Question

At \(700 \mathrm{~K}\) the equilibrium constant for the reaction $$ \mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(\mathrm{s})+2 \mathrm{Cl}_{2}(g) $$ is \(K_{p}=0.76\). A flask is charged with \(2.00 \mathrm{~atm}\) of \(\mathrm{CCl}_{4}\), which then reaches equilibrium at \(700 \mathrm{~K}\). (a) What fraction of the \(\mathrm{CCl}_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?\) (b) What are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

Short answer

The fraction of CCl4 converted into C and Cl2 is approximately \(0.258\). The equilibrium partial pressures are approximately \(1.4845 \mathrm{~atm}\) for CCl4 and \(1.031 \mathrm{~atm}\) for Cl2.

Step-by-step solution

(Step 1: Write the given reaction and Kp expression)

The given reaction is: $$\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(\mathrm{s})+2\mathrm{Cl}_{2}(g)$$ The Kp expression for this reaction is: $$K_p = \frac{[\mathrm{Cl}_2]^2}{[\mathrm{CCl}_4]}$$

(Step 2: Set up an ICE table to determine equilibrium pressures)

An ICE table is used to help understand the changes that occur during a chemical reaction. The letters in the acronym represent: I: Initial state C: Change E: Equilibrium We will create an ICE table for partial pressures of CCl4 and Cl2: | | CCl4 | Cl2 | |-------|------|-----| |Initial| 2.00 | 0 | | Change| -x | +2x | |Equil. |(2-x) | 2x | Here, x represents the partial pressure of CCl4 that reacts to form C and Cl2. Since solid C does not exert pressure, we do not include it in our ICE table.

(Step 3: Write the Kp expression using the ICE table values)

Insert the Equilibrium values from the ICE table into the Kp expression: $$0.76 = \frac{(2x)^2}{(2 - x)} $$

(Step 4: Solve the equation for x)

Solve the quadratic equation to determine x: $$0.76(2 - x) = 4x^2$$ $$1.52 - 0.76x = 4x^2$$ $$4x^2 + 0.76x - 1.52 = 0$$ The solutions for x are approximately -0.7295 and 0.5155. Since partial pressure must be positive, we reject the negative value and accept x=0.5155 atm.

(Step 5: Answer the questions: fraction of CCl4 converted and equilibrium pressures)

(a) To find the fraction of CCl4 converted into C and Cl2, divide x by the initial pressure of CCl4: Fraction of CCl4 converted = \(\frac{x}{2.00\, \text{atm}}\) Fraction of CCl4 converted ≈ \(\frac{0.5155}{2.00} ≈ 0.258\) (b) To find the equilibrium partial pressures of CCl4 and Cl2, plug x into our equilibrium expressions from the ICE table: Partial pressure of CCl4 at equilibrium = 2 - x ≈ 2 - 0.5155 ≈ 1.4845 atm Partial pressure of Cl2 at equilibrium = 2x ≈ 2(0.5155) ≈ 1.031 atm

Key concepts

Equilibrium Constant

Chemical reactions reach a state where the concentrations of reactants and products remain constant over time; this is known as chemical equilibrium. The equilibrium constant \( K_p \) is a crucial factor in understanding reactions that involve gases. It helps to quantify the ratio of the partial pressures of the reactants and products at this equilibrium state.

In our example, the reaction is \[\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(\mathrm{s})+2 \mathrm{Cl}_{2}(g),\] and the equilibrium constant given is \( K_p = 0.76 \). This equilibrium constant is calculated using the partial pressures, which are represented in a simplified form where solids like carbon don't affect the pressure and are, therefore, excluded.

The expression for \( K_p \) gives you a clear idea of the relationship between the pressures of the gases involved, and finding this constant helps you predict how the system behaves at equilibrium.

ICE Table

An ICE table is a problem-solving tool that helps us visualize the changes in concentrations or pressures in a chemical system as it reaches equilibrium. ICE stands for Initial, Change, and Equilibrium.

Let's break it down:

• Initial: Represents the starting partial pressures of reactants and products before any reaction occurs.
• Change: Denotes the amount by which the pressures shift as the reaction proceeds toward equilibrium.
• Equilibrium: Shows the pressures after the reaction has reached equilibrium.

For our reaction, the initial pressure of \( \mathrm{CCl}_4 \) is 2.00 atm, and \( \mathrm{Cl}_2 \) starts at 0 atm. As the reaction proceeds, \( \mathrm{CCl}_4 \) decreases by \( x \) amount, and \( \mathrm{Cl}_2 \) increases by \( 2x \) because of the reaction stoichiometry, giving equilibrium pressures of \((2 - x)\) atm for \( \mathrm{CCl}_4 \) and \( 2x \) atm for \( \mathrm{Cl}_2 \). This structured approach helps manage the data needed to solve equilibrium problems.

Partial Pressures

Partial pressure is the pressure that a gas in a mixture would exert if it alone occupied the entire volume. It's a measure crucially related to the concentrations of gases in a reaction. In equilibrium calculations involving gases, partial pressures are often preferred over concentrations, particularly when using the equilibrium constant \( K_p \).

For instance, in the given problem, you start with a total pressure contributed entirely by \( 2.00 \) atm of \( \mathrm{CCl}_4 \). As the reaction progresses to equilibrium, a part of this is converted into \( \mathrm{Cl}_2 \), changing the system's dynamics. Accurately determining the partial pressures of these gases at equilibrium is fundamental to calculate other parameters like the equilibrium constant, and it helps predict how shifts in initial conditions affect the equilibrium state.

Kp Expression

The \( K_p \) expression is a mathematical representation that helps calculate and understand the pressures of gases in a chemical equilibrium song. In gaseous reactions, knowing \( K_p \) can tell us about the extent and direction of the reaction. Higher \( K_p \) values indicate a greater formation of products at equilibrium.For the reaction \[\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(\mathrm{s})+2 \mathrm{Cl}_{2}(g),\]the \( K_p \) is expressed as:\[ K_p = \frac{(\mathrm{Cl}_2)^2}{(\mathrm{CCl}_4)}.\]This formula is derived from the balanced chemical equation and shows the ratio of the product's partial pressures to the reactant's partial pressures, specifically considering only the gaseous components. Solving this \( K_p \) expression using the pressures obtained from the ICE table provides insight into how much of the \( \mathrm{CCl}_4 \) has reacted and how much \( \mathrm{Cl}_2 \) has been produced, crucial for determining equilibrium dynamics.