Suppose 15.00 g of solid ammonium hydrogen sulfide is introduced into a 500.-mL flask at 25°C, the flask is sealed, and the system is allowed to reach equilibrium. What is the partial pressure of ammonia in this flask if K_P = 0.108 at 25°C for the following reaction? NH₄HS(s) NH₃(g) + H₂S(g)

Question

Quizplus · Accepted Answer

We can start by writing the equilibrium expression using the concentrations (or partial pressures) of the species involved:

$$K_\mathrm{p}=\frac{P_{\ce{NH3}}\cdot P_{\ce{H2S}}}{P_{\ce{NH4HS}}}$$

We are given the value of $K_\mathrm{p}$, so we just need to find the partial pressures of the gases at equilibrium. Let's call $x$ the amount (in moles) of $\ce{NH4HS}$ that dissociates to form $\ce{NH3}$ and $\ce{H2S}$.

At equilibrium, we will have: 
- $15.00-x$ g of $\ce{NH4HS}$ left in the flask, and therefore a partial pressure of $P_{\ce{NH4HS}}=\frac{n}{V}\cdot RT=\frac{m/M}{V}\cdot RT$, where $M$ is the molar mass of $\ce{NH4HS}$ and $n$ is the amount in moles. 
- $x$ moles of $\ce{NH3}$, with a partial pressure of $P_{\ce{NH3}}=\frac{n}{V}\cdot RT=\frac{x\cdot RT}{V}$. 
- $x$ moles of $\ce{H2S}$, with a partial pressure of $P_{\ce{H2S}}=\frac{n}{V}\cdot RT=\frac{x\cdot RT}{V}$.

Substituting these expressions into the equilibrium expression and solving for $x$, we get:

$$K_\mathrm{p}=\frac{x^2\cdot RT^2}{V^2\cdot(m/M-x)\cdot RT}=\frac{x^2}{(m/M-x)}$$

$$x^2=K_\mathrm{p}\cdot(m/M-x)$$

$$x^2+K_\mathrm{p}\cdot x-m/M\cdot K_\mathrm{p}=0$$

Solving this quadratic equation for $x$, we obtain:

$$x=\frac{-K_\mathrm{p}+\sqrt{K_\mathrm{p}^2+4\cdot K_\mathrm{p}\cdot(m/M)}}{2}$$

$$x=\frac{-0.108+\sqrt{0.108^2+4\cdot 0.108\cdot (15.00/68.12)}}{2}$$

$$x=0.456	ext{ mol}$$

Therefore, the partial pressure of $\ce{NH3}$ in the flask is:

$$P_{\ce{NH3}}=\frac{x\cdot RT}{V}=\frac{0.456	ext{ mol}\cdot 0.0821	ext{ L}\cdot	ext{atm/mol}\cdot	ext{K}\cdot (25+	ext{273.15 K})}{0.500	ext{ L}}=0.329	ext{ atm}$$

So the answer is (C).