Question

At \(100^{\circ} \mathrm{C}, K_{\mathrm{p}}=60.6\) for the reaction $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ In a given experiment, 0.10 atm of each component is placed in a container. Is the system at equilibrium? If not, in which direction will the reaction proceed?

Short answer

The system is not at equilibrium and the reaction will proceed in the forward direction.

Step-by-step solution

Write the expression for the equilibrium constant

For the reaction \[ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Br}_{2}(g) \] the expression for the equilibrium constant in terms of partial pressures, \(K_p\), is given by: \[ K_p = \frac{(P_{\mathrm{NO}})^2 (P_{\mathrm{Br}_2})}{(P_{\mathrm{NOBr}})^2} \]

Substitute initial partial pressures into the reaction quotient \( Q_p \)

Given initially: \( P_{\mathrm{NOBr}} = 0.10 \) atm, \( P_{\mathrm{NO}} = 0.10 \) atm, and \( P_{\mathrm{Br}_2} = 0.10 \) atm. The reaction quotient \( Q_p \) is calculated as: \[ Q_p = \frac{(P_{\mathrm{NO}})^2 (P_{\mathrm{Br}_2})}{(P_{\mathrm{NOBr}})^2} = \frac{(0.10)^2 (0.10)}{(0.10)^2} = 0.10 \]

Compare \( Q_p \) with \( K_p \)

Compare the value of the reaction quotient \( Q_p \) with the equilibrium constant \( K_p \). In this case, \( Q_p = 0.10 \) and \( K_p = 60.6 \).

Determine the direction of the reaction

Since \( Q_p < K_p \), the reaction will proceed in the forward direction to form more products (\( \mathrm{NO} \) and \( \mathrm{Br}_2 \)) until equilibrium is reached.

Key concepts

reaction quotient

The reaction quotient, denoted as \( Q \), is a value that helps determine the direction in which a chemical reaction will proceed to reach equilibrium. It is calculated using the same expression as the equilibrium constant but with the initial concentrations or pressures of the reactants and products. Unlike the equilibrium constant \( K \), which only uses equilibrium concentrations, the reaction quotient uses the current state of the reaction.

In the given exercise, the reaction is:
2\( \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Br}_{2}(g) \)

The reaction quotient \( Q_p \) for this reaction in terms of partial pressures is:

\[ Q_p = \frac{(P_{\mathrm{NO}})^2 (P_{\mathrm{Br}_2})}{(P_{\mathrm{NOBr}})^2} \]

By plugging in the initial partial pressures given in the problem,
\( P_{\mathrm{NOBr}} = 0.10 \) atm, \( P_{\mathrm{NO}} = 0.10 \) atm, and \( P_{\mathrm{Br}_2} = 0.10 \) atm, we calculate:
\[ Q_p = \frac{(0.10)^2 (0.10)}{(0.10)^2} = 0.10 \]

Comparing \( Q_p \) with \( K_p \) tells us whether the reaction will proceed forward or reverse to achieve equilibrium.

equilibrium constant

The equilibrium constant, denoted as \( K \), is a fixed value that indicates the ratio of the concentrations or partial pressures of the products to the reactants when a chemical reaction is at equilibrium. For reactions involving gases, the equilibrium constant is written as \( K_p \) to reflect that it is expressed in terms of partial pressures.

In this case, the equilibrium constant for the reaction
\[ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Br}_{2}(g) \]
is given by the expression:
\[ K_p = \frac{(P_{\mathrm{NO}})^2 (P_{\mathrm{Br}_2})}{(P_{\mathrm{NOBr}})^2} \]

At \( 100^{\circ} \mathrm{C} \), the value of \( K_p \) is 60.6. The equilibrium constant is a fundamental indicator of the relative amounts of reactants and products at equilibrium:

• If \( K_p \) is much greater than 1, the reaction favors products at equilibrium.
• If \( K_p \) is much less than 1, the reaction favors reactants at equilibrium.
• If \( K_p \) is around 1, neither reactants nor products are strongly favored.

When comparing the reaction quotient \( Q_p \) to \( K_p \), we can determine how the system will adjust to reach equilibrium.

direction of reaction

Understanding the direction of the reaction involves comparing the reaction quotient \( Q_p \) with the equilibrium constant \( K_p \). This comparison indicates whether the system needs to produce more products, more reactants, or if it is already at equilibrium.

There are three possible outcomes:

• If \( Q_p < K_p \): The reaction will proceed in the forward direction, forming more products until equilibrium is reached. In our problem, \( Q_p = 0.10 \) and \( K_p = 60.6 \), so the reaction moves forward.
• If \( Q_p > K_p \): The reaction will proceed in the reverse direction, forming more reactants until equilibrium is achieved.
• If \( Q_p = K_p \): The reaction is already at equilibrium, and no net change will occur.

By comparing \( Q_p = 0.10 \) with \( K_p = 60.6 \), it's clear that since \( Q_p < K_p \), the reaction will move forward. This means more \( \mathrm{NO} \) and \( \mathrm{Br}_2 \) will form from \( \mathrm{NOBr} \) until the system reaches equilibrium.