Problem 56 - Chemistry The Central Science In Si Units

Question

At \(80^{\circ} \mathrm{C}, K_{c}=1.87 \times 10^{-3}\) for the reaction $$ \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g) $$ (a) Calculate the equilibrium concentrations of \(\mathrm{PH}_{3}\) and \(\mathrm{BCl}_{3}\) if a solid sample of \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is placed in a closed vessel at \(80^{\circ} \mathrm{C}\) and decomposes until equilibrium is reached. (b) If the flask has a volume of \(0.250 \mathrm{~L}\), what is the minimum mass of \(\mathrm{PH}_{3} \mathrm{BCl}_{3}(s)\) that must be added to the flask to achieve equilibrium?

Step-by-Step Solution

Answer

(a) PH₃ and BCl₃ have equilibrium concentrations of 0.0433 M each. (b) Minimum mass of PH₃BCl₃ required is 1.40 grams.

Step 1: Understanding the Problem

We need to calculate the equilibrium concentrations of both \(\mathrm{PH}_3\) and \(\mathrm{BCl}_3\) given the equilibrium constant \(K_c = 1.87 \times 10^{-3}\) at \(80^{\circ} \mathrm{C}\). Additionally, for a flask with volume \(0.250 \mathrm{~L}\), we need to determine the minimum mass of \(\mathrm{PH}_3 \mathrm{BCl}_3(s)\) required to establish equilibrium.

Step 2: Setup the Equilibrium Expression

The equilibrium expression for the decomposition of \(\mathrm{PH}_3 \mathrm{BCl}_3\) is:\[K_c = \frac{[\mathrm{PH}_3][\mathrm{BCl}_3]}{1} = [\mathrm{PH}_3][\mathrm{BCl}_3]\]Since \(K_c = 1.87 \times 10^{-3}\), we can set this equal to the product of the concentrations of the gases at equilibrium.

Step 3: Define Variables for Equilibrium Concentrations

Assume that \(x\) moles of \(\mathrm{PH}_3 \mathrm{BCl}_3\) decompose, producing \(x\) moles each of \(\mathrm{PH}_3\) and \(\mathrm{BCl}_3\). Thus, at equilibrium, \([\mathrm{PH}_3] = x\) and \([\mathrm{BCl}_3] = x\). Substituting these into the expression, we have:\[K_c = x^2\]

Step 4: Solve for x

From Step 3, we have:\[x^2 = 1.87 \times 10^{-3}\]Take the square root of both sides to solve for \(x\):\[x = \sqrt{1.87 \times 10^{-3}} = 0.0433\]Thus, \([\mathrm{PH}_3] = 0.0433 \text{ M}\) and \([\mathrm{BCl}_3] = 0.0433 \text{ M}\).

Step 5: Calculate Moles at Equilibrium

Use the relation \(\text{moles} = \text{concentration} \times \text{volume}\) to find the moles of each gas in the \(0.250 \mathrm{~L}\) flask:\[\text{Moles of } \mathrm{PH}_3 = \text{Moles of } \mathrm{BCl}_3 = 0.0433 \times 0.250 = 0.010825\]

Step 6: Minimum Mass of PH₃BCl₃ Required

Since \(0.010825\) moles of \(\mathrm{PH}_3\) are produced, the same amount of \(\mathrm{PH}_3 \mathrm{BCl}_3\) must have decomposed. The molar mass of \(\mathrm{PH}_3 \mathrm{BCl}_3\) is approximately \(129.33 \mathrm{~g/mol}\). Therefore, the minimum mass needed is:\[\text{mass} = 0.010825 \times 129.33 = 1.40 \text{ grams}\]

Key Concepts

Equilibrium Constant

The concept of the equilibrium constant, denoted as \( K_c \), is a crucial aspect of chemical equilibrium. It provides a quantitative measure of the position of the equilibrium for a chemical reaction.
At a specific temperature, \( K_c \) remains constant for a particular reaction, no matter the initial concentrations of the reactants and products. This constancy makes it an invaluable tool for chemists.

In our example, the equilibrium constant is given by:

• \( K_c = 1.87 \times 10^{-3} \) for the reaction at \( 80^{\circ} \mathrm{C} \).

To understand what this value means, it's important to note that it emanates from the ratio of the concentrations of products to reactants at equilibrium.
Since solid compounds do not appear in the equilibrium expression, only the gaseous products \( [\mathrm{PH}_3] \) and \([\mathrm{BCl}_3] \) are considered:

\[ K_c = \frac{[\mathrm{PH}_3][\mathrm{BCl}_3]}{1} \]
The low value of \( K_c \) suggests that at equilibrium, the concentrations of the products are relatively low compared to the reactants, indicating the reaction does not completely favor the production of gas at this temperature.

Concentration Calculations

Concentration calculations are pivotal in determining how much of a substance is present in a given volume, and this is essential for solving equilibrium problems.
In equilibrium scenarios, we often start by assuming a change in concentration using an unknown variable, typically denoted as \( x \).

For our reaction, when \( x \) moles of \( \mathrm{PH}_3 \mathrm{BCl}_3 \) decompose, they produce \( x \) moles each of \( \mathrm{PH}_3 \) and \( \mathrm{BCl}_3 \). Thus, at equilibrium, we have:

• \([\mathrm{PH}_3] = x\)
• \([\mathrm{BCl}_3] = x\)

Using the expression \( K_c = x^2 \), we solve for \( x \) as follows:
\[ x = \sqrt{1.87 \times 10^{-3}} = 0.0433 \] M
This calculated \( x \) value provides the equilibrium concentrations of both \( \mathrm{PH}_3 \) and \( \mathrm{BCl}_3 \) in the reaction. To find the number of moles present in a specific volume, you multiply this concentration by the volume, leading to the equation:

\[ \text{moles} = \text{concentration} \times \text{volume} \]
Through this calculation, you can determine the moles required to establish equilibrium in a given volume, such as a \(0.250 \mathrm{~L}\) flask.

Reaction Chemistry

Understanding the basics of reaction chemistry is essential for comprehending how substances interact and change into different substances. The core aspects of reaction chemistry involve reactants, products, and the balance between them at equilibrium. This balance is what defines the extent of a reaction.

For the decomposition reaction of \( \mathrm{PH}_3 \mathrm{BCl}_3 \), knowing that it forms \( \mathrm{PH}_3 \) and \( \mathrm{BCl}_3 \) allows us to consider how the substances convert under specific conditions. Here, the reaction has been given a particular equilibrium constant, indicating it reaches a specific point where the forward and reverse reactions occur at the same rate.
In this balanced state, despite the ongoing nature of microscopic changes, the macroscopic concentrations remain unchanged.

Moreover, mass conservation is a key principle, as shown by the need to calculate the amount of \( \mathrm{PH}_3 \mathrm{BCl}_3 \) needed to reach equilibrium. This is done by linking moles to their respective molar masses, such as using the molar mass of \( \mathrm{PH}_3 \mathrm{BCl}_3 \), approximately \( 129.33 \mathrm{~g/mol} \).
This approach ensures you understand how reactants must translate into the exact stoichiometric quantities required for products at equilibrium, emphasizing the intricate balance achieved in reaction chemistry.