Problem 95 - Chemistry

Question

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ 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

The concentration of $\mathrm{Cl}_{2}$ under these conditions is: $$[\mathrm{Cl}_{2}] = 9 \times 10^{-3} \mathrm{M}$$

Step 1: Write the equilibrium constant expression

For the given reaction: $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g)$$ We can write the equilibrium constant expression (K) as: $$K = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]}$$

Step 2: Use the given information to create a relationship between the concentrations

We are informed that the concentration of PCl5 is twice the concentration of PCl3, which can be represented as: $$[\mathrm{PCl}_{5}] = 2\cdot [\mathrm{PCl}_{3}]$$

Step 3: Replace the relationship in the equilibrium constant expression

Substitute the relationship between PCl5 and PCl3 concentrations in the equilibrium constant expression: $$K = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{2\cdot[\mathrm{PCl}_{3}]}$$

Step 4: Simplify and solve for the concentration of Cl2

Simplify the expression and solve for the concentration of Cl2: $$4.5 \times 10^{-3} = \frac{[\mathrm{Cl}_{2}]}{2}$$ Multiply both sides by 2 to isolate the concentration of Cl2: $$[\mathrm{Cl}_{2}] = 2 \times 4.5 \times 10^{-3}$$ Calculate the concentration of Cl2: $$[\mathrm{Cl}_{2}] = 9 \times 10^{-3} \mathrm{M}$$

Step 5: Write the final answer

The concentration of Cl2 under these conditions is: $$[\mathrm{Cl}_{2}] = 9 \times 10^{-3} \mathrm{M}$$

Key Concepts

Chemical Equilibrium

In a chemical reaction, reaching a state of equilibrium means the rates of the forward and reverse reactions are equal. This doesn't mean the concentrations of the reactants and products are the same, but that they remain constant over time. The chemical equilibrium is dynamic, meaning that reactions continue to occur, but they do so at the same rate forward and backward. During equilibrium, reactants are being converted to products and vice versa at the same rate, which stabilizes both. This stabilization is the essence of chemical equilibrium. An essential tool in understanding equilibrium in a reaction is the equilibrium constant, denoted as \(K\). It helps predict the direction the reaction will proceed to reach equilibrium. A low \(K\) value, like \(4.5 \times 10^{-3}\), indicates that, at equilibrium, the concentration of reactants is higher than that of the products. In the given reaction involving \(\mathrm{PCl}_5 \) dissociating into \(\mathrm{PCl}_3 \) and \(\mathrm{Cl}_2\), we observe this process of reaching equilibrium by measuring and comparing these concentrations at equilibrium.

Concentration Expressions

The role of concentration expressions in chemical reactions is crucial for determining various reaction characteristics, such as equilibrium. An equilibrium constant expression involves concentrations of the products and reactants raised to the power of their coefficients in the balanced equation. For example, in the expression \(K = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]}\), each concentration is measured in molarity (M), which represents moles per liter. This concise form of demonstrating equilibrium also tells us how changes in concentration can shift the balance of a reaction.For our specific exercise, the concentration of \(\mathrm{PCl}_5\) is described to be twice the concentration of \(\mathrm{PCl}_3\). This relationship allows for the simplification of the equilibrium expression, as viewing the ratio provides insights into how the concentrations of all species are interdependent. Understanding these expressions and their use is fundamental to solving problems involving equilibrium constants.

Reaction Quotient

Before reaching equilibrium, the reaction quotient, denoted \(Q\), can provide immediate insight into which direction a reaction will move to reach that balanced state. The mathematical formulation of \(Q\) uses the same expression as \(K\), but it can be calculated at any point in the reaction, not just at equilibrium.In the context of our problem, viewing the expression \(K = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]}\) allows us to determine when the reaction is at equilibrium (\(Q = K\)). If \(Q < K\), then the reaction will proceed forward, forming more products. Conversely, if \(Q > K\), the reaction will shift backwards, favoring the reactants. Understanding the role of \(Q\) helps predict shifts and dynamics in concentration before a stable equilibrium is reached. This guides the adjustment of conditions like temperature and pressure to control and drive a reaction toward equilibrium under specific circumstances.