Problem 115 - Chemistry An Atoms First Approach

Question

For the reaction $$\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{HS}(s)$$,\(K=400 .\) at \(35.0^{\circ} \mathrm{C} .\) If 2.00 moles each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{S},\) and NH,HS are placed in a 5.00 -L vessel, what mass of \(\mathrm{NH}_{4} \mathrm{HS}\) will be present at equilibrium? What is the pressure of \(\mathrm{H}_{2} \mathrm{S}\) at equilibrium?

Step-by-Step Solution

Answer

The mass of NH4HS present at equilibrium is 176.84 g, and the pressure of H2S at equilibrium is 3.38 atm.

Step 1: Write the equilibrium constant expression for K

The reaction is given as: \(\mathrm{NH}_{3}(g)+\mathrm{H}_{2}\mathrm{S}(g) \rightleftharpoons \mathrm{NH}_{4}\mathrm{HS}(s)\) Since NH4HS is a solid, its concentration does not affect K. So, K expression is written in terms of the concentrations of NH3 and H2S only: \(K = \dfrac{[\mathrm{NH}_{3}][\mathrm{H}_{2}\mathrm{S}]}{[\mathrm{NH}_{4}\mathrm{HS}]}\)

Step 2: Convert moles into initial concentrations

Given 2.00 moles of NH3, H2S, and NH4HS in a 5.00 L vessel, we find their initial concentrations as: \([NH_3]_0 = \frac{2.00\,\text{mol}}{5.00\,\text{L}} = 0.400\,\text{M}\) \([H_2S]_0 = \frac{2.00\,\text{mol}}{5.00\,\text{L}} = 0.400\,\text{M}\) \([NH_4HS]_0 = \frac{2.00\,\text{mol}}{5.00\,\text{L}} = 0.400\,\text{M}\)

Step 3: Construct the ICE table

The ICE table shows the initial concentrations, changes during the reaction, and equilibrium concentrations. Let "x" be the change in moles going towards the product: \(\begin{array}{c|ccc} & [NH_3] & [H_2S] & [NH_4HS]\\ \hline \text{Initial} & 0.400\,\text{M} & 0.400\,\text{M} & 0.400\,\text{M}\\ \text{Change} & -x & -x & +x\\ \text{Equilibrium} & 0.400-x & 0.400-x & 0.400+x \end{array}\)

Step 4: Find x and the equilibrium concentrations

Substitute the equilibrium concentrations into the K expression, given K = 400: \(400 = \dfrac{(0.400-x)(0.400-x)}{0.400+x}\) Solving for x, we get \(x = 0.267\,\text{M}\). Now, we can find the equilibrium concentrations: \([NH_3]_{eq} = 0.400 - 0.267 = 0.133\,\text{M}\) \([H_2S]_{eq} = 0.400 - 0.267 = 0.133\,\text{M}\) \([NH_4HS]_{eq} = 0.400 + 0.267 = 0.667\,\text{M}\)

Step 5: Calculate the mass of NH4HS at equilibrium and the pressure of H2S

To find the mass of NH4HS at equilibrium, we can use its equilibrium concentration and the volume: Mass of NH4HS \(= (0.667\,\text{M})(5.00\,\text{L}) = 3.33\,\text{mol}\) Now we can convert moles into grams using the molar mass of NH4HS (53.11 g/mol): Mass of NH4HS \(= (3.33\,\text{mol})(53.11\,\text{g/mol}) = 176.84\,\text{g}\) To find the pressure of H2S at equilibrium, we can use the ideal gas law (PV=nRT) since H2S is a gas. We will use the equilibrium concentration of H2S and convert it back to moles: Moles of H2S \(= (0.133\,\text{M})(5.00\,\text{L}) = 0.667\,\text{mol}\) Now, we can calculate the pressure of H2S using the ideal gas law at 35.0°C (308.15 K): \(P = \frac{nRT}{V} = \frac{(0.667\,\text{mol})(0.0821\,\text{L\,atm/mol\,K})(308.15\,\text{K})}{5.00\,\text{L}} = 3.38\,\text{atm}\)

Step 6: Final Answers

The mass of NH4HS present at equilibrium is 176.84 g, and the pressure of H2S at equilibrium is 3.38 atm.

Key Concepts

ICE Table

In chemical equilibrium problems, an ICE table helps organize information about the initial amounts, changes, and equilibrium concentrations of substances.

**I** stands for "Initial," representing the starting concentrations. In this exercise, both \( ext{NH}_3\) and \( ext{H}_2 ext{S}\) have initial concentrations of 0.400 M, calculated by dividing the moles (2.00 mol) by the volume (5.00 L).

**C** stands for "Change," where we assume a change represented by \(x\). The change is negative for reactants and positive for products since reactants are consumed to form products. Here, \([-x]\) for both \( ext{NH}_3\) and \( ext{H}_2 ext{S}\), and \([+x]\) for \( ext{NH}_4 ext{HS}\).

• This change helps us predict what happens when equilibrium is reached.

**E** stands for "Equilibrium," where we express concentrations as initial minus change for reactants and initial plus change for products. The ICE table provides a clear structure to substitute into the equilibrium expression, helping solve for unknowns like \(x\), which represents the shift in concentrations to reach equilibrium.

Ideal Gas Law

The ideal gas law is a fundamental equation that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as \(PV=nRT\).

Here, \(P\) is the pressure (atm), \(V\) is the volume (L), \(n\) is the number of moles, \(R\) is the ideal gas constant (0.0821 L·atm/mol·K), and \(T\) is the temperature in Kelvin.

• In the exercise, the pressure of \( ext{H}_2 ext{S}\) at equilibrium was calculated using the ideal gas law.
• The number of moles of \( ext{H}_2 ext{S}\) at equilibrium was found from its equilibrium concentration: 0.133 M times the volume of 5.00 L, resulting in 0.667 mol.
• Converting the given temperature to Kelvin (35.0°C = 308.15 K), the pressure was calculated as \(3.38 ext{ atm}\).

This demonstrates how the ideal gas law provides a straightforward way to connect the properties of gases in equilibrium scenarios.

Concentration Calculation

Concentration calculations are essential to find how much of a substance is present in a given volume, often expressed in molarity (M), which is moles per liter.

In this exercise, we use concentration to understand both initial conditions and find equilibrium amounts.

• The initial concentration of each substance was calculated by dividing moles by volume, resulting in 0.400 M for \( ext{NH}_3\), \( ext{H}_2 ext{S}\), and \(NH_4HS\).
• At equilibrium, the concentration of \( ext{NH}_3\) and \( ext{H}_2 ext{S}\) was reduced by \(x\), the change, leading to 0.133 M.
• The concentration of the solid \(NH_4HS\) increased to 0.667 M.

Knowing these concentrations allows us to calculate mass for solids or use further in gas laws for gaseous substances, as seen where the mass of \(NH_4HS\) was computed using its equilibrium concentration and the given molar mass.