Problem 83 - Chemistry: An Atoms-Focused Approach

Question

Consider the following reaction: \(\operatorname{PCl}_{5}(g) \rightleftharpoons \operatorname{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \quad K_{\mathrm{p}}=23.6 \mathrm{at} 500 \mathrm{K}\) a. Calculate the equilibrium partial pressures of the reactants and products if the initial pressures are \(P_{\mathrm{RC}_{5}}=0.560\) atm and \(P_{\mathrm{PC}_{2}}=0.500 \mathrm{atm}\) b. If more chlorine is added after cquilibrium is reached, how will the concentrations of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) change?

Step-by-Step Solution

Answer

Answer: The equilibrium pressures are PCl₅ = 0.344 atm, PCl₃ = 0.216 atm, and Cl₂ = 0.716 atm. When more Cl₂ is added after equilibrium is reached, the concentration of PCl₅ increases and the concentration of PCl₃ decreases.

Step 1: Understand the equilibrium expression

We are given the reaction: \(\operatorname{PCl}_{5}(g) \rightleftharpoons \operatorname{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) The equilibrium expression for this reaction can be written as: \(K_p = \frac{P_{PCl_3} P_{Cl_2}}{P_{PCl_5}}\)

Step 2: Calculate the change in pressures

Let's denote the change in the pressures of \(\operatorname{PCl}_{5}\), \(\operatorname{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) with x. At equilibrium, we will obtain the following changes in partial pressures: \(P_{PCl_5} = 0.560 - x\) \(P_{PCl_3} = x\) \(P_{Cl_2} = 0.500 + x\)

Step 3: Substituting values into the equilibrium expression and solving for x

Using the above changes in pressures, substitute the values into the equilibrium constant expression, and solve for x: \(K_p = \frac{P_{PCl_3} P_{Cl_2}}{P_{PCl_5}} = \frac{x(0.500+x)}{0.560-x} = 23.6\) After solving for x, we get: \(x = 0.216\)

Step 4: Calculate the equilibrium pressures

Now, let's determine the equilibrium pressures using the value of x: \(P_{PCl_5} = 0.560 - 0.216 = 0.344\) atm \(P_{PCl_3} = 0.216\) atm \(P_{Cl_2} = 0.500 + 0.216 = 0.716\) atm

Step 5: Determining the effect of adding Cl₂

If more chlorine is added after equilibrium is reached, we can use Le Châtelier's Principle to predict the effect on the pressures of \(\operatorname{PCl}_{5}\) and \(\operatorname{PCl}_{3}\). The addition of Cl₂ will shift the equilibrium to the left to minimize the disturbance. Therefore, the concentration of \(\operatorname{PCl}_{5}\) will increase, and the concentration of \(\operatorname{PCl}_{3}\) will decrease. So the answers are: a. \(P_{PCl_5} = 0.344\) atm, \(P_{PCl_3} = 0.216\) atm, and \(P_{Cl_2} = 0.716\) atm b. If more chlorine is added, the concentration of \(\operatorname{PCl}_{5}\) increases and the concentration of \(\operatorname{PCl}_{3}\) decreases.

Key Concepts

Equilibrium Constant

In a chemical reaction, the equilibrium constant, denoted as \(K_p\) when dealing with gases, is a measure of the extent of a reaction at equilibrium. It provides a ratio of the concentration or partial pressures of the products to the reactants, with each raised to the power of their stoichiometric coefficients from the balanced chemical equation. For instance, in the reaction: \[ \operatorname{PCl}_{5}(g) \rightleftharpoons \operatorname{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \] the equilibrium constant expression is given by: \[ K_p = \frac{P_{PCl_3} P_{Cl_2}}{P_{PCl_5}} \]Knowing the value of \(K_p\) helps us understand whether a reaction favors the formation of products or reactants at equilibrium. When \(K_p \gg 1\), it signifies that the reaction heavily favors products. Conversely, if \(K_p \ll 1\), the reaction predominantly holds onto its reactants.

Le Châtelier's Principle

This principle is a tool for predicting how a change in conditions affects chemical equilibrium. It states that if an external change is applied to a system at equilibrium, the system will adjust to partly counteract the change and re-establish equilibrium. One can use this principle to predict the effect of adding or removing chemicals and changing pressure or temperature. In our reaction:: \(\operatorname{PCl}_{5}(g) \rightleftharpoons \operatorname{PCl}_{3}(g) + \mathrm{Cl}_{2}(g)\),adding more chlorine gas \(Cl_2\) after equilibrium would disturb the system. According to Le Châtelier's Principle, the equilibrium will shift to the left, thus increasing the concentration of \(PCl_5\) and decreasing the concentrations of \(PCl_3\). This is because the reaction will try to reduce the increased concentration of \(Cl_2\) by forming more \(PCl_5\).

Partial Pressure

Partial pressure refers to the pressure exerted by an individual gas in a mixture of gases. For reactions involving gases at equilibrium, it is important because it influences how we set up and calculate equilibrium expressions. Each gas in a mixture contributes to the total pressure, according to its fraction in the volume of the mixture. The total pressure of the system is the sum of partial pressures of the gases present. In our case, \[ P_{PCl_5} = 0.560 - x, \hspace{5pt} P_{PCl_3} = x, \hspace{5pt} P_{Cl_2} = 0.500 + x \]represents changes in pressures that occur as the system reaches equilibrium. Each gas's partial pressure contributes to the overall dynamics of how equilibrium is achieved and maintained in the system.

Gaseous Reactions

Gaseous reactions involve substances in their gas state participating in the reaction. These reactions are sensitive to changes in pressure and volume due to the intrinsic properties of gases. Therefore, understanding how gases behave is essential. For example, according to Avogadro's law, at constant temperature and pressure, equal volumes of gases contain an equal number of molecules. In the equilibrium context for reactions like \(\operatorname{PCl}_{5}(g) \rightleftharpoons \operatorname{PCl}_{3}(g) + \mathrm{Cl}_{2}(g)\), the stoichiometry and initial pressures decide the position of equilibrium, which is quantitatively expressed using partial pressures. Calculating these pressures helps in understanding how much of each reactant and product gas is present under equilibrium conditions. Strong comprehension of these concepts empowers us to manipulate reaction conditions effectively to favor the formation of desired products.