Problem 7 - Chemistry and Chemical Reactivity

Question

Calculating an Equilibrium Constant The reaction $$ \mathrm{PCl}_{5}(\mathrm{g}) \rightleftarrows \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ was examined at \(250^{\circ} \mathrm{C}\). At equilibrium, \(\left[\mathrm{PCl}_{5}\right]=4.2 \times\) \(10^{-5} \mathrm{mol} / \mathrm{L},\left[\mathrm{PCl}_{3}\right]=1.3 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) and \(\left[\mathrm{Cl}_{2}\right]=\) \(3.9 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Calculate \(K\) for the reaction.

Step-by-Step Solution

Answer

The equilibrium constant \( K \) is 1.21.

Step 1: Understand the Reaction and Principle

First, identify the balanced chemical equation for the reaction: \( \mathrm{PCl}_{5}( ext{g}) \rightleftharpoons \mathrm{PCl}_{3}( ext{g}) + \mathrm{Cl}_{2}( ext{g}). \) This is a reversible reaction, and at equilibrium, the rates of the forward and reverse reactions are equal.

Step 2: Write the Equilibrium Expression

For the given reaction, the equilibrium constant \( K \) is expressed as: \( K = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]}. \) This expression relates the concentrations of the reactants and products at equilibrium.

Step 3: Substitute the Given Concentrations

Insert the given equilibrium concentrations into the equilibrium expression:\[[\mathrm{PCl}_{5}] = 4.2 \times 10^{-5} \, \text{mol/L}, \quad [\mathrm{PCl}_{3}] = 1.3 \times 10^{-2} \, \text{mol/L}, \quad [\mathrm{Cl}_{2}] = 3.9 \times 10^{-3} \, \text{mol/L}.\]Substitute these values into the equation: \( K = \frac{(1.3 \times 10^{-2})(3.9 \times 10^{-3})}{4.2 \times 10^{-5}}. \)

Step 4: Calculate the Equilibrium Constant

Perform the calculations:\[K = \frac{(1.3 \times 3.9) \times 10^{-5}}{4.2 \times 10^{-5}} = \frac{5.07 \times 10^{-5}}{4.2 \times 10^{-5}}.\]Simplify the expression to get \( K = 1.21. \) This is the equilibrium constant for the reaction at the given temperature.

Key Concepts

Reversible Reactions

In chemistry, many reactions can proceed in both forward and reverse directions. These are known as reversible reactions. It is like a two-way street for chemicals, connecting reactants and products via a dynamic process.

• In the forward direction, reactants form products.
• In the reverse direction, products can revert back to reactants.

These types of reactions do not go to completion. Instead, they reach a balance, or equilibrium, where the speed (or rate) of the forward reaction equals the rate of the reverse reaction. This means the quantities of reactants and products remain stable over time, although molecules continue to react back and forth. This was observed in the reaction of \( \mathrm{PCl}_5\) breaking down into \( \mathrm{PCl}_3\) and \( \mathrm{Cl}_2\) at \(250^{\circ} \text{C}\).

Chemical Equilibrium

Chemical equilibrium is a fascinating state in a reversible reaction. At this point, there's no further change in the concentrations of reactants and products. But it's important to know that equilibrium doesn't mean the reaction has stopped. Instead, there is a continuous, balanced exchange of materials between reactants and products.

At equilibrium, both reactants and products are present, and neither is used up completely. The familiar process of reaching this state can be likened to a game of tug-of-war where both teams end up perfectly balanced. For the reaction \( \mathrm{PCl}_5(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) \), equilibrium was achieved when the concentrations were

• \([\mathrm{PCl}_5] = 4.2 \times 10^{-5} \text{ mol/L}\)
• \([\mathrm{PCl}_3] = 1.3 \times 10^{-2} \text{ mol/L} \)
• \([\mathrm{Cl}_2] = 3.9 \times 10^{-3} \text{ mol/L} \)

Understanding equilibrium helps us to control chemical reactions in industrial applications, ensuring the desired output, such as when manufacturing chemicals.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a measure that helps in understanding where a reaction is in relation to equilibrium. It's calculated using the same formula as the equilibrium constant. However, \(Q\) uses the actual concentrations or pressures at any given time, not just at equilibrium.

For the transition involving \(\mathrm{PCl}_5\), the expression is given by:\[Q = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]}\]By comparing the value of \(Q\) with the equilibrium constant \(K\):

• If \(Q < K\), the reaction will proceed forward to reach equilibrium.
• If \(Q > K\), the reaction will go in reverse.
• If \(Q = K\), the system is at equilibrium.

In summary, the reaction quotient is a handy tool for predicting which way a reaction needs to proceed to reach equilibrium.

Molar Concentrations

Molar concentration is a simple concept but plays a crucial role in chemistry. It refers to the amount of a substance (moles) contained in a given volume (liters). It is measured in \( \text{mol/L} \), commonly called molarity.

When calculating the equilibrium constant, knowing the molar concentrations of reactants and products is essential. For the reaction involving \( \mathrm{PCl}_5 \):

• \([\mathrm{PCl}_5] = 4.2 \times 10^{-5} \text{ mol/L}\)
• \([\mathrm{PCl}_3] = 1.3 \times 10^{-2} \text{ mol/L} \)
• \([\mathrm{Cl}_2] = 3.9 \times 10^{-3} \text{ mol/L} \)

These values help in determining the equilibrium constant \(K\) for the system using the equilibrium expression. Molar concentrations provide insight into how chemical reactions proceed and the balance of species in a mixture. They are key in adjusting reaction conditions in the lab.