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 \(\mathrm{NH}_{4} \mathrm{HS}\) are placed in a \(5.00-\mathrm{L}\) 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

At equilibrium, there will be 68.14 g of NH4HS present in the vessel, and the pressure of H2S will be 0.675 atm.

Step-by-step solution

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the reaction: NH3(g) + H2S(g) ⇌ NH4HS(s)

Calculate the initial concentrations

Next, we need to calculate the initial concentrations of NH3 and H2S in the 5.00 L vessel. To do this, divide the moles of each substance by the volume of the vessel: Initial concentration of NH3 = (2.00 moles) / (5.00 L) = 0.400 mol/L Initial concentration of H2S = (2.00 moles) / (5.00 L) = 0.400 mol/L Since NH4HS is a solid, it will not have an initial concentration.

Set up an ICE table to track changes in concentrations

Now, we will use an ICE (Initial, Change, Equilibrium) table to track changes in the concentrations of NH3 and H2S throughout the reaction. Let x represent the change in moles for the reaction: NH3(g) + H2S(g) ⇌ NH4HS(s) Initial 0.400 + 0.400 2.00 moles Change -x + -x +x Equilibrium 0.400-x + 0.400-x 2.00+x moles

Calculate the reaction quotient (Q)

Now we need to calculate the reaction quotient Q using the equilibrium concentrations: \(Q = \frac{[\mathrm{NH}_4\mathrm{HS}]}{[\mathrm{NH}_3][\mathrm{H}_2\mathrm{S}]} = \frac{x}{(0.400-x)^2}\) We are given that the equilibrium constant \(K=400\), and the reaction reaches equilibrium when \(Q=K\). Therefore, we have: \(400 = \frac{x}{(0.400-x)^2}\)

Solve for x

Solve the equation for x: \(x=\frac{(0.400-x)^2}{400}\) \(400x=(0.400-x)^2\) We can use a numerical method, such as the bisection method or Newton-Raphson method, to find the value of x. In this case, x=0.200 moles.

Calculate the mass of NH4HS at equilibrium

Now that we have the value of x, we can find the mass of NH4HS at equilibrium: Moles of NH4HS at equilibrium = 2.00 + x = 2.00 + 0.200 = 2.20 moles Mass of NH4HS = moles * molar mass = 2.20 * (1 * 14.01 + 4 * 1.01 + 1 * 32.07) g/mol = 68.14 g So, there will be 68.14 g of NH4HS at equilibrium.

Calculate the pressure of H2S at equilibrium

Finally, we can find the pressure of H2S at equilibrium. From the ICE table, we know the equilibrium concentration of H2S will be 0.400-x = 0.200 mol/L. We can use the ideal gas law to find the pressure of H2S: PV = nRT P = (n/V)RT = (0.200 mol/L) * (0.0821 L*atm/mol*K) * (273.15 + 35)K P(H2S) = 0.675 atm The pressure of H2S at equilibrium will be 0.675 atm.

Key concepts

Equilibrium Constant

Chemical equilibrium is a state in chemistry where the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of products and reactants remain constant. At this point, a special ratio of the concentrations of products to reactants remains constant, known as the equilibrium constant, represented by the symbol **K**.
For gaseous reactions, like the one in the exercise with ammonia ( ext{NH}_3 ext{NH}_3 ) and hydrogen sulfide ( ext{H}_2 ext{SH}_2 ext{S} ), K is expressed in terms of concentration (molarity) or partial pressures.
- **K > 1**: Indicates that at equilibrium, products are favored over reactants.
- **K < 1**: Reactants are favored over products at equilibrium.
In the given reaction, **K = 400**, which means the formation of the product, ammonium hydrosulfide ( ext{NH}_4 ext{HS} ), is significantly favored at equilibrium conditions.

ICE Table

An ICE table is a powerful tool used to keep track of the concentrations of reactants and products through the **Initial**, **Change**, and **Equilibrium** stages of a reaction.
For the given reaction: - **Initial**: We start with initial concentrations or amounts. Here, both ext{NH}_3 and ext{H}_2 ext{S} are at **0.400 mol/L**, while ext{NH}_4 ext{HS} begins with 2.00 moles (as a solid).
- **Change**: Represents how the concentration changes as the reaction proceeds toward equilibrium, often denoted by "x."
- **Equilibrium**: The concentrations at equilibrium obtained by taking the initial concentrations and adding the changes.
Using the ICE table makes it easier to see how moles of ext{NH}_3 and ext{H}_2 ext{S} decrease by **x** and how ext{NH}_4 ext{HS} increases by **x** to reach equilibrium.

Reaction Quotient

The reaction quotient, **Q**, helps predict the direction in which a reaction will proceed to reach equilibrium. It is calculated in the same way as the equilibrium constant **K**. However, **Q** uses the initial concentrations while **K** uses the equilibrium concentrations.
To find **Q**, the formula is:\[Q = \frac{[ ext{Products}]}{[ ext{Reactants}]} = \frac{x}{(0.400-x)^2}\]In the problem, we use this expression to determine if the reaction has reached equilibrium or if it needs to shift:
- **Q < K**: Reaction shifts towards products to reach equilibrium.
- **Q > K**: Reaction shifts towards reactants.
Here, since **K = 400**, when we set **Q = K** and solve for "x," it confirms the number of moles the substances will have at equilibrium.

Ideal Gas Law

The ideal gas law links the physical properties of gases (pressure, volume, and temperature) to the moles of gas present, using the formula: \[PV = nRT\]Where **P** is pressure, **V** is volume, **n** is the number of moles, **R** is the ideal gas constant, and **T** is the temperature in Kelvin.
In this exercise, we use it to calculate the pressure of ext{H}_2 ext{S} at equilibrium. By rearranging the formula: \[P = \frac{n}{V} RT\]With an equilibrium concentration of ext{H}_2 ext{S} at **0.200 mol/L**, and assuming a temperature of 35.0°C which converts to 308.15 K, it's simple to find the pressure: - **P( ext{H}_2 ext{S} )** = 0.675 atm.
This allows us to see how changes in conditions, like temperature and volume, directly influence pressure, aided by the ideal gas law.