Problem 7 - Chemistry Matter and Change

Question

Challenge The reaction \(\mathrm{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) reaches equilibrium at 900 \(\mathrm{K}\). \(K_{\mathrm{eq}}\) is \(8.2 \times 10^{-2} .\) If the equilibrium concentrations of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) are \(0.150 \mathrm{M},\) what is the equilibrium concentration of \(\mathrm{COCl}_{2} ?\)

Step-by-Step Solution

Answer

The equilibrium concentration of \(\mathrm{COCl}_2\) is approximately 0.274 M.

Step 1: Write the Equilibrium Expression

For the given reaction, the equilibrium expression is: \[ K_{eq} = \frac{[CO][Cl_2]}{[COCl_2]} \] where \([CO]\), \([Cl_2]\), and \([COCl_2]\) are the concentrations of CO, Cl2, and COCl2 at equilibrium, respectively.

Step 2: Substitute Known Values

We know the equilibrium concentrations of CO and Cl2 are both 0.150 M. Also, \( K_{eq} = 8.2 \times 10^{-2} \). Substituting these values into the equilibrium expression gives: \[ 8.2 \times 10^{-2} = \frac{(0.150)(0.150)}{[COCl_2]} \] Simplifying the numerator: \( (0.150)(0.150) = 0.0225 \).

Step 3: Solve for [COCl2]

Now, solve for \([COCl_2]\) by rearranging the equation: \[ [COCl_2] = \frac{0.0225}{8.2 \times 10^{-2}} \]. Calculating this gives: \[ [COCl_2] = \frac{0.0225}{0.082} \approx 0.274 \]

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_{eq} \), is a numerical value that characterizes the ratio between the concentrations of products and reactants of a reversible chemical reaction at equilibrium. It provides valuable insights into the position of equilibrium. A higher \( K \) value indicates that products are favored at equilibrium, while a lower \( K \) suggests that reactants are favored. This is a fixed value under a specific set of conditions, particularly temperature.
For the given reaction: \[\mathrm{COCl}_{2} (g) \rightleftharpoons \mathrm{CO} (g) + \mathrm{Cl}_{2} (g)\]The equilibrium constant expression is:\[K_{eq} = \frac{[CO][Cl_2]}{[COCl_2]}\]- The square brackets indicate the concentrations of each species at equilibrium.- For this problem, \( K_{eq} = 8.2 \times 10^{-2} \) at 900 K, which tells us that at this temperature the reaction does not vastly favor products nor reactants, as it is a moderate value.Understanding \( K \) helps predict the direction in which a reaction needs to shift to reach equilibrium, especially when starting conditions are far from it.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of reactants and products that remain constant over time once a chemical reaction has reached equilibrium. In the case of our example reaction, we are focusing on the gases - CO- Cl₂- COCl₂ Each has an equilibrium concentration that balances out according to the equilibrium constant.In mathematical terms, these are represented as \([CO]\), \([Cl_2]\), and \([COCl_2]\). The key is that the concentration values adjust until they satisfy the equilibrium constant expression:\[K_{eq} = \frac{[CO][Cl_2]}{[COCl_2]}\]In our step-by-step solution, we were given the equilibrium concentrations of CO and Cl₂ as 0.150 M each and found that the concentration of COCl₂ at equilibrium was approximately 0.274 M by rearranging and solving the equilibrium expression.To find any missing concentration, substitute known values into the equilibrium expression and solve for the unknown. This method ensures that the calculated concentrations support the set equilibrium constant.

Reaction Quotient

The reaction quotient \( Q \) provides a snapshot of a reaction's status at any point in time by relating to the same expression used for \( K_{eq} \), but unlike \( K_{eq} \), it's not restricted to equilibrium conditions. Instead, it helps determine the direction in which a non-equilibrium system will proceed to achieve equilibrium.For the reaction:\[\mathrm{COCl}_{2} (g) \rightleftharpoons \mathrm{CO} (g) + \mathrm{Cl}_{2} (g)\]The expression for \( Q \) is:\[Q = \frac{[CO][Cl_2]}{[COCl_2]}\]By comparing \( Q \) to \( K_{eq} \):- If \( Q < K_{eq} \), the forward reaction is favored to form more products, moving towards equilibrium.- If \( Q > K_{eq} \), the reverse reaction is favored to form more reactants, moving towards equilibrium.- If \( Q = K_{eq} \), the reaction is at equilibrium.This concept is useful when predicting how changes in concentration affect the position of equilibrium, particularly in dynamic systems where concentrations are manipulated or altered.