32$), and another Hydrogen ($1$).

With the molar mass in hand, finding the initial number of moles is straightforward:

So, our starting point is $0.1\text{ mol}$ of solid ${\text{NH}}_4\text{HS}$ resting at the bottom of the flask.

The solid doesn't just sit there; it begins to decompose. The balanced chemical equation for this process is:

Notice the phase states. We have a solid reactant and two gaseous products. This is a heterogeneous equilibrium.

The problem states that $20\%$ of the solid decomposes. This percentage gives us our degree of dissociation, $\alpha =0.2$. It tells us exactly how much of our initial stash reacts.

Let's set up an ICE (Initial, Change, Equilibrium) table to track the moles of each species.

Initially, we have $0.1\text{ mol}$ of the solid and $0\text{ mol}$ of the gases. The change is dictated by the $20\%$ decomposition. Twenty percent of $0.1\text{ mol}$ is $0.02\text{ mol}$. Therefore, $0.02\text{ mol}$ of the solid will disappear. Because the stoichiometry is $1:1:1$, the decomposition of $0.02\text{ mol}$ of solid will produce exactly $0.02\text{ mol}$ of ${\text{NH}}_3$ and $0.02\text{ mol}$ of ${\text{H}}_2\text{S}$.

At equilibrium, the moles are: - ${\text{NH}}_4\text{HS(s)}=0.1-0.02=0.08\text{ mol}$ - ${\text{NH}}_3\text{(g)}=0+0.02=0.02\text{ mol}$ - ${\text{H}}_2\text{S(g)}=0+0.02=0.02\text{ mol}$

The equilibrium constant in terms of concentration, $K_c$, requires molarities, not just moles. Molarity is moles per unit volume. Since our flask has a volume of $2\text{ L}$, we divide the equilibrium moles of our gases by $2$.

Now, we write the expression for $K_c$. Here is a critical rule of heterogeneous equilibrium: the active mass (concentration) of a pure solid or pure liquid is considered to be constant and is conventionally taken as $1$. Therefore, the solid ${\text{NH}}_4\text{HS}$ does not appear in our $K_c$ expression.

The question ultimately asks for $K_p$, the equilibrium constant in terms of partial pressures. While we could calculate the partial pressures of the gases using the ideal gas law and then find $K_p$, there is a much more elegant route. We use the fundamental relationship between $K_p$ and $K_c$:

Here, $R$ is the universal gas constant, $T$ is the absolute temperature, and $\Delta n_g$ is the change in the number of moles of gas between products and reactants.

Looking at our balanced equation, we have $1$ mole of ${\text{NH}}_3$ gas and $1$ mole of ${\text{H}}_2\text{S}$ gas on the product side, totaling $2$ moles. On the reactant side, we only have a solid, so there are $0$ moles of gas.

Now, we substitute all our known values into the equation. The problem provides $R=0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}$ and we know $T=300\text{ K}$.

First, let's compute the term inside the parentheses:

Next, we square this value:

Finally, we multiply by $K_c$:

The problem states that $K_p$ is in the form $x\times 10^{-2}$ and asks for the nearest integer value of $x$. Comparing our result, $6.0516\times 10^{-2}$, with the given format, we can clearly see that $x=6.0516$.

Rounding to the nearest integer, we arrive at our final destination:

This problem beautifully illustrates the interconnectedness of stoichiometry, gas laws, and equilibrium principles. By carefully tracking the physical states and applying the correct relationships, even seemingly complex heterogeneous systems can be unraveled with precision.