Question

The following reaction is at equilibrium: \(\mathrm{CO}+\mathrm{Cl}_{2} \rightleftarrows \mathrm{CoCl}_{2}\) The equilibrium [CO] and [Cl \(_{2}\) ] are \(0.088 \mathrm{M}\) and \(0.103 \mathrm{M}\), respectively. If the \(K_{\mathrm{eq}}\) is 0.225 , what is the equilibrium [COCl \(_{2}\) ]?

Short answer

The equilibrium [COCl"] is 0.00204 M.

Step-by-step solution

Write the Expression for Equilibrium Constant

The equilibrium constant expression for the reaction \( \mathrm{CO} + \mathrm{Cl}_2 \rightleftarrows \mathrm{COCl}_2 \) is given by: \[ K_{\mathrm{eq}} = \frac{[\mathrm{COCl}_2]}{[\mathrm{CO}][\mathrm{Cl}_2]} \] where \([\mathrm{COCl}_2]\), \([\mathrm{CO}]\), and \([\mathrm{Cl}_2]\) are the concentrations of the respective species at equilibrium.

Substitute Known Values into the Expression

Substitute the known values into the expression for the equilibrium constant. We know \(K_{\mathrm{eq}} = 0.225\), \([\mathrm{CO}] = 0.088 \mathrm{M}\), and \([\mathrm{Cl}_2] = 0.103 \mathrm{M}\). Let \(x\) be the equilibrium concentration of \([\mathrm{COCl}_2]\). The expression becomes: \[ 0.225 = \frac{x}{0.088 \times 0.103} \]

Solve for x

First, calculate \(0.088 \times 0.103 = 0.009064\). Substitute this back into the equation: \[ 0.225 = \frac{x}{0.009064} \]. Now solve for \(x\): \[ x = 0.225 \times 0.009064 \]. Calculating this gives \(x = 0.0020394 \mathrm{M}\).

Round Answer to Appropriate Significant Figures

Since the given concentrations \(0.088 \mathrm{M}\) and \(0.103 \mathrm{M}\) each have three significant figures, and the constant \(0.225\) also has three significant figures, we round \(x\) to three significant figures. Therefore, \(x = 0.00204 \mathrm{M}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a key concept in chemistry, describing the state where the concentrations of reactants and products remain constant over time in a closed system. This happens because the forward and reverse reactions occur at the same rate. In our exercise, the reaction between carbon monoxide (\(\mathrm{CO}\)) and chlorine (\(\mathrm{Cl}_2\)) to form \(\mathrm{COCl}_2\) exemplifies this concept.

Equilibrium does not mean that the amounts of reactants and products are equal, but rather that their concentrations remain steady. This is important because it allows the prediction of reaction outcomes and manipulates conditions to favor desired products in chemical manufacturing.

Understanding equilibrium helps in studying how changes in concentration, pressure, or temperature can shift the position of equilibrium according to Le Châtelier's Principle. In general:

• If more reactants are added, the system shifts to produce more products.
• Increasing the concentration of products shifts the equilibrium towards more reactants.
• Changing temperature or pressure can also alter equilibrium, depending on the characteristics of the reaction.

These shifts can be predicted and measured using equilibrium constants, providing a quantitative way to understand reaction dynamics at equilibrium.

Reaction Concentrations

In chemical reactions, the concentration of reactants and products is a crucial factor in determining the rate and outcome of the reaction. In our example, we analyze the equilibrium concentrations of \(\mathrm{CO}\), \(\mathrm{Cl}_2\), and \(\mathrm{COCl}_2\) to understand the state of the reaction. Concentrations are typically expressed in molarity (M), which measures the moles of solute per liter of solution.

The concentration values given in the exercise—\([\mathrm{CO}] = 0.088 \mathrm{M}\) and \([\mathrm{Cl}_2] = 0.103 \mathrm{M}\)—are key to calculating the equilibrium concentration of the product \(\mathrm{COCl}_2\). They provide the necessary framework for plugging into the equilibrium expression.

Reaction concentrations directly impact the equilibrium state and determine how much of each substance is present once the system reaches equilibrium. These concentrations are used to calculate the equilibrium constant (\(K_{\mathrm{eq}}\)), which indicates the product-to-reactant concentration ratio when the reaction is at equilibrium.

Knowing this offers insights into how strongly the reaction leans towards products or reactants, guiding predictions and adjustments for desired chemical outcomes.

Equilibrium Calculations

Equilibrium calculations are vital for predicting the concentrations of chemicals involved in a reaction at equilibrium. In our given exercise, these calculations help find the concentration of \(\mathrm{COCl}_2\) when equilibrium is established.

The equilibrium constant expression is set up based on the balanced chemical equation. For the reaction \(\mathrm{CO} + \mathrm{Cl}_2 \rightleftarrows \mathrm{COCl}_2\), it is expressed as:
\[ K_{\mathrm{eq}} = \frac{[\mathrm{COCl}_2]}{[\mathrm{CO}][\mathrm{Cl}_2]} \]
Substituting known values for \([\mathrm{CO}]\), \([\mathrm{Cl}_2]\), and \(K_{\mathrm{eq}}\), the equation reforms to solve for \(x\), representing the equilibrium concentration of \(\mathrm{COCl}_2\).

The mathematical steps involve:

• Calculating the product of the reactant concentrations.
• Rearranging the equation to solve for \(x\).
• Multiplying both sides by the computed product value.
• Carrying out the final multiplication to find \(x\), ensuring results reflect significant figures based on initial data.

Successful equilibrium calculations enable us to accurately determine and alter reaction conditions in laboratory and industrial settings, paving the way for effective chemical processes.