Problem 96 - Chemistry An Atoms First Approach

Question

For the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g), K_{\mathrm{p}}=0.25\) at a certain temperature. If 0.040 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is reacted initially, calculate the equilibrium partial pressures of \(\mathrm{NO}_{2}(g)\) and \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\).

Step-by-Step Solution

Answer

The equilibrium partial pressures of \(\mathrm{N}_{2}\mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) are 0.02 atm and 0.04 atm, respectively.

Step 1: Write the balanced chemical equation

The balanced chemical equation given in the problem is \(\mathrm{N}_{2}\mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\).

Step 2: Set up Qp

The reaction quotient, Qp, is defined as the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients to the partial pressures of the reactants raised to their stoichiometric coefficients. In this case, Qp is \(\frac{(\mathrm{NO}_2)^2}{(\mathrm N_2O_4)}\).

Step 3: Find change in pressure at equilibrium

Let x represent the change in pressure at equilibrium. Then, the equilibrium pressure of N2O4 will be (0.040 - x) atm and the pressure of NO2 will be 2x atm.

Step 4: Set up the Kp equation

Given, \(K_{p} = 0.25\). We can set up the Kp equation using the equilibrium pressures of NO2 and N2O4: \[\frac{(2x)^2}{(0.040-x)}=0.25\]

Step 5: Solve for x

Solve the equation for x: \[(2x)^2 = 0.25(0.040-x)\] \[4x^2 = 0.01 - 0.25x\] Rearrange the equation into a quadratic: \[4x^2 + 0.25x - 0.01 = 0\] Now, use the quadratic formula to find the value of x: \[x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\] \[x = \frac{-0.25\pm\sqrt{(0.25)^2 - 4(4)(-0.01)}}{2(4)}\] Find two possible values of x: \[x_1 \approx 0.02 \, \text{atm}\] \[x_2 \approx -0.005 \, \text{atm}\] We can ignore x2 since it gives a negative pressure, which is not possible in this context.

Step 6: Calculate the equilibrium partial pressures

Now that we have the value of x, we can find the equilibrium partial pressures of N2O4 and NO2 as follows: \[\mathrm{N}_{2}\mathrm{O}_{4}: 0.040 - x \approx 0.040 - 0.02 = 0.02 \, \text{atm}\] \[\mathrm{NO}_{2}: 2x \approx 2(0.02) = 0.04 \, \text{atm}\] The equilibrium partial pressures of N2O4 and NO2 are 0.02 atm and 0.04 atm, respectively.

Key Concepts

Equilibrium Constant

In a chemical reaction, the equilibrium constant ( K_ { p} ) provides valuable information about the balance between reactants and products at equilibrium. It is calculated using the partial pressures of the gaseous substances involved in the reaction. For reactions involving gases, K_ { p} closely relates to the concept of the reaction quotient, Q_ { p} , which compares the measured product and reactant pressures. When these pressures do not change further, the system has reached equilibrium, making K_ { p} an essential parameter.

• For a balanced chemical equation, the equilibrium constant expression can be set up where the products are in the numerator and the reactants are in the denominator, each raised to the power of their stoichiometric coefficients.
• In the original exercise, K_ { p} was given as 0.25, meaning at equilibrium, products were present in certain amounts relative to reactants in the gaseous form.

Understanding K_ { p} allows us to predict the direction in which a reaction will shift if reactant or product concentrations are altered, aiming to restore equilibrium by adjusting these concentrations accordingly.

Partial Pressure

Partial pressure refers to the pressure exerted by an individual gas within a mixture of gases, contributing to the total pressure. When dealing with reactions involving gases, each gas has its partial pressure, which can considerably influence the reaction's equilibrium state.

• The concept of partial pressure is rooted in Dalton's Law of Partial Pressures, which states that the total pressure in a gas mixture is the sum of the partial pressures of each component gas.
• In the original problem, the partial pressure of N_ {2} O_ {4} at the start was 0.040 atm. Over time, as the reaction proceeded towards equilibrium, the partial pressures of N_ {2} O_ {4} and NO_ {2} changed, impacting the reaction quotient and ultimately establishing equilibrium.

Establishing the partial pressures at equilibrium helps us compute specific values such as K_ { p} , ensuring we understand the dynamic behavior of gaseous reactions.

Stoichiometry

Stoichiometry plays a crucial role in understanding chemical reactions because it relates to the quantitative relationship between reactants and products. It's like a recipe that lets you know how much of each ingredient (reactant) is needed to make a product.

• The coefficients in a balanced chemical equation show the molar ratios of reactants and products, guiding us in calculating the amounts consumed or produced during the reaction.
• In the example reaction N_ {2} O_ {4} ⇌ 2 NO_ {2}, for every mole of N_ {2} O_ {4} consumed, two moles of NO_ {2} are generated, illustrating a 1:2 stoichiometric ratio.

Understanding stoichiometry ensures accuracy in predicting the outcome of a reaction, allowing us to solve equilibrium problems and calculate changes in pressure like in the provided exercise.

Quadratic Equation

Using a quadratic equation is a common method when solving for unknown values in chemistry, especially in equilibrium problems that deal with changes in concentration or pressure. These equations are necessary when equilibrium expressions lead to polynomial equations with powers higher than one.

• In the exercise, the equation 4x^2 + 0.25x - 0.01 = 0 was derived from the equilibrium expression for K_ { p} by accounting for the changes in pressures due to the reaction's stoichiometry and initial conditions.
• The quadratic formula is employed to solve such equations: x = (-b±√(b^2 - 4ac))/(2a).

This formula allows us to compute possible x values (change in pressure) leading us to determine equilibrium partial pressures. Solving quadratic equations is essential, as in our problem, it results in finding realistic, applicable solutions for pressure in the context of gas equilibria.