Problem 7 - Chemistry and Chemical Reactivity

Question

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, \(\left[\mathrm{PC} 1_{5}\right]=\) \(4.2 \times 10^{-5} \mathrm{mol} / \mathrm{L},\left[\mathrm{PCl}_{3}\right]=1.3 \times 10^{-2} \mathrm{mol} / \mathrm{I},\) and \(\left[\mathrm{Cl}_{2}\right]=3.9 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Calculate \(K_{\mathrm{c}}\) for the reaction.

Step-by-Step Solution

Answer

The equilibrium constant \(K_{\mathrm{c}}\) is 1.21.

Step 1: Write the Expression for Equilibrium Constant

The equilibrium constant for the reaction \(\mathrm{PCl}_{5}(\mathrm{g}) \rightleftarrows \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) is given by \(K_{\mathrm{c}} = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]}\). This equation will be used to calculate \(K_{\mathrm{c}}\).

Step 2: Substitute the Concentration Values into the Expression

Substitute the equilibrium concentrations into the above expression: \([\mathrm{PCl}_{5}] = 4.2 \times 10^{-5} \; \mathrm{mol/L},\; [\mathrm{PCl}_{3}] = 1.3 \times 10^{-2} \; \mathrm{mol/L},\; [\mathrm{Cl}_{2}] = 3.9 \times 10^{-3} \; \mathrm{mol/L}\). The expression becomes: \( K_{\mathrm{c}} = \frac{(1.3 \times 10^{-2})(3.9 \times 10^{-3})}{4.2 \times 10^{-5}}\).

Step 3: Perform the Calculation

Calculate the value of \(K_{\mathrm{c}}\) by first computing the numerator: \((1.3 \times 10^{-2}) \times (3.9 \times 10^{-3}) = 5.07 \times 10^{-5}\). Then divide this by the concentration of \(\mathrm{PCl}_{5}\): \(\frac{5.07 \times 10^{-5}}{4.2 \times 10^{-5}} = 1.21\).

Step 4: Confirm and Report the Calculation

Make sure the calculations were performed correctly, verify each step for any potential errors, and then report the final result. The equilibrium constant \(K_{\mathrm{c}}\) for the reaction is 1.21.

Key Concepts

Chemical Equilibrium

In a chemical reaction, chemical equilibrium is reached when the rates of the forward and reverse reactions are equal. This means that the concentration of reactants and products remain constant over time. Chemical equilibrium doesn't mean that the reactants and products are present in equal amounts, but that their proportions stay unchanged as long as the system is undisturbed.
For example, consider the reaction: \[\mathrm{PCl}_{5} \rightleftarrows \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \]At equilibrium, the rate of formation of \(\mathrm{PCl}_{5}\) from \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) equals the rate at which \(\mathrm{PCl}_{5}\) decomposes. Thus, the amounts of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) stay the same, which is quantified using the equilibrium constant. The equilibrium constant \(K_c\) is a crucial factor here. It helps us predict the direction of the reaction and understand the ratio of products to reactants at equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle offers insight into the behavior of a reaction system when disturbances occur. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system responds by counteracting the disturbance to restore equilibrium. Here's how it can impact a reaction:

• Change in Concentration: If you add more reactants, say more \(\mathrm{PCl}_{5}\), the system will shift to the right, meaning more \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) will be produced until a new equilibrium is established.
• Change in Temperature: For the given reaction, increasing temperature might favor the endothermic direction, thus a shift towards more \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) if the reaction is endothermic.
• Change in Pressure: Since this reaction involves gases, an increase in pressure would shift the equilibrium towards the side with fewer moles of gas, affecting the balance of species involved.

Understanding these shifts is essential for predicting how reactions behave under different conditions, helping in industrial applications and chemical synthesis.

Equilibrium Concentration

Equilibrium concentration refers to the concentrations of the reactants and products in a chemical reaction at equilibrium. These concentrations can be calculated and are used to determine the equilibrium constant \(K_c\). By using the reaction\[\mathrm{PCl}_{5} \rightleftarrows \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \]we establish the following expression for \(K_c\):\[K_c = \frac{[\mathrm{PCl}_{3}] [\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]}\]To find \(K_c\), you insert the known equilibrium concentrations into this formula. This helps not only calculate \(K_c\) but also give insights into the reaction's dynamics. For example, if the concentrations of \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) are known to be higher than \(\mathrm{PCl}_{5}\), the reaction favors products. Knowing the equilibrium concentration also allows chemists to control and optimize reactions that are crucial in forming desired chemicals in various industrial processes.