Problem 44 - Chemistry An Atoms First Approach

Question

The reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ has \(K_{\mathrm{p}}=109\) at \(25^{\circ} \mathrm{C}\). If the equilibrium partial pressure of \(\mathrm{Br}_{2}\) is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm, calculate the partial pressure of NO at equilibrium.

Step-by-Step Solution

Answer

The equilibrium partial pressure of NO is \(0.0584\: atm\).

Step 1: Write down the balanced chemical equation

The given chemical equation is: \(2 NO(g) + Br_2(g) \rightleftharpoons 2 NOBr(g)\)

Step 2: Write the K_p expression

Based on the balanced chemical equation, the K_p expression for the reaction at equilibrium can be written as: \(K_p = \frac{[NOBr]^2}{[NO]^2 \cdot [Br_2]}\)

Step 3: Substitute the given values of K_p, [Br2], and [NOBr] into the K_p expression

We are given: \(K_p = 109\) Equilibrium partial pressure of \(NOBr = 0.0768\: atm\) Equilibrium partial pressure of \(Br_2 = 0.0159\: atm\) Substituting these values into the K_p expression: \[109 = \frac{(0.0768)^2}{[NO]^2 \cdot (0.0159)}\]

Step 4: Solve for equilibrium partial pressure of NO

To calculate the equilibrium partial pressure of NO, we need to rearrange the equation: \[ [NO]^2 = \frac{(0.0768)^2}{(0.0159) \cdot 109}\] Now, calculate the value of \([NO]^2\): \[ [NO]^2 = \frac{0.00591}{1.733}\] \[ [NO]^2 = 0.00341\] Lastly, take the square root of the result to determine the equilibrium partial pressure of NO: \[ [NO] = \sqrt{0.00341}\] \[ [NO] = 0.0584\: atm\] Therefore, the equilibrium partial pressure of NO is \(0.0584\: atm\).

Key Concepts

Kp Expression

Understanding the equilibrium constant expressed in terms of pressure, commonly denoted as \( K_p \), is a fundamental element in grasping chemical equilibrium in gaseous systems. The \( K_p \) expression is derived directly from the balanced chemical equation.

For instance, given the reaction \( 2 NO(g) + Br_2(g) \rightleftharpoons 2 NOBr(g) \), the \( K_p \) expression is written as \( K_p = \frac{[NOBr]^2}{[NO]^2 \cdot [Br_2]} \). This representation encapsulates the stoichiometry of the reaction where the equilibrium pressures of the products and reactants are raised to the power of their coefficients in the balanced equation.

Moreover, the \( K_p \) expression does not include solids or liquids because their concentrations don't change under equilibrium conditions and hence have no effect on the pressure-based equilibrium constant.

Equilibrium Partial Pressure

The concept of equilibrium partial pressure is pivotal for understanding the behavior of gases in a reaction at equilibrium. In a mixture of gases, each gas contributes a pressure, called partial pressure, which is the pressure that the gas would exert if it alone occupied the entire volume.

When a reaction has reached equilibrium, the partial pressures of the gases involved remain constant. The equilibrium partial pressure is determined using the ideal gas law and is denoted as \( [Gas] \) in the \( K_p \) expression. Knowing the equilibrium partial pressures of some of the reactants or products allows us to calculate others, as demonstrated in the exercise where the equilibrium partial pressures of \( Br_2 \) and \( NOBr \) allowed for the determination of the partial pressure of \( NO \).

Le Chatelier's Principle

Le Chatelier's principle helps us understand how a system at equilibrium responds to changes in concentration, temperature, volume, or pressure. It states that if a system at equilibrium is subjected to a change, the system will adjust itself to partially oppose the effect of that change.

If the pressure is increased by reducing volume, the equilibrium shifts to favor the side with fewer moles of gas. Conversely, if pressure is decreased, the equilibrium favors the side with more moles. Temperature changes can shift the equilibrium depending on whether the reaction is exothermic or endothermic. Adding or removing a reactant or product will also shift the equilibrium to the side that counteracts the change.

Nevertheless, Le Chatelier's principle provides a qualitative insight and doesn't give the quantitative change, which is where \( K_p \) and reaction quotient play a critical role.

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a measure that tells us how close a system is to reaching equilibrium. Like the \( K_p \) expression, \( Q \) uses the actual partial pressures of the gases involved in the reaction at any given moment, not just at equilibrium.

When \( Q = K_p \), the system is at equilibrium. If \( Q < K_p \), the system will shift to the right, favoring the production of more products to reach equilibrium. Alternatively, if \( Q > K_p \), the system will shift to the left, favoring the reactants. By comparing \( Q \) to \( K_p \), one can predict which direction the system will move to achieve equilibrium.

This becomes exceptionally useful when dealing with changes in reaction conditions and analyzing how a reaction mixture will respond when it is not initially at equilibrium.