Problem 11 - Chemistry Matter and Change

Question

Calculate Determine the value of \(K_{\mathrm{eq}}\) at 400 \(\mathrm{K}\) for this equation: \(\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) if \(\left[\mathrm{PCl}_{5}\right]=0.135 \mathrm{mol} / \mathrm{L},\left[\mathrm{PCl}_{3}\right]=0.550 \mathrm{mol} / \mathrm{L}\) , and \(\left[\mathrm{Cl}_{2}\right]=0.550 \mathrm{mol} / \mathrm{L} .\)

Step-by-Step Solution

Answer

The equilibrium constant \(K_{eq}\) for the given reaction at 400 K is 2.24.

Step 1: Identify the reactants and products

The given chemical reaction is: \[ \mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \] Reactants: \(\mathrm{PCl}_{5}\) Products: \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\)

Step 2: Write the expression for the equilibrium constant

We can write the expression for the equilibrium constant as: \[ K_{eq} = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]} \]

Step 3: Substitute given concentrations into the expression

Now, we can plug in the given concentrations: \[ K_{eq} = \frac{(0.550\,\mathrm{mol}\,/\,\mathrm{L})(0.550\,\mathrm{mol}\,/\,\mathrm{L})}{(0.135\,\mathrm{mol}\,/\,\mathrm{L})} \]

Step 4: Calculate the value of \(K_{eq}\)

Finally, perform the calculation: \[ K_{eq} = \frac{(0.550\,\mathrm{mol}\,/\,\mathrm{L})(0.550\,\mathrm{mol}\,/\,\mathrm{L})}{(0.135\,\mathrm{mol}\,/\,\mathrm{L})} = 2.24074\dots \] Rounding to three significant figures, we get: \[ K_{eq} = 2.24 \] Therefore, the equilibrium constant \(K_{eq}\) for the given reaction at 400 K is 2.24.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating state that occurs in reversible chemical reactions. It happens when the rates of the forward and backward reactions are equal. At this point, the concentrations of the reactants and products remain constant over time, not because the reactions stop, but because they happen simultaneously at the same rate. Think of it like a balanced seesaw; both sides continue to move, but the overall position stays level. In the context of the reaction \[\mathrm{PCl}_5(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})\], the system reaches equilibrium when the decomposition of \(\mathrm{PCl}_5\) into \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) equals the rate at which \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) recombine to form \(\mathrm{PCl}_5\). Understanding chemical equilibrium helps us predict how a change in conditions, such as temperature or pressure, could shift the balance of the reaction in one direction or another.

Reaction Quotient

The reaction quotient \(Q\) is a crucial concept that is similar to the equilibrium constant \(K_{eq}\), but with a key difference: \(Q\) is calculated using the current concentrations of the reactants and products, not just at equilibrium. By comparing \(Q\) to \(K_{eq}\), we can determine which direction a reaction will shift to reach equilibrium:

• If \(Q < K_{eq}\), the reaction will proceed forward to produce more products.
• If \(Q > K_{eq}\), the reaction will proceed in reverse to produce more reactants.
• If \(Q = K_{eq}\), the system is already at equilibrium.

For the reaction \(\mathrm{PCl}_5(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})\), calculating \(Q\) using the provided concentrations can indicate whether the system is already at equilibrium or how it may need to adjust to reach it.

Concentration Calculation

Concentration calculations are fundamental in determining how much of a substance is present in a given volume. This is critical for calculating the equilibrium constant \(K_{eq}\). To calculate \(K_{eq}\) for the given reaction, the concentrations of each substance in moles per liter \((\mathrm{mol}/\mathrm{L})\) are plugged into the equilibrium expression:\[K_{eq} = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]}\]Plugging in the values from the exercise:\[K_{eq} = \frac{(0.550\,\mathrm{mol}/\mathrm{L})(0.550\,\mathrm{mol}/\mathrm{L})}{(0.135\,\mathrm{mol}/\mathrm{L})}\]Executing this calculation gives the equilibrium constant \(K_{eq} = 2.24\), once rounded to three significant figures. Accurate concentration calculations ensure precise \(K_{eq}\) values, vital for predicting the extent of reactions under various conditions.

Chemical Reaction

A chemical reaction involves a process in which substances, called reactants, transform into different substances, known as products. Reactions can be classified as either irreversible or reversible. In the context of chemical equilibrium, reversible reactions are of prime interest. They can proceed in both directions: forward and backward.The given reversible reaction\[\mathrm{PCl}_5(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})\]shows \(\mathrm{PCl}_5\) breaking down into \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\). Understanding chemical reactions is essential to grasp how concentration changes and temperature shifts can impact the balance between reactants and products. This balance determines the value of \(K_{eq}\), reflecting the conditions of equilibrium for a specific reaction. Recognizing the dynamic nature of chemical reactions helps in predicting how equilibrium can be influenced by various factors.