Question

What is the equilibrium partial pressure of \(\mathrm{COBr}_{2}\) if the equilibrium partial pressures of \(\mathrm{CO}\) and \(\mathrm{Br}_{2}\) are 0.666 atm and 0.235 atm and the \(K_{\mathrm{P}}\) for this equilibrium is \(4.08 ?\) \(\mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{COBr}_{2}(\mathrm{~g})\)

Short answer

The equilibrium partial pressure of \( COBr_2 \) is 0.639 atm.

Step-by-step solution

Understand the Reaction and Equilibrium

The chemical reaction of interest is \( \mathrm{CO(g) + Br_2(g) \rightleftharpoons COBr_2(g)} \). Here, \( K_{\mathrm{P}} \) is the equilibrium constant expressed in terms of partial pressures for the reaction at a given temperature. We are provided with \( K_{\mathrm{P}} = 4.08 \), and the partial pressures of CO and \( \mathrm{Br}_2 \) at equilibrium.

Write the Expression for K_p

The expression for \( K_{\mathrm{P}} \) is given by the equation:\[ K_{\mathrm{P}} = \frac{{P_{\mathrm{COBr_2}}}}{{P_{\mathrm{CO}} \cdot P_{\mathrm{Br_2}}}} \]where \( P_{\mathrm{COBr_2}} \), \( P_{\mathrm{CO}} \), and \( P_{\mathrm{Br_2}} \) are the partial pressures of \( COBr_2 \), \( CO \), and \( Br_2 \), respectively.

Substitute Known Values into K_p Expression

Substitute the known equilibrium partial pressures of \( CO \) and \( \mathrm{Br}_2 \) in the \( K_{\mathrm{P}} \) expression:\[ 4.08 = \frac{{P_{\mathrm{COBr_2}}}}{{0.666 \times 0.235}} \]This reflects the problem's values for partial pressures of \( CO \) and \( \mathrm{Br}_2 \).

Solve for Equilibrium Partial Pressure of COBr2

Rearrange the equation to solve for \( P_{\mathrm{COBr_2}} \):\[ P_{\mathrm{COBr_2}} = 4.08 \times (0.666 \times 0.235) \]Calculate:\[ P_{\mathrm{COBr_2}} = 4.08 \times 0.15651 = 0.63853 \text{ atm} \]

Key concepts

Equilibrium Constant

The equilibrium constant, often represented as \( K \), is a numerical value that provides insight into the position of equilibrium in a chemical reaction. It tells us the ratio of product concentrations to reactant concentrations when a reaction is at equilibrium. In this particular exercise, we focus on \( K_{\mathrm{P}} \), which is the equilibrium constant in terms of partial pressures of gases involved in the reaction.

The equilibrium constant depends on the specific reaction and the temperature, but not on the initial concentrations of reactants and products. It remains constant for a given reaction at a specific temperature. A larger \( K \) value indicates a greater concentration of products at equilibrium, suggesting the reaction favors the products. Conversely, a smaller \( K \) value implies the reaction favors the reactants.

Understanding \( K_{\mathrm{P}} \) is crucial for predicting how a reaction will behave under different conditions and for calculating unknown quantities, like the partial pressures of substances involved in the reaction at equilibrium.

Partial Pressure

In gas phase reactions, partial pressure is an essential concept, particularly when dealing with equilibrium systems. It represents the pressure exerted by a single gas in a mixture of gases, as if it were alone in the container. Each gas in a mixture contributes to the total pressure in proportion to its mole fraction and the total pressure of the gas mixture.

For our reaction, the partial pressures of carbon monoxide (CO), bromine (Br₂), and carbonyl bromide (COBr₂) play a key role in determining the position of the equilibrium. Given the partial pressures for CO and Br₂, the task is to compute the partial pressure of COBr₂ using the equilibrium constant. These values are crucial because they appear in the equilibrium expression, providing the exact measure of how the pressure of each gas balances the others in the equilibrium state."

Knowing the partial pressures at equilibrium helps us solve for unknown values using mathematical expressions, ensuring we can predict the dynamics of the reaction.

Equilibrium Expression

An equilibrium expression is a mathematical representation of the equilibrium state of a chemical reaction. It allows us to compute various compounds' equilibrium concentrations or partial pressures based on the equilibrium constant. For the given reaction:

• \(\mathrm{CO}(\mathrm{g}) + \mathrm{Br}_2(\mathrm{g}) \rightleftharpoons \mathrm{COBr}_2(\mathrm{g})\)

The equilibrium expression for \( K_{\mathrm{P}} \) is given by:\[K_{\mathrm{P}} = \frac{{P_{\mathrm{COBr_2}}}}{{P_{\mathrm{CO}} \times P_{\mathrm{Br_2}}}}\]
This equation uses the partial pressures of the reactants and products, demonstrating how they influence the equilibrium state. By substituting known values into this expression, we can solve for unknowns, such as the equilibrium partial pressure of COBr₂.

The equilibrium expression is pivotal as it enables chemists to relate changes in conditions (such as pressure and temperature) to shifts in equilibrium, which can help predict how a reaction system might respond in different scenarios. This involves understanding that the products are on the numerator while the reactants are on the denominator in the fraction, illustrating their relative pressures in a balanced reaction.