Problem 35 - Chemistry The Molecular Science

Question

This reaction was examined at \(250^{\circ} \mathrm{C}\). $$ \begin{array}{c} \mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \\ \text { At equilibrium, }\left[\mathrm{PCl}_{5}\right]=4.2 \times 10^{-5} \mathrm{M},\left[\mathrm{PCl}_{3}\right]= \end{array} $$ \(1.3 \times 10^{-2} \mathrm{M},\) and \(\left[\mathrm{Cl}_{2}\right]=3.9 \times 10^{-3} \mathrm{M} .\) Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

Step-by-Step Solution

Answer

The equilibrium constant \( K_c \) is approximately 1.21.

Step 1: Write the Expression for Equilibrium Constant

The equilibrium constant expression for the given reaction \( \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \) can be written using the concentrations of the products divided by the concentration of the reactants: \[ K_c = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]} \]

Step 2: Substitute Given Concentrations into the Expression

Substitute the given equilibrium concentrations into the expression for \( K_c \):- \( [\mathrm{PCl}_{5}] = 4.2 \times 10^{-5} \; M \)- \( [\mathrm{PCl}_{3}] = 1.3 \times 10^{-2} \; M \)- \( [\mathrm{Cl}_{2}] = 3.9 \times 10^{-3} \; M \).Therefore,\[ K_c = \frac{(1.3 \times 10^{-2})(3.9 \times 10^{-3})}{4.2 \times 10^{-5}} \]

Step 3: Calculate the Numerator of the Expression

Multiply the concentrations of \( \mathrm{PCl}_{3} \) and \( \mathrm{Cl}_{2} \) to find the value of the numerator:\[ (1.3 \times 10^{-2})(3.9 \times 10^{-3}) = 5.07 \times 10^{-5} \]

Step 4: Divide by the Reactant Concentration

Divide the value found in Step 3 by the concentration of \( \mathrm{PCl}_{5} \):\[ \frac{5.07 \times 10^{-5}}{4.2 \times 10^{-5}} \approx 1.21 \]

Step 5: Conclusion

The calculated equilibrium constant \( K_c \) for the reaction at \(250^{\circ} \mathrm{C}\) is approximately \(1.21\).

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \) for reactions in solutions, is a fundamental concept in chemical equilibrium. It provides a numeric representation of the balance between reactants and products in a chemical reaction at equilibrium. At equilibrium, although the concentrations of reactants and products remain constant, they are not necessarily equal to each other. Their ratio, however, is constant at a given temperature.
This equilibrium constant is calculated based on the concentrations of products divided by the concentrations of reactants, with each concentration raised to the power of its stoichiometric coefficient. For the reaction considered here, the expression for \( K_c \) is:

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

This relationship highlights how the concentrations are interdependent at equilibrium. A high \( K_c \) value indicates that at equilibrium, the reaction mixture contains more products than reactants, suggesting that the reaction favors product formation. In this exercise, we calculated \( K_c \) to be approximately 1.21, which means the reaction slightly favors the formation of \( \mathrm{PCl}_{3} \) and \( \mathrm{Cl}_{2} \).

Le Chatelier's Principle

Le Chatelier's Principle is an important guideline that helps predict how a change in conditions can affect the position of equilibrium in a chemical reaction. This principle states that if an external change is applied to a system at equilibrium, the system will adjust itself to partially counteract the effect of that change.
Consider our reaction, \( \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \), and several possible changes:

• Temperature: If the external temperature increases, the equilibrium will shift in the direction that absorbs heat. This shift depends on whether the reaction is exothermic or endothermic.
• Pressure: Changing pressure affects gaseous equilibria. An increase in pressure will drive the equilibrium toward the side with fewer gas molecules.
• Concentration: Adding more \( \mathrm{PCl}_{5} \) would shift the equilibrium towards producing more \( \mathrm{PCl}_{3} \) and \( \mathrm{Cl}_{2} \) to re-establish equilibrium.

Understanding this principle helps in predicting how various alterations can be made to optimize reaction conditions and yields in real-world applications.

Dynamic Equilibrium

Dynamic equilibrium is a fascinating state where a chemical reaction continues to occur at the molecular level, but there is no overall change in the concentrations of reactants and products. This dynamic process means that the forward and reverse reactions are occurring at equal rates, keeping the system stable and balanced.
In our equilibrium involving \( \mathrm{PCl}_{5} \), the decomposition and combination of molecules are constant but equal, resulting in no net change. This balance is achieved after a period when the reaction starts, and conditions are established to enable both reactions to proceed simultaneously. The appearance of no change is misleading as molecules continue to transform; it's just that they're continuously offset by their reverse changes.
Moreover, reaching this state depends on factors like temperature and concentration. For instance, changes to these parameters will result in adjustments to the rates of forward and backward reactions until a new balance is achieved. Understanding dynamic equilibrium is crucial, as it underpins the stable yet responsive nature of chemical processes, crucial in industrial applications where continuous reaction conditions are often sought.