Problem 89 - Chemistry The Central Science In Si Units

Question

At \(700 \mathrm{~K}\), the equilibrium constant for the reaction $$ \mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g) $$ is \(K_{p}=77\). A flask is charged with \(202.7 \mathrm{kPa}\) of \(\mathrm{CCl}_{4}\), which then reaches equilibrium at 700 K. (a) What fraction of the \(\mathrm{CCl}_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?(\mathbf{b})\) What are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

Step-by-Step Solution

Answer

(a) Approximately 0.37 fraction of \( \mathrm{CCl}_4 \) is converted. (b) At equilibrium, \( P_{\mathrm{CCl}_4} \approx 127.7 \mathrm{~kPa} \) and \( P_{\mathrm{Cl}_2} \approx 150 \mathrm{~kPa} \).

Step 1: Initial Setup

We start by defining the change in pressure. Let the change in pressure for the conversion of \( \mathrm{CCl}_4 \) be \( x \). Initially, only \( \mathrm{CCl}_4 \) is present at a pressure of \( 202.7 \mathrm{~kPa} \). At equilibrium, the pressure of \( \mathrm{CCl}_4 \) decreases by \( x \), \( \mathrm{Cl}_2 \) is formed at \( 2x \), and \( \mathrm{C} \) solid does not contribute to pressure.

Step 2: Write the Expression for Equilibrium Constant

The equilibrium expression in terms of partial pressure is given by \( K_p = \frac{{P_{\mathrm{Cl}_2}^2}}{{P_{\mathrm{CCl}_4}}} \). Substitute the equilibrium expressions: \( P_{\mathrm{Cl}_2} = 2x \) and \( P_{\mathrm{CCl}_4} = 202.7 - x \). Thus, the equilibrium expression becomes: \( K_p = \frac{{(2x)^2}}{{202.7 - x}} \).

Step 3: Set Up the Equation

Using the value \( K_p = 77 \), we have the equation \( 77 = \frac{{4x^2}}{{202.7 - x}} \). This is a quadratic equation in terms of \( x \). Multiply through by \( 202.7 - x \) to clear the denominator: \( 77(202.7 - x) = 4x^2 \).

Step 4: Solve the Quadratic Equation

Expand and rearrange the equation: \( 77 \times 202.7 - 77x = 4x^2 \). This becomes \( 4x^2 + 77x - 15610.9 = 0 \). Use the quadratic formula: \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \), where \( a = 4 \), \( b = 77 \), \( c = -15610.9 \). Calculating the discriminant \( b^2 - 4ac \) and then \( x \).

Step 5: Calculate Partial Pressures at Equilibrium

After finding \( x \), calculate the partial pressure of \( \mathrm{CCl}_4 \) at equilibrium: \( 202.7 - x \), and \( \mathrm{Cl}_2 \) is produced in twice the amount of \( x \), so its pressure is \( 2x \).

Step 6: Calculate the Fraction of \( \mathrm{CCl}_4 \) Converted

The fraction of \( \mathrm{CCl}_4 \) converted to products is given by \( \frac{x}{202.7} \), where \( x \) is the change in pressure from the initial \( \mathrm{CCl}_4 \) amount.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted by \( K_p \) when dealing with gases, is a value that quantifies the ratio of the products to reactants at equilibrium in a chemical reaction. It is essential in understanding how far a reaction proceeds before reaching equilibrium. The general form of the equilibrium constant expression depends on the stoichiometry of the reaction.
In the given exercise, we have the reaction:

• \( \mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g) \)

In terms of partial pressures, the equilibrium constant expression becomes:

• \( K_p = \frac{{P_{\mathrm{Cl}_2}^2}}{{P_{\mathrm{CCl}_4}}} \)

Remember that solids, like the carbon in this reaction, do not alter the equilibrium constant, as their concentration is considered constant. The equilibrium constant provides essential insights into the position of equilibrium, indicating whether reactants or products are favored at equilibrium.
The magnitude of \( K_p \) indicates the extent of the reaction. In this case, a \( K_p \) value of 77 signals that products (\( \mathrm{Cl}_2 \)) are favored over reactants (\( \mathrm{CCl}_4 \)). Understanding these concepts helps in predicting the direction and extent of a reaction under specific conditions.

Partial Pressure

Partial pressure refers to the pressure exerted by a particular gas in a mixture. Each gas contributes proportionally to the total pressure based on its concentration and the overall temperature of the system. When analyzing chemical equilibria in gaseous systems, understanding partial pressures is crucial.
In our exercise, the system begins with \( 202.7 \mathrm{kPa} \) of \( \mathrm{CCl}_4 \). As the reaction proceeds towards equilibrium, \( \mathrm{CCl}_4 \) decomposes, thereby reducing its partial pressure while \( \mathrm{Cl}_2 \), the product gas, exerts its own partial pressure by contributing to the overall pressure.
It is useful to express partial pressures using an ICE (Initial, Change, Equilibrium) table, making it easier to visualize the changes as the system reaches equilibrium:

• Initial: \( P_{\text{CCl}_4} = 202.7 \mathrm{kPa} \), \( P_{\text{Cl}_2} = 0 \).
• Change: \( \mathrm{CCl}_4 \rightarrow x \), \( \mathrm{Cl}_2 \rightarrow 2x \).
• Equilibrium: \( P_{\text{CCl}_4} = 202.7 - x \), \( P_{\text{Cl}_2} = 2x \).

Recognizing these relationships allows for the calculation of equilibrium conditions using the provided equilibrium constant.

Quadratic Equation

Quadratic equations often arise in chemistry when dealing with polynomial expressions, particularly in equilibrium problems. Solving these equations is essential to find quantities such as changes in concentration or pressure.
In this exercise, to determine the equilibrium pressures, we set up the equation based on \( K_p = 77 \):

• \( 77 = \frac{{4x^2}}{{202.7 - x}} \)

This simplifies to a quadratic form:

• \( 4x^2 + 77x - 15610.9 = 0 \)

To solve it, we can use the quadratic formula:

• \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \), where \( a = 4 \), \( b = 77 \), and \( c = -15610.9 \).

Calculating the discriminant \( b^2 - 4ac \) will show if real solutions exist and their nature. Solving these equations provides the change in pressure \( x \), which further allows us to calculate equilibrium conditions of the reactants and products.

Chemical Kinetics

Chemical kinetics studies the rate at which chemical reactions occur. It's interconnected with chemical equilibrium since it defines how fast a reaction approaches equilibrium. While the exercise primarily focuses on equilibrium, understanding the kinetics behind it adds depth.
Kinetics explores the pathways the reactions take and how variables like temperature and pressure influence reaction speed. While not explicitly calculated in equilibrium problems, knowing that the equilibrium in the exercise is achieved at \( 700 \mathrm{~K} \) gives insight into the reaction speed and product formation.
Equilibrium and kinetics together tell us:

• How fast equilibrium is reached.
• The dominant reaction pathway.
• The influence of temperature or pressure on speed.

Understanding chemical kinetics enhances our grasp of equilibrium dynamics, allowing for a better appreciation of how systems behave over time.