Problem 67 - Chemistry: The Central Science

Question

Solid \(\mathrm{NH}_{4} \mathrm{HS}\) is introduced into an evacuated flask at \(24^{\circ} \mathrm{C}\). The following reaction takes place: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ At equilibrium the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) taken together) is \(0.614 \mathrm{~atm}\). What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The \(K_p\) value for this equilibrium at \(24^{\circ} \mathrm{C}\) is approximately \(0.094\ \mathrm{atm^2}\).

Step 1: 1. Assume initial amounts of gases

Initially, we will assume that there are no moles of \(\mathrm{NH}_{3}(g)\) and \(\mathrm{H}_{2}\mathrm{S}(g)\) present.

Step 2: 2. Write the reaction expression

$\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$

Step 3: 3. Determine the change in moles at equilibrium

At equilibrium, let x moles of \(\mathrm{NH}_{4}\mathrm{HS}(s)\) dissociate into x moles of \(\mathrm{NH}_{3}(g)\) and x moles of \(\mathrm{H}_{2}\mathrm{S}(g)\).

Step 4: 4. Write the equilibrium expression for partial pressures

\(K_p = \frac{P_{\mathrm{NH}_{3}} \cdot P_{\mathrm{H}_{2} \mathrm{S}}}{(1-x^2)}\) (Note that since \(\mathrm{NH}_{4} \mathrm{HS}\) is a solid, it is not included in the equilibrium constant expression.)

Step 5: 5. Find the mole fractions

Mole fraction of \(\mathrm{NH}_{3}=\frac{x}{x+x}=\frac{1}{2}\) and Mole fraction of \(\mathrm{H}_{2} \mathrm{S}=\frac{x}{x+x}=\frac{1}{2}\)

Step 6: 6. Calculate the partial pressures

Using given total pressure \(0.614 \ \mathrm{atm}\), we can find the partial pressures. \(P_{\mathrm{NH}_{3}} = \text{mole fraction of}\ \mathrm{NH}_{3}\times \text{total pressure} = \frac{1}{2}(0.614\ \mathrm{atm}) =0.307\ \mathrm{atm}\) \(P_{\mathrm{H}_{2}\mathrm{S}} = \text{mole fraction of}\ \mathrm{H}_{2}\mathrm{S}\times \text{total pressure}=\frac{1}{2}(0.614\ \mathrm{atm}) =0.307\ \mathrm{atm}\)

Step 7: 7. Calculate \(K_{p}\) at equilibrium

Now we can substitute the partial pressures in the equilibrium constant expression. \(K_p = \frac{P_{\mathrm{NH}_{3}} \cdot P_{\mathrm{H}_{2} \mathrm{S}}}{(1-x^2)}= \frac{0.307\ \mathrm{atm} \cdot 0.307\ \mathrm{atm}}{(1-x^2)}\) Since x is very small, we can assume \(1-x^2 \approx 1\), so we have \(K_p = 0.307\ \mathrm{atm} \cdot 0.307\ \mathrm{atm} \approx 0.094\ \mathrm{atm^2}\) The \(K_p\) value for this equilibrium at \(24^{\circ} \mathrm{C}\) is approximately \(0.094\ \mathrm{atm^2}\).

Key Concepts

Chemical Reaction

A chemical reaction is a process where reactants are transformed into products through the breaking and forming of chemical bonds. In our given problem, solid ammonium hydrosulfide \(\mathrm{NH}_{4} \mathrm{HS}\) decomposes into ammonia \(\mathrm{NH}_{3}(g)\) and hydrogen sulfide \(\mathrm{H}_{2} \mathrm{S}(g)\) when introduced into a flask. This reaction is reversible, indicated by the double-headed arrow, and can reach a state of dynamic equilibrium.

At equilibrium, the rate of the forward reaction, where \(\mathrm{NH}_{4} \mathrm{HS}\) dissociates into gases, equals the rate of the reverse reaction, where the gases recombine to form the solid. This condition gives rise to a constant composition of reactants and products in the system, which can be quantitively expressed by the equilibrium constant \(K_p\).

Equilibrium Constant (Kp)

The equilibrium constant \(K_p\) is a dimensionless number representing the ratio of the products' partial pressures to the reactants' partial pressures, each raised to the power of their stoichiometric coefficients in a chemical reaction involving gases. We use \(K_p\) when dealing with partial pressures. The expression for \(K_p\) is derived from the balanced chemical equation and does not include solid or liquid species, as their concentrations do not change.

In our exercise, the solid \(\mathrm{NH}_{4} \mathrm{HS}\) is not included in the \(K_p\) expression, which is given by \(K_p = \frac{P_{\mathrm{NH}_{3}} \cdot P_{\mathrm{H}_{2} \mathrm{S}}}{(1-x^2)}\). The importance of \(K_p\) lies in its ability to predict the position of equilibrium and determine reactant or product dominance under given conditions, which in turn assists in understanding the feasibility of the chemical process.

Partial Pressure

Partial pressure is the pressure exerted by an individual gas in a mixture of gases. It is proportional to its mole fraction in the mixture and the total pressure. Dalton’s Law of Partial Pressures states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

For instance, in the exercise provided, the total pressure at equilibrium is given as \(0.614 \mathrm{~atm}\). With the assumption that \(\mathrm{NH}_{3}(g)\) and \(\mathrm{H}_{2} \mathrm{S}(g)\) are present in equal amounts, their mole fractions are both \(\frac{1}{2}\), yielding equal partial pressures of \(0.307 \mathrm{~atm}\). These partial pressures are crucial to determining the \(K_p\) of the system. Understanding partial pressures helps students solve problems related to gas-phase equilibria and predict the behavior of gases in different conditions.