Question

For the reaction

$\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{S}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{N}{\mathit{H}}_{\mathbf{4}}\mathit{H}\mathit{S}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$.

. at 35.0⁰C. If 2.00 moles each of NH₃, H₂S, andNH₄HS are placed in a 5.00-L vessel, what mass ofNH₄HS will be present at equilibrium? What is the pressure of H₂S at equilibrium?

Short answer

192 g of NH₄HS will be present at equilibrium. The pressure of H₂S at equilibrium will be 1.3 atm.

Step-by-step solution

Step 1:Determinethe amount of NH3 reacted

$\begin{array}{l}N{H}_3\left(g\right)+H_{2}S\left(s\right)\rightleftharpoons N{H}_{4}HS\left(s\right)K=400\\ Initial:\frac{2.00mol}{5.00L}\frac{2.00mol}{5.00L}-\\ Change:-x-x-\\ Equil:0.400-x0.400-x-\end{array}$

Let the amount of NH₃that reacts to reach equilibrium be x mol/L.

Therefore, we can write

$\begin{array}{c}K=\frac{1}{\left[N{H}_3\right]\left[H_{2}S\right]}\\ K=400\\ 400=\frac{1}{\left[N{H}_3\right]\left[H_{2}S\right]}\\ 400=\frac{1}{(0.400-x)(0.400-x)}\end{array}$

$\begin{array}{c}{(0.400-x)}^2=\frac{1}{400}\\ (0.400-x)={\left(\frac{1}{400}\right)}^{\frac{1}{2}}\\ 0.400-x=0.0500\\ x=0.35M\end{array}$

Determine the mass of NH4HS present at equilibrium

Number of moles of NH₄HS(s) produced

$\begin{array}{l}=5.00L\times \frac{0.35molN{H}_3}{L}\times \frac{1molN{H}_{4}HS}{molN{H}_3}\\ =1.75mol\end{array}$

Total number of moles of NH₄HS(s) produced=2.00 mol initially+ 1.75 mol produced=3.75 mol

Total mass ofNH₄HS(s) at equilibrium

$\begin{array}{l}3.75molN{H}_{4}HS\times \frac{51.12gN{H}_{4}HS}{molN{H}_{4}HS}\\ =192gN{H}_{4}HS\end{array}$

Determine the pressure of H2S at equilibrium

Amount of $H_{2}S$at equilibrium

$\begin{array}{c}{\left[H_{2}S\right]}_e=0.400M-x\\ =0.400M-0.350M\\ =0.05M{H}_{2}S\end{array}$

$\begin{array}{c}{P}_{H_{2}S}=\frac{n_{H_{2}S}RT}{V}\\ =\frac{n_{H_{2}S}}{V}\times R\times T\\ =\frac{0.050mol}{1L}\times \frac{0.08206Latm}{Kmol}\times 308.2K\\ =1.3atm\end{array}$