Problem 76 - Chemistry and Chemical Reactivity

Question

Pure \(\mathrm{PCl}_{5}\) gas is placed in a 2.00 -L. flask. After heating to \(250^{\circ} \mathrm{C}\) the pressure of \(\mathrm{PCl}_{5}\) is initially 2.000 atm. However, the gas slowly but only partially decomposes to gaseous \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\). When equilibrium is reached, the partial pressure of \(\mathrm{Cl}_{2}\) is 0.814 atm. Calculate \(K_{\mathrm{p}}\) for the decomposition.

Step-by-Step Solution

Answer

The equilibrium constant \(K_p\) is approximately 0.558.

Step 1: Write the Balanced Equation

The decomposition reaction of \(\mathrm{PCl}_{5}\) is: \[ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \] This shows that each mole of \(\mathrm{PCl}_{5}\) decomposes into one mole of \(\mathrm{PCl}_{3}\) and one mole of \(\mathrm{Cl}_{2}\).

Step 2: Initial Conditions and Changes

Initially, the pressure of \(\mathrm{PCl}_5\) is 2.000 atm, and \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) are both 0 atm. Let the change in pressure at equilibrium for \(\mathrm{Cl}_2\) be \(x = 0.814\) atm. Hence, the changes for the other gases are: \(\mathrm{PCl}_{5}: -x\) and \(\mathrm{PCl}_{3}: +x\) atm.

Step 3: Calculate Equilibrium Pressures

The pressure of \(\mathrm{PCl}_{5}\) at equilibrium is \(2.000 - x\) atm. Since \(x = 0.814\), the equilibrium pressure of \(\mathrm{PCl}_{5}\) is \(2.000 - 0.814 = 1.186\) atm. The pressure of \(\mathrm{PCl}_{3}\) at equilibrium is \(x\), which equals 0.814 atm.

Step 4: Write Expression for \(K_p\)

The expression for the equilibrium constant \(K_p\) in terms of pressure is: \[ K_p = \frac{P_{\mathrm{PCl}_{3}} \times P_{\mathrm{Cl}_{2}}}{P_{\mathrm{PCl}_{5}}} \] Using the equilibrium pressures calculated: \[ K_p = \frac{0.814 \times 0.814}{1.186} \]

Step 5: Solve for \(K_p\)

Plug the values into the equation and solve for \(K_p\): \[ K_p = \frac{0.814 \times 0.814}{1.186} = \frac{0.662396}{1.186} \approx 0.558 \] Therefore, \(K_p\) for the decomposition reaction is approximately 0.558.

Key Concepts

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \( K_p \), is essential in understanding chemical reactions that involve gases. It describes the ratio of the partial pressures of products to reactants at equilibrium. For the decomposition reaction of \( \mathrm{PCl}_5 \) described in the exercise, \( K_p \) evaluates how far the reaction proceeds before equilibrium is met. The expression for \( K_p \) is derived from the balanced chemical equation. In this case, it's given by the formula:\[ K_p = \frac{P_{\mathrm{PCl}_3} \times P_{\mathrm{Cl}_2}}{P_{\mathrm{PCl}_5}} \]Here, \( P \) stands for the partial pressures of the respective gases at equilibrium. This calculated value of \( K_p \) tells us about the extent of the reaction; a larger \( K_p \) would mean more products are formed when equilibrium is reached.

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. When dealing with chemical equilibrium and gas reactions, the concept of partial pressure becomes crucial. In the context of the decomposition of \( \mathrm{PCl}_5 \), initially, the entire 2.000 atm belongs to \( \mathrm{PCl}_5 \) since no reaction products are present at the start. As \( \mathrm{PCl}_5 \) decomposes, it converts into \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \), and equilibrium is established when the partial pressures of \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \) stabilize at 0.814 atm each. Thus, understanding the partial pressures of each gas tells us how the total pressure at equilibrium is distributed among them, highlighting the dynamics of the system under specified conditions.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances. In the problem, \( \mathrm{PCl}_5 \) dissociates into \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \). This is signified by:\[ \mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g) + \mathrm{Cl}_2(g) \]Such reactions are distinguished by a tendency of the compound to break its chemical bonds under specific conditions, such as heat, into smaller fragments. Here, the heating to \( 250^{\circ} \mathrm{C} \) provides the necessary energy for \( \mathrm{PCl}_5 \) to decompose. Recognizing this reaction type helps in predicting how the reaction components will behave when subject to specific conditions.

Gas Laws

Gas laws help explain the behavior of gases under various conditions of temperature, volume, and pressure, integral for calculations involving chemical equilibria in gaseous reactions.Boyle's Law, Charles's Law, Avogadro's Law, and the Ideal Gas Law collectively allow prediction of these behaviors. In the equilibrium setting of the \( \mathrm{PCl}_5 \) decomposition:- Boyle’s Law (\( P \propto \frac{1}{V} \) at constant \( T \)) shows that pressure changes with volume.- Charles’s Law (\( V \propto T \) at constant \( P \)) explains how gas expands on heating.Understanding these principles provides the theoretical basis for determining how changes in conditions, like temperature (as seen with heating to \( 250^{\circ} \text{C} \)), can drastically influence reaction equilibrium and partial pressures. This thorough grasp of gas laws and their practical applications enables accurate predictions in chemical reactions involving gases.