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 \(2.00 \mathrm{~mol}\) each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{~S}\), and \(\mathrm{NH}_{4} \mathrm{HS}\) 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?

Short answer

In conclusion, at equilibrium, there will be approximately 198.46 grams of NH4HS and the pressure of H2S will be approximately 1.47 atm.

Step-by-step solution

Calculate initial concentrations

Given the initial amounts of reactants and products, we can calculate the initial concentrations by dividing the moles by the volume of the vessel (5.00 L). Initial concentrations: [NH3] = 2.00 mol / 5.00 L = 0.400 M [H2S] = 2.00 mol / 5.00 L = 0.400 M [NH4HS] = 2.00 mol / 5.00 L = 0.400 M

Set up an ICE table

The ICE table represents the initial molar concentrations of reactants and products (I), the change in concentrations due to the reaction (C), and the equilibrium concentrations (E). NH3(g) + H2S(g) ⇌ NH4HS(s) Initial 0.400 M 0.400 M 0.400 M Change -x -x +x Equilibrium (0.400-x) (0.400-x) (0.400+x) Note that in the Change row, we consider the forward reaction decreases the NH3 and H2S molar concentrations (-x) and increases the NH4HS molar concentration (+x).

Write the equilibrium constant expression and solve for x

The reaction's equilibrium constant expression, K, is the ratio of the product molar concentration raised to its stoichiometric coefficients to the reactants' molar concentrations also raised to their stoichiometric coefficients. K = \(\frac{[NH4HS]}{[NH3][H2S]}\) We know K = 400. Plug in the equilibrium concentrations from the ICE table into the equilibrium constant expression and solve for x. 400 = \(\frac{0.400+x}{(0.400-x)(0.400-x)}\)

Find the equilibrium concentrations

After solving the equation for x, we can plug this value back into the equilibrium concentrations (E) and find the equilibrium concentrations of NH3, H2S, and NH4HS. x ≈ 0.342 M [NH3] = 0.400 - 0.342 = 0.058 M [H2S] = 0.400 - 0.342 = 0.058 M [NH4HS] = 0.400 + 0.342 = 0.742 M

Calculate the mass of NH4HS at equilibrium

Now that we have the equilibrium concentration of NH4HS, we can convert it to moles and then to mass using the molar mass of NH4HS. Moles of NH4HS = (0.742 M)(5.00 L) = 3.71 mol Mass of NH4HS = (3.71 mol)(53.49 g/mol) = 198.46 g (approximately)

Calculate the pressure of H2S at equilibrium

To find the pressure of H2S at equilibrium, use the equilibrium concentration and the ideal gas law in the form: P = n/V × R × T. Pressure of H2S = [H2S] × R × T R (Ideal gas constant) = 0.0821 L⋅atm/mol⋅K T (Temperature) = 35°C + 273.15 = 308.15 K Pressure of H2S = (0.058 M)(0.0821 L⋅atm/mol⋅K)(308.15 K) = 1.47 atm (approximately) In conclusion, at equilibrium, there will be approximately 198.46 grams of NH4HS and the pressure of H2S will be approximately 1.47 atm.

Key concepts

Equilibrium Constant

When a chemical reaction reaches a state where the rates of the forward and reverse reactions are equal, it is said to be at equilibrium. The equilibrium constant, denoted as 'K', is a value that expresses the ratio of the concentrations of products to reactants at this point, each raised to the power of their respective stoichiometric coefficients. Imagine it as a snapshot capturing the balance of a seesaw, with reactants on one side and products on the other.

The magnitude of 'K' provides insight into the position of equilibrium. A high 'K' value suggests a reaction mixture rich in products, whereas a low 'K' indicates a mixture with more reactants. This constant is only affected by temperature; changes in concentrations, pressures, or the presence of a catalyst do not alter it. In the given exercise, 'K' is 400, which strongly favors the formation of products, in this case, NH4HS. In an equilibrium expression, pure solids and liquids are omitted because their concentrations do not change.

ICE Tables

To tackle questions involving reaction equilibria, chemists often use an ICE table, which stands for Initial, Change, and Equilibrium. This is a tool that systematically tracks changes in molar concentrations or partial pressures of reactants and products as a reaction progresses from its initial state to equilibrium.

The 'Initial' row lists the starting concentrations or pressures. The 'Change' row specifies the adjustments in these values as the system moves towards equilibrium, often expressed using a variable like 'x'. Lastly, the 'Equilibrium' row combines the initial and change rows to reveal the final state of the system. Utilizing the equilibrium constant, 'K', and the ICE table, one can calculate the unknown changes in concentrations to find the equilibrium concentrations or pressures. This process was elegantly demonstrated in the step-by-step solution, where an ICE table elucidated the behavior of the reaction components under equilibrium conditions.

Ideal Gas Law

The ideal gas law is a cornerstone of chemical and physical sciences. It relates the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas through the equation PV=nRT, where R is the universal gas constant. This simplistic yet powerful equation assumes gases consist of particles with no volume and no intermolecular forces.

Applying this law helps to determine the physical behavior of gases under various conditions. In our exercise, the ideal gas law was used to calculate the pressure of H2S gas at equilibrium. It is important to understand that the 'T' in the ideal gas law is always in Kelvin, which involves converting from Celsius by adding 273.15. Also, ensuring units are consistent - here involving liters (L), atmospheres (atm), and mol - is critical for accurate calculations. This concept tied beautifully into Step 6 of the solution, effectively securing the equilibrium pressure of H2S.