Problem 37 - General Chemistry: Principles and Modern Applications

Question

Equilibrium is established in a 2.50 L flask at \(250^{\circ} \mathrm{C}\) for the reaction $$\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=3.8 \times 10^{-2}$$ How many moles of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present at equilibrium, if (a) 0.550 mol each of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are initially introduced into the flask? (b) \(0.610 \mathrm{mol} \mathrm{PCl}_{5}\) alone is introduced into the flask?

Step-by-Step Solution

Answer

The solutions to equation at Step 3 and Step 6 will give the equilibrium mole values in Cases A and B, respectively. You need to solve this equation to get the numerical answer.

Step 1: Analyzing Scenario (A)

Case (a) begins with 0.550 mol each of \( \mathrm{PCl}_{5} \) and \( \mathrm{PCl}_{3} \). At equilibrium, \( \mathrm{PCl}_{5} \) decreases by \( x \), \( \mathrm{PCl}_{3} \) increases by \( x \), and \( \mathrm{Cl}_{2} \) also increases by \( x \). The equilibrium concentration of each compound can be calculated by dividing these moles by the volume of the flask (2.50 L).

Step 2: Setting Up Equilibrium Expression for Scenario (A)

Set up an equation for \( K_c \) using the concentrations from Step 1. \(\ K_c = [\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]/[\mathrm{PCl}_{5}] = (0.550+x)(0+x)/(0.550-x)\). This simplifies to \( K_c = \frac{x^2}{0.550-x}\) as \( 0 + x \) just becomes \( x \).

Step 3: Solving for 'x' in Scenario (A)

Now solve this equation for \( x \). Substitute known \( K_c = 3.8 \times 10^{-2} \) into the equation obtained from Step 2 and solve for \( x \). Therefore, \( 3.8 \times 10^{-2} = \frac{x^2}{0.550-x}\).

Step 4: Analyzing Scenario (B)

In case (b), only \(0.610 \, \mathrm{mol} \, \mathrm{PCl}_{5} \) is initially introduced. By a similar reasoning as in Scenario A, at equilibrium, \( \mathrm{PCl}_{5} \) decreases by \( x \), \( \mathrm{PCl}_{3} \) and \( \mathrm{Cl}_{2} \) both increase by \( x \).

Step 5: Setting Up Equilibrium Expression for Scenario (B)

Similarly to step 2, write the equilibrium equation. This time, \(K_c = [\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]/[\mathrm{PCl}_{5}] = (0 +x)(0+x)/(0.610 -x)\). This simplifies to \( K_c = \frac{x^2}{0.610-x} \).

Step 6: Solving for 'x' in Scenario (B)

Substitute the known \( K_c = 3.8 \times 10^{-2} \) into the equation and solve for \( x \) to find the equilibrium number of moles. \( 3.8 \times 10^{-2} = \frac{x^2}{0.610-x} \).

Key Concepts

Equilibrium Constant (Kc)

The equilibrium constant, represented as \(K_c\), is a crucial concept in understanding chemical equilibrium. It measures the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced chemical equation. In our case, the equilibrium reaction is \( \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \). For this reaction, the equilibrium constant is given by the expression:

• \[K_c = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]}\]

Here, the square brackets denote the molar concentrations at equilibrium.
The value \(K_c = 3.8 \times 10^{-2}\) implies a relatively small extent of reaction under the given conditions \((250^{\circ} \mathrm{C})\).
Understanding \(K_c\) helps predict the direction of the reaction: if the initial reaction quotient (similar to \(K_c\), but with initial concentrations) is greater than \(K_c\), the reaction will proceed towards the reactants. Conversely, if it is smaller, the reaction shifts towards the products.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is similar in form to the equilibrium constant \(K_c\), yet it applies to reaction mixtures that might not be at equilibrium. It is calculated using the same formula as \(K_c\) but with the initial concentrations:

• \[Q = \frac{[\mathrm{PCl}_{3}]_0[\mathrm{Cl}_{2}]_0}{[\mathrm{PCl}_{5}]_0}\]

For Scenario (a), where we begin with 0.550 mol each of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\), \(Q\) initially would be zero as \([\mathrm{Cl}_{2}]_0 = 0\).
In Scenario (b), with only \(\mathrm{PCl}_{5}\) present, \(Q = 0\) because the concentration of \([\mathrm{PCl}_{3}]\) and \([\mathrm{Cl}_{2}]\) starts at zero, which implies immediate redirection towards product formation to reach equilibrium.
Therefore, \(Q\) is an essential tool in predicting how the reaction will proceed to achieve equilibrium conditions.

Le Chatelier's Principle

Le Chatelier's principle is a helpful concept that predicts how a chemical system at equilibrium responds to changes in concentration, temperature, or pressure. When an external change is imposed on a system at equilibrium, the system adjusts to minimize that disturbance and restore a new equilibrium state.
For instance, if more \(\mathrm{PCl}_{5}\) is added to our equilibrium system, the reaction counterbalances by shifting towards \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) production, minimizing the effect of added \(\mathrm{PCl}_{5}\).
If the temperature is increased, and assuming our reaction is endothermic in the forward direction, more heat would favor the formation of products \(\mathrm{PCl}_{3} + \mathrm{Cl}_{2}\).
Other factors like changes in pressure tend to affect reactions involving gases, especially when there are different numbers of moles of gas on either side of the equation.
In our reaction, equal numbers of gas moles are on each side \((1:1)\), so changes in pressure might not significantly affect the reaction position. Le Chatelier's principle thus helps in grasping how equilibrium disruptions prompt shifts in reaction dynamics.