Question

Carbonylbromide, \(\mathrm{COBr}_{2}\), can be formed by combining carbon monoxide and bromine gas. $$ \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{COBr}_{2}(\mathrm{~g}) $$ When equilibrium is established at \(346 \mathrm{~K},\) the partial pressures (in atm) of \(\mathrm{COBr}_{2}, \mathrm{CO},\) and \(\mathrm{Br}_{2}\) are 0.12,1.00 , and \(0.65,\) respectively. (a) Calculate \(K_{\mathrm{p}}\) at \(346 \mathrm{~K}\). (b) Enough bromine condenses to decrease its partial pressure to 0.50 atm. Calculate the equilibrium partial pressures of all gases after equilibrium is re-established.

Short answer

(a) \( K_p = 0.185 \); (b) New pressures: CO=0.967 atm, Br2=0.467 atm, COBr2=0.153 atm.

Step-by-step solution

Write the Equilibrium Expression for Kp

The equilibrium expression for the reaction \( \mathrm{CO} + \mathrm{Br}_2 \rightleftharpoons \mathrm{COBr}_2 \) in terms of partial pressures is given by:\[ K_p = \frac{P_{\mathrm{COBr}_2}}{P_{\mathrm{CO}} \times P_{\mathrm{Br}_2}} \]

Plug in Values and Calculate Kp

Insert the given equilibrium partial pressures into the equilibrium expression:\[ K_p = \frac{0.12}{1.00 \times 0.65} \]Solve for \( K_p \):\[ K_p = \frac{0.12}{0.65} = 0.1846 \approx 0.185 \]

Establish New Equilibrium Conditions

When bromine partial pressure is reduced to 0.50 atm, changes will occur in the system to re-establish equilibrium. Let the change in pressure of \( \mathrm{CO} \) and \( \mathrm{Br}_2 \) be \( -x \) and \( \mathrm{COBr}_2 \) be \( +x \).

Set up the New Equilibrium Expression

Substitute the new conditions into the expression:\[ K_p = \frac{0.12 + x}{(1.00 - x)(0.50 - x)} \]The known value of \( K_p \) is 0.185. So:\[ 0.185 = \frac{0.12 + x}{(1.00 - x)(0.50 - x)} \]

Solve for x

Solve the equilibrium equation for \( x \). After some algebraic manipulations, this becomes:\[ 0.185 (1.00 - x)(0.50 - x) = 0.12 + x \]Expand and solve the quadratic equation. Assume \( x \) is small for simplification, leading to the pressure changes:\[ x = 0.033 \] (approximated by iteration or quadratic formula).

Calculate New Equilibrium Pressures

Substitute \( x \) back to find the new equilibrium pressures:- \( P_{\mathrm{CO}} = 1.00 - x = 1.00 - 0.033 = 0.967 \text{ atm} \)- \( P_{\mathrm{Br}_2} = 0.50 - x = 0.50 - 0.033 = 0.467 \text{ atm} \)- \( P_{\mathrm{COBr}_2} = 0.12 + x = 0.12 + 0.033 = 0.153 \text{ atm} \)

Key concepts

Equilibrium Constant (Kp)

In chemical equilibrium, the equilibrium constant, denoted as \( K_p \), is a crucial concept, especially when dealing with gases. This constant provides a snapshot of the ratio of product and reactant pressures at equilibrium. It is defined specifically in terms of partial pressures for reactions involving gases. When we consider the formation of carbonyl bromide \((\mathrm{COBr}_2)\) from carbon monoxide and bromine, the reaction reaches a state where the rate of the forward reaction equals that of the reverse reaction. This is equilibrium. At this point, the partial pressures become stable, and their relationship can be expressed using \( K_p \).

• The general form of \( K_p \) for a reaction \( aA + bB \rightleftharpoons cC + dD \) is:
  \[ K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \]
• For our specific example with CO, Br₂, and COBr₂, it simplifies to:
  \[ K_p = \frac{P_{\mathrm{COBr}_2}}{P_{\mathrm{CO}} \times P_{\mathrm{Br}_2}} \]
By calculating \( K_p \), we gain insights into the position of equilibrium and how product-favored or reactant-favored a system is under the given conditions.

Partial Pressure

Partial pressure is vital in the context of reactions involving gases. It is the pressure that a single gas component in a mixture would exert if it occupied the entire volume by itself. Understanding partial pressure helps in solving equilibrium problems
because it enables us to quantify each gas's contribution to the total pressure. For instance, when carbon monoxide and bromine gas react to form carbonyl bromide, the equilibrium state reflects the balancing of their partial pressures.

• The total pressure is a sum of all individual gas partial pressures:
  \( P_{\text{total}} = P_{\mathrm{CO}} + P_{\mathrm{Br}_2} + P_{\mathrm{COBr}_2} \).
• Changes in these pressures can affect how many products or reactants are present at equilibrium.

For example, if bromine condenses and its partial pressure decreases, as in the problem presented, the system will adjust its partial pressures to achieve equilibrium again. By maintaining this balance, partial pressures can therefore offer insight into the shifting dynamics of chemical equilibria.

Le Chatelier's Principle

Le Chatelier's Principle is a helpful guideline for predicting the behavior of a system at equilibrium when subjected to changes (like pressure, concentration, or temperature). This principle essentially states that an equilibrium system will adjust to counteract any imposed change, aiming to restore its state of balance.
In our scenario where gaseous bromine's partial pressure decreases, Le Chatelier's Principle helps explain the system's response. Since the partial pressure of Br₂ decreases:

• The equilibrium will shift to increase the production of Br₂ to compensate for its loss.
• Re-establishing equilibrium involves adjusting the concentrations or pressures of other reactants and products (i.e., CO and COBr₂, respectively).

The resulting shift will often manifest as a change in partial pressures among the reactants and products, as seen in the step-by-step solution provided.
Hence, Le Chatelier's Principle offers a predictive tool for understanding how equilibrium adapts in response to external changes, enabling us to anticipate such shifts in chemical reactions easily.