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?

Short answer

(a) The fraction of CCl4 converted into C and Cl2 is approximately 0.133. (b) The partial pressures at equilibrium are: CCl4 ≈ 1.734 atm C ≈ 0.266 atm Cl2 ≈ 0.532 atm

Step-by-step solution

Write the balanced chemical equation and the expression for Kp

The balanced chemical equation is given as: \[ \mathrm{CCl_{4}(g) \rightleftharpoons C(s) + 2 Cl_{2}(g)} \] The expression for Kp is given by: \[ K_p = \frac{\left[P(\mathrm{Cl_{2}})\right]^2}{P(\mathrm{CCl_{4}})} \]

Set up the ICE table

The ICE table for this reaction looks like: | | CCl4 | C | Cl2 | |---------|--------------|-------------|--------------| | Initial (atm) | 2.00 | 0 | 0 | | Change (atm) | -x | +x | +2x | | Equilibrium (atm) | \(2.00 - x\) | x | 2x | Where x is the amount of CCl4 that gets converted into C and Cl2.

Use the expression for Kp to solve for x

Replacing the pressure terms in the Kp expression with the equilibrium values, we have: \[ 0.76 = \frac{(2x)^2}{2.00 - x} \]

Solve for x

Solve the equation for x: \[ 0.76(2.00 - x) = 4x^2 \] Expanding and rearranging the equation to form a quadratic equation: \[ 4x^2 + 0.76x - 1.52 = 0 \] Solve the quadratic equation using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where a = 4, b = 0.76, and c = -1.52 Solving for x, we get x ≈ 0.266 atm (only consider the positive value, since negative pressure doesn't make sense).

Calculate the fraction of CCl4 converted

The fraction of CCl4 converted into the products is given by the formula: \[ \text{Fraction converted} = \frac{x}{\text{Initial pressure of CCl4}} \] Substitute the values and calculate the fraction: \[ \text{Fraction converted} = \frac{0.266}{2.00} \approx 0.133 \]

Calculate the partial pressures at equilibrium

Using the values of x, we can calculate the partial pressures at equilibrium: \[ P(\mathrm{CCl_{4}}) = 2.00 - x \approx 1.734 \text{ atm} \] \[ P(\mathrm{C}) = x \approx 0.266 \text{ atm} \] \[ P(\mathrm{Cl_{2}}) = 2x \approx 0.532 \text{ atm} \] To summarize: (a) The fraction of CCl4 converted into C and Cl2 is approximately 0.133 (b) The partial pressures at equilibrium are: CCl4 ≈ 1.734 atm C ≈ 0.266 atm Cl2 ≈ 0.532 atm

Key concepts

Equilibrium Constant

The equilibrium constant (\( K_p \)) is a vital concept in understanding chemical equilibrium. It defines the ratio of product pressures to reactant pressures at equilibrium. For a generic reaction \( aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g) \), the equilibrium constant expression is given by:\[K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}\]where \(P_C, P_D, P_A,\) and \(P_B\) are the partial pressures of the gases involved.
For our specific reaction involving tetrachloromethane (\( \text{CCl}_4 \)), carbon (\( \text{C} \)), and chlorine (\( \text{Cl}_2 \)), the equilibrium constant equation simplifies to only include the gases:\[K_p = \frac{[P(\text{Cl}_2)]^2}{P(\text{CCl}_4)}\]This is because the partial pressure of the solid carbon is not included in the expression as its activity is constant. The value of \( K_p \) indicates how far the reaction tends to go towards the products side under equilibrium conditions. In this case, \( K_p = 0.76 \) suggests a moderate conversion of reactants to products.

ICE Table

The ICE table is a useful organizational tool for understanding changes in concentrations or pressures during a reaction that achieves equilibrium. "ICE" stands for Initial, Change, and Equilibrium.

• Initial: Represents the starting pressures or concentrations of the reactants and products.
• Change: Indicates how much the concentrations or pressures change as the system approaches equilibrium.
• Equilibrium: Shows the final pressures or concentrations when the reaction is at equilibrium.

For our reaction, initially, there are only reactants, so \( P(\text{CCl}_4) = 2.00 \) atm and both \( P(\text{C}) \) and \( P(\text{Cl}_2) \) are 0 atm. Once the reaction proceeds:

• \( \Delta \text{CCl}_4 = -x \)
• \( \Delta \text{C} = +x \)
• \( \Delta \text{Cl}_2 = +2x \)

This ultimately leads to the equilibrium pressures:

\( P(\text{CCl}_4) = 2.00 - x \)

\( P(\text{C}) = x \)

\( P(\text{Cl}_2) = 2x \)

Partial Pressures

Partial pressures are a crucial concept when dealing with gases in chemical equilibrium, representing the pressure exerted by a single type of gas in a mixture. In any system at equilibrium, the total pressure is the sum of the partial pressures of the gases present.
When calculating partial pressures during chemical reactions like the one in our exercise, we track the pressure changes using the ICE table's equilibrium row. For example:

• Before reaction: \( P(\text{CCl}_4) = 2.00 \) atm, \( P(\text{Cl}_2) = 0 \) atm
• At equilibrium: \( P(\text{CCl}_4) = 2.00 - x \) atm, and \( P(\text{Cl}_2) = 2x \) atm

These partial pressures help us to write the equilibrium expression:\( K_p = \frac{(2x)^2}{2.00-x} \)which allows solving for \( x \)and finding the equilibrium pressures.

Quadratic Equation

In many instances, chemical equilibrium problems involving \( K_p \)and partial pressures require solving quadratic equations. This emerges because the equilibrium expression is a rational equation that can turn into a quadratic when rearranged, as shown below.
From our example, starting from:\[0.76 = \frac{(2x)^2}{2.00 - x} \]Reorganizing gives:\[0.76(2.00-x) = 4x^2\]Expand and rearrange the equation into standard quadratic form:\[4x^2 + 0.76x - 1.52 = 0\]To solve it, apply the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \( a=4 \), \( b=0.76 \), and \( c=-1.52 \). Calculate the square root and solve for the positive \( x \), since negative pressure isn’t realistic:\[x \approx 0.266 \,\text{atm}\]Solving quadratic equations often aids in finding unknowns crucial for determining equilibrium states.