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?

Short answer

The concentration of \(\mathrm{Cl}_{2}\) under these conditions is \(9 \times 10^{-3}\) mol/L.

Step-by-step solution

Write the equilibrium expression for the reaction

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

Set up the equilibrium expression with given information to solve for Cl2 concentration

We are given that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3}\). Let the concentration of \(\mathrm{PCl}_{3}\) be x. So, the concentration of \(\mathrm{PCl}_{5}\) = 2x. We substitute these in the equilibrium constant expression: \[ K = \frac{(x)([\mathrm{Cl}_{2}])}{(2x)} \]

Solve for Cl2 concentration using the given K value

The given K value is 4.5 × 10^{-3}. Substitute K in the equation and solve for the concentration of \(\mathrm{Cl}_{2}\): \[ 4.5 \times 10^{-3} = \frac{(x)([\mathrm{Cl}_{2}])}{(2x)} \] Now, solve for the concentration of \(\mathrm{Cl}_{2}\) by multiplying both sides by 2x: \[ 2x(4.5 \times 10^{-3}) = (x)([\mathrm{Cl}_{2}])\] Since x is present on both sides, we can divide both sides by x: \[ 2(4.5 \times 10^{-3}) = [\mathrm{Cl}_{2}] \] Now, multiply to get the concentration of Cl2: \[ [\mathrm{Cl}_{2}] = 9 \times 10^{-3} \text{ mol/L}\] Thus, the concentration of \(\mathrm{Cl}_{2}\) under these conditions is \(9 \times 10^{-3}\) mol/L.

Key concepts

Equilibrium Constant

In the world of chemistry, the equilibrium constant, often symbolized as \(K\), is a crucial concept. It defines the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation, at a chemical equilibrium state.
For the reaction

• \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\)

the equilibrium constant expression would be:\[ K = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]} \]Here’s how it works:

• The concentrations of the products \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) appear in the numerator.
• The concentration of the reactant \(\mathrm{PCl}_{5}\) is in the denominator.
• Note that each concentration is raised to the power of 1, as per the stoichiometry of this particular reaction.

This constant gives us a snapshot of the balance between reactants and products at equilibrium but doesn’t indicate the speed of reaching equilibrium. A lower \(K\) value implies fewer products at equilibrium, while a higher \(K\) means more products.

Le Chatelier's Principle

Le Chatelier’s Principle tells us how a system at equilibrium responds to external changes. If a change is made in the conditions of a chemical equilibrium, the system will shift to counteract that change.
This principle helps us predict the direction in which a reaction will shift when:

• We alter concentrations of reactants or products.
• The temperature or pressure of the system changes.

For example, if we increase the concentration of \(\mathrm{PCl}_{5}\), the system will adjust to form more \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) to re-establish equilibrium. Conversely, removing some \(\mathrm{Cl}_{2}\) would encourage the production of more \(\mathrm{Cl}_{2}\) from \(\mathrm{PCl}_{5}\), shifting the reaction in the forward direction.
This adjustment occurs until a new equilibrium state is established. Le Chatelier’s Principle thus provides insight into reaction dynamics and helps us control product yields in industrial processes.

Reaction Quotient

The reaction quotient, represented as \(Q\), is similar to the equilibrium constant \(K\) but can be calculated at any point in the reaction, not just at equilibrium. The reaction quotient uses the same formula as \(K\):\[ Q = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]} \]Evaluating \(Q\) helps determine the state of a reaction:

• If \(Q = K\), the system is at equilibrium.
• If \(Q > K\), there are more products than at equilibrium, prompting a shift towards reactants.
• If \(Q < K\), there are fewer products than at equilibrium, and the reaction will move towards forming more products.

Knowing where \(Q\) stands relative to \(K\) is essential for understanding the progression of a reaction and predicting the changes involved to reach equilibrium. Such insights are valuable when adjusting reaction conditions to optimize yield in experimental and industrial settings.