Problem 87 - Chemistry The Central Science

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 \(K_{p}=0.76 .\) A flask is charged with 2.00 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?

Step-by-Step Solution

Answer

(a) The fraction of the CCl4 converted into C and Cl2 is approximately 0.0556 or 5.56%. (b) The partial pressures of CCl4 and Cl2 at equilibrium are approximately 1.89 atm and 0.2224 atm, respectively.

Step 1: 1. Expression for equilibrium constant Kp in terms of partial pressures

For the reaction: \(\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s) + 2 \mathrm{Cl}_{2}(g)\) Let's initially assign the partial pressure of CCl4 as \(P_{\mathrm{CCl}_4}\) and that of Cl2 as \(P_{\mathrm{Cl}_2}\). As stated in the problem, the equilibrium constant, Kp, is given as 0.76. The expression for Kp is \(K_{p} = \frac{P_{\mathrm{Cl}_2}^2}{P_{\mathrm{CCl}_4}}\)

Step 2: 2. Relationship between change in partial pressures

At the beginning, partial pressure of CCl4 is 2.00 atm and there is no Cl2. Let x be the pressure of CCl4 that reacts and converts to Cl2. Thus, at equilibrium, the new partial pressures are: \(P_{\mathrm{CCl}_4} = 2.00 - x\) \(P_{\mathrm{Cl}_2} = 2x\)

Step 3: 3. Substitute pressures in the Kp expression and solve for x

Plug the partial pressures into the Kp expression and solve for x: \(0.76 = \frac{(2x)^2}{(2.00 - x)}\) Solve the quadratic equation for x. Multiplying both sides by (2.00-x): \(1.52x - 0.76x^2 = 4x^2\) Rearrange terms: \(5.28x^2 - 1.52x = 0\) Now we can solve for x: \(x = \frac{-1.52 \pm \sqrt{1.52^2-4(5.28)(0)}}{2(5.28)}\) At this point, we have a positive and a negative value of x. However, the negative value has no physical meaning in this problem since it would imply an increase in the partial pressure of CCl4, contradicting the reaction. Thus, we must choose the positive value: \(x = \frac{-1.52 + \sqrt{1.52^2}}{2(5.28)} \approx 0.1112\)

Step 4: 4. Calculate the fraction of CCl4 converted

The fraction of CCl4 converted is given by the ratio of x to the initial partial pressure of CCl4: Fraction converted = \(\frac{x}{2.00}\) Fraction converted = \(\frac{0.1112}{2.00} \approx 0.0556\)

Step 5: Part (a) Answer

(a) The fraction of the CCl4 converted into C and Cl2 is approximately 0.0556 or 5.56%.

Step 6: 5. Determine partial pressures of CCl4 and Cl2 at equilibrium

Now we will calculate the equilibrium partial pressures of CCl4 and Cl2 using the value of x: \(P_{\mathrm{CCl}_4} = 2.00 - x = 2.00 - 0.1112 \approx 1.89 \mathrm{~atm} \) \(P_{\mathrm{Cl}_2} = 2x = 2(0.1112) \approx 0.2224\mathrm{~atm} \)

Step 7: Part (b) Answer

(b) The partial pressures of CCl4 and Cl2 at equilibrium are approximately 1.89 atm and 0.2224 atm, respectively.

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented as \(K_p\) for gas-phase reactions, is a numerical value that depicts the ratio of the concentrations (or partial pressures) of the products to the reactants at equilibrium. It's unique to each reaction at a given temperature.

• Defined mathematically, the equilibrium constant expression for a gaseous reaction is given as \(K_p = \frac{P_{ ext{products}}^{ ext{coefficients}}}{P_{ ext{reactants}}^{ ext{coefficients}}}\).
• It only depends on the pressures of gaseous components involved in the chemical equation.
• For example, in the reaction \( \text{CCl}_4(g) \rightleftharpoons \text{C}(s) + 2 \text{Cl}_2(g)\), the equilibrium constant expression is \(K_p = \frac{P_{ ext{Cl}_2}^2}{P_{ ext{CCl}_4}}\).
• \(K_p\) indicates the extent to which a reaction proceeds before reaching equilibrium; a higher \(K_p\) means more products are formed while a lower \(K_p\) suggests fewer products and more reactants residual at equilibrium.

Understanding and using \(K_p\) allows chemists to predict how a system will behave under various conditions, making it an essential tool in chemical equilibrium analysis.

Partial Pressure

Partial pressure is the pressure that a gas in a mixture would exert if it occupied the entire volume by itself. It's crucial in calculating equilibrium positions for gaseous reactions.

• The partial pressure of a gas is directly proportional to its mole fraction in the gas mixture, affected by the same factors that affect the pressure of the overall gas: temperature, volume, and amount of gas present.
• In reactions, the change in partial pressures can help determine shifts in equilibrium or conversion rates of reactants and products.

In the context of our exercise, the initial partial pressure of \(\text{CCl}_4\) is given as 2.00 atm. With this information and the alteration caused by the reaction, we adjust the pressures of reactants and products until they precisely fit the equilibrium constant equation.

Gaseous Reaction

Gaseous reactions involve gas molecules as reactants and products, and they are particularly sensitive to changes in pressure and temperature. In our specific example, we are looking at a decomposition reaction where \(\text{CCl}_4(g)\) transforms into solid carbon and \(\text{Cl}_2(g)\).

• These reactions are governed by Le Chatelier's principle, which predicts the shift in equilibrium due to changes in pressure, volume, temperature, or concentration.
• When analyzing gaseous reactions, we focus on the partial pressures of gases, as they help us understand how much of each component is present at equilibrium.
• This means calculating the change in pressure for each component as the reaction progresses to its equilibrium state.

For example, starting with 2.00 atm of \(\text{CCl}_4\), the gaseous Cl2 generated at equilibrium can be determined by applying the equilibrium constant equation, allowing us to solve for conversion rates.

Quadratic Equation

In certain chemical equilibrium problems, solving for unknowns involves setting up and solving a quadratic equation. The quadratic equation arises from balancing the changes in concentration or pressure terms according to the balanced chemical equation and equilibrium expression.

• The standard form of the quadratic equation is \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.
• The solutions for \(x\) can be found using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• In chemical contexts, after substituting equilibrium concentrations into the \(K_p\) expression, the resulting equation often fits the quadratic form.
• Recall that physically meaningful solutions must satisfy positive concentrations or pressures only, so sometimes one solution is discarded.

In the exercise, using \[0.76 = \frac{(2x)^2}{(2.00 - x)}\,\] leads to a quadratic equation: \(5.28x^2 - 1.52x = 0\). Solving it gives the value of x, indicating the equilibrium progression of the reaction.