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 \(700 \mathrm{~K}\). (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 conversion of \(\mathrm{CCl}_4\) into \(\mathrm{C}\) and \(\mathrm{Cl}_2\) is approximately \(0.24\), and the equilibrium partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) are \(154.1\mathrm{kPa}\) and \(97.2\mathrm{kPa}\), respectively.

Step-by-step solution

Write the equilibrium expressions

Let \(x\) be the moles of \(\mathrm{CCl}_4\) that decompose. The initial pressure of \(\mathrm{CCl}_4\) is \(202.7\mathrm{kPa}\). So, the equilibrium pressure of \(\mathrm{CCl}_4\) will be \((202.7-x)\mathrm{kPa}\). Since the molecular coefficient of \(\mathrm{Cl}_2\) in the balanced chemical reaction is 2, each mole of decomposed \(\mathrm{CCl}_4\) will produce 2 moles of \(\mathrm{Cl}_2\). Therefore, the equilibrium pressure of \(\mathrm{Cl}_2\) will be \((2x)\mathrm{kPa}\). Note that solid carbon does not contribute to the equilibrium expression.

Write the expression for \(K_p\)

The equilibrium constant for a given reaction is defined as the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients to the product of the partial pressures of the reactants raised to their stoichiometric coefficients. Thus, for the given reaction, the expression for \(K_p\) can be written as: $$K_p=\frac{P_{\mathrm{Cl}_2}^2}{P_{\mathrm{CCl}_4}}$$ Substitute the equilibrium expressions of the partial pressures in terms of \(x\): $$77=\frac{(2x)^2}{202.7-x}$$

Solve for \(x\)

Solve the equation for \(x\): $$77(202.7-x)=(2x)^2$$ $$30.5(202.7-x)=2x^2$$ $$6163.5-30.5x=2x^2$$ Rearrange the equation: $$2x^2+30.5x-6163.5=0$$ Solve this quadratic equation using the quadratic formula: $$x = \frac{-30.5\pm\sqrt{(-30.5)^2-4(2)(-6163.5)}}{2(2)}$$ After solving, we get two possible values for \(x\), but only one value is physically meaningful (i.e., \(x < 202.7\)): $$x\approx48.6$$

Calculate the fraction of conversion and equilibrium pressures

(a) Calculate the fraction of conversion of \(\mathrm{CCl}_4\) into \(\mathrm{C}\) and \(\mathrm{Cl}_2\): Fraction of conversion \(=\frac{x}{202.7}=\frac{48.6}{202.7}\approx0.24\) (b) Calculate the equilibrium partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\): \(\mathrm{P_{CCl}_4}=202.7-x=202.7-48.6\approx154.1\mathrm{kPa}\) \(\mathrm{P_{Cl}_2}=2x=2(48.6)\approx97.2\mathrm{kPa}\) Therefore, the fraction of conversion of \(\mathrm{CCl}_4\) into \(\mathrm{C}\) and \(2\mathrm{Cl}_2\) is approximately \(0.24\), and the equilibrium partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) are \(154.1\mathrm{kPa}\) and \(97.2\mathrm{kPa}\), respectively.

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, often denoted as \(K_p\) for gaseous reactions, provides a relationship between the concentrations of reactants and products when a reaction reaches equilibrium. For the given decomposition of \(\mathrm{CCl}_4\), the equilibrium constant is \(77\) at \(700\, \mathrm{K}\). This equilibrium constant helps us understand how much of the reactants convert to products.

• An equilibrium constant greater than 1 indicates that the products are favored at equilibrium.
• Conversely, a value less than 1 means reactants are favored.

In this case, because \(K_p = 77\), it shows a significant conversion of \(\mathrm{CCl}_4\) to \(\mathrm{C}\) and \(\mathrm{Cl}_2\). Remember, the equilibrium constant only changes with temperature, so \(K_p = 77\) is strictly valid only at \(700\, \mathrm{K}\). By calculating the equilibrium constant, chemists can predict the extent of the reaction.

Partial Pressure

Partial pressure is the pressure that a particular gas in a mixture of gases would exert if it occupied the entire volume by itself. This concept is crucial in understanding how gases behave in chemical reactions, particularly those involving equilibrium.

• In our example, \(\mathrm{CCl}_4\) starts with an initial pressure of \(202.7\, \mathrm{kPa}\).
• As the reaction progresses to equilibrium, the pressure changes according to how much \(\mathrm{CCl}_4\) remains and how much \(\mathrm{Cl}_2\) is formed.

Accordingly, the equilibrium partial pressures are determined by the amount of reactant that decomposes. This affects calculations using the equation for \(K_p\). The pressures facilitate understanding of the quantitative aspects of the reaction composition at equilibrium.

Quadratic Equation

When dealing with equilibrium reactions like this one, we often need to solve a quadratic equation to find unknown variables, such as the amount of a reactant or product. The quadratic formula is used when the equation is in the standard form: \[ax^2 + bx + c = 0\]
In this scenario:- The resulting equation to solve for \(x\) is \(2x^2 + 30.5x - 6163.5 = 0\).- By applying the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]we find both potential solutions.- Only values of \(x\) that make sense in context (like being less than the initial pressure) are considered physically relevant.
The quadratic equation emerges naturally from the relationship between pressures and equilibrium constants, offering a systematic way to solve for unknown variables.

Stoichiometry

Stoichiometry refers to the quantitative relationships between reactants and products in a chemical reaction. It is essential for understanding how much of each substance participates in the reaction.

• In this case, stoichiometry determines that one mole of \(\mathrm{CCl}_4\) produces two moles of \(\mathrm{Cl}_2\).
• This molar relationship is reflected in the change in pressure at equilibrium.

By adding these stoichiometric coefficients into the equilibrium expression, we ensure the reaction's pressure changes are accurately captured. Stoichiometry ensures that our mathematical expressions reflect the true nature of chemical changes.