Problem 85 - Chemistry An Atoms First Approach

Question

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction,$$\operatorname{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$.If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3},\) what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?

Step-by-Step Solution

Answer

Under these equilibrium conditions, the concentration of \(Cl_2\) can be expressed in terms of the concentration of \(PCl_3\) as follows: \([Cl_2] = 9 \times 10^{-3} x\), where x is the concentration of \(PCl_3\). To find the exact concentration of \(Cl_2\), more information on the concentrations of either \(PCl_3\) or \(PCl_5\) would be needed.

Step 1: Write the equilibrium expression for the given reaction

The equilibrium expression for the reaction is given by: $$K = \frac{[PCl_3][Cl_2]}{[PCl_5]}$$

Step 2: Use given information to form equations

Let the concentration of PCl3 be x. According to the given conditions, the concentration of PCl5 is twice that of PCl3, so its concentration is 2x. The equilibrium expression becomes: $$K= \frac{x[Cl_2]}{2x}$$

Step 3: Solve for the concentration of Cl2

Now, we will solve for the concentration of Cl2. To do this, we will isolate [Cl2] in the equilibrium expression: $$[Cl_2] = 2Kx$$ We are given the value of K to be \(4.5 \times 10^{-3}\). So the expression for [Cl2] becomes: $$[Cl_2] = 2(4.5 \times 10^{-3})(x)$$

Step 4: Finding the concentration of Cl2 in terms of the concentration of PCl3

Under these equilibrium conditions, the concentration of Cl2 will be: $$[Cl_2] = 9 \times 10^{-3} x$$ This expression gives the concentration of Cl2 in terms of the concentration of PCl3. To find the exact concentration of Cl2, we would need more information on the concentrations of either PCl3 or PCl5. However, we now have a mathematical relationship between the concentration of Cl2 and PCl3 under the given equilibrium conditions.

Key Concepts

Equilibrium Expression

Understanding chemical equilibrium is crucial for studying chemical reactions. At equilibrium, a chemical system reaches a state where neither the reactants nor the products have a tendency to change over time, given that temperature and pressure remain constant. The quantitative measurement of this state is captured in the equilibrium expression, which is derived from the balanced chemical equation.

For the reaction \[\begin{equation} \text{PCl}_5 (g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g), \end{equation}\]we derive the equilibrium expression as:\[\begin{equation} K = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}. \end{equation}\]In this expression, \( K \) stands for the equilibrium constant, and the square brackets represent the concentrations of the substances in moles per liter. The concentrations of the products, \( \text{PCl}_3 \) and \( \text{Cl}_2 \), are in the numerator, and the concentration of the reactant, \( \text{PCl}_5 \), is in the denominator. It's important to note that pure solids and liquids are omitted from the equilibrium expression because their concentrations do not change.

In the given problem, using the provided equilibrium constant and the relationship between the concentrations of \( \text{PCl}_3 \) and \( \text{PCl}_5 \), we form equations to calculate the unknown concentration of one of the reactants or products.

Chemical Reaction

The transformation of substances during a chemical reaction involves breaking chemical bonds in the reactants and the formation of new bonds to create the products. For the reaction in our example, \( \text{PCl}_5 \) breaks down into \( \text{PCl}_3 \) and \( \text{Cl}_2 \). This is a reversible reaction, as indicated by the double-headed arrow. It means the reaction can proceed in both the forward and reverse direction until equilibrium is reached.

At this point, the rate at which \( \text{PCl}_5 \) breaks down into \( \text{PCl}_3 \) and \( \text{Cl}_2 \) is equal to the rate at which \( \text{PCl}_3 \) and \( \text{Cl}_2 \) combine to form \( \text{PCl}_5 \). Distinguishing between different types of reactions and understanding their unique equilibrium expressions are critical for predicting the behavior of chemical systems under various conditions.

The example given is a classic representation of a decomposition reaction within the realm of equilibrium chemistry, highlighting how initial concentrations can influence the system's dynamic.

Reaction Quotient

The reaction quotient, denoted by \( Q \), plays a pivotal role in predicting the direction in which a reaction will proceed to reach chemical equilibrium. It has the same form as the equilibrium constant expression but uses the initial concentrations of the reactants and products rather than the equilibrium concentrations.

For any moment in time before the system reaches equilibrium, the reaction quotient is given as:\[\begin{equation} Q = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}. \end{equation}\]Comparing the reaction quotient \( Q \) with the equilibrium constant \( K \) provides valuable insights. If \( Q < K \), the forward reaction will be favored for the system to reach equilibrium. Conversely, if \( Q > K \), the reverse reaction will proceed until the equilibrium is established. When \( Q = K \), the system is already at equilibrium, and no net change will occur.

In the exercise provided, we use the known equilibrium constant and the stoichiometric relationships to solve for the concentration of \( \text{Cl}_2 \) when the system is at equilibrium. Understanding how to manipulate and interpret the reaction quotient is a vital skill in working with chemical equilibria.