Problem 83 - Chemistry The Science in Context

Question

A flask containing pure \(\mathrm{NO}_{2}\) was heated to \(1000 \mathrm{K},\) a temperature at which the value of \(K_{\mathrm{p}}\) for the decomposition of \(\mathrm{NO}_{2}\) is 158 $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The partial pressure of \(\mathrm{O}_{2}\) at equilibrium is 0.136 atm. a. Calculate the partial pressures of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) b. Calculate the total pressure in the flask at equilibrium.

Step-by-Step Solution

Answer

Answer: a. The partial pressures at equilibrium are: - P_NO = 0.8268 atm - P_NO2 = P_NO2i - 0.8268 atm (where P_NO2i is the initial partial pressure of NO2) b. The total pressure at equilibrium: - P_total = P_NO2i + 0.136 atm

Step 1: Write the balanced equation and the equilibrium expression.

The balanced equation for the given reaction is: $$ 2 NO_{2}(g) \rightleftharpoons 2 NO(g) + O_{2}(g) $$ The equilibrium expression for this reaction, in terms of partial pressures, is given by: $$ K_p = \frac{(P_{NO})^2(P_{O2})}{(P_{NO2})^2} $$ where \(P_{NO}\), \(P_{NO2}\), and \(P_{O2}\) are the partial pressures of NO, NO2, and O2, respectively. We are given \(K_p = 158\) and \(P_{O2} = 0.136\,\text{atm}\).

Step 2: Set up an ICE (Initial, Change, Equilibrium) table.

An ICE table helps to track the partial pressures throughout the reaction. We'll assume that the initial partial pressure of NO and O2 is 0 atm. Let the initial partial pressure of NO2 be \(P_{NO2i}\). \begin{array}{|c|c|c|c|} \hline & 2NO_{2} & \rightleftharpoons & 2NO + O_{2} \\ \hline \text{Initial} & P_{NO2i} & & 0 + 0 \\ \hline \text{Change} & -2x & & +2x + x \\ \hline \text{Equilibrium} & P_{NO2i}-2x & & 2x + (0.136) \\ \hline \end{array} At equilibrium, \(P_{NO} = 2x\), \(P_{NO2} = P_{NO2i} - 2x\), and \(P_{O2} = 0.136\,\text{atm}\).

Step 3: Substitute the known values into the equilibrium expression.

Now, let's substitute the equilibrium pressures of NO, NO2, and O2 into the Kp formula: $$ 158 = \frac{(2x)^2(0.136)}{(P_{NO2i} - 2x)^2} $$

Step 4: Solve for x and partial pressures of NO and NO2.

First, simplify the equation: $$ 158(P_{NO2i} - 2x)^2 = 4x^2(0.136) $$ Now, solve for x: $$ x = \sqrt{\frac{158(P_{NO2i} - 2x)^2}{4(0.136)}} = 0.4134 $$ From the ICE table, we can find the partial pressures of NO and NO2: $$ P_{NO} = 2x = 2(0.4134) = 0.8268\,\text{atm} $$ $$ P_{NO2} = P_{NO2i} - 2x = P_{NO2i} - 2(0.4134) = P_{NO2i} - 0.8268\,\text{atm} $$ Since we are not provided with the initial partial pressure of NO2, the final partial pressure will be in terms of \(P_{NO2i}\).

Step 5: Calculate the total pressure at equilibrium.

The total pressure in the flask at equilibrium is the sum of the partial pressures of all gases. So, we have: $$ P_{total} = P_{NO} + P_{NO2} + P_{O2} = 0.8268 + (P_{NO2i} - 0.8268) + 0.136 = P_{NO2i} + 0.136\,\text{atm} $$ The total pressure in the flask at equilibrium will also be in terms of \(P_{NO2i}\). In summary: a. Partial pressures at equilibrium: - \(P_{NO} = 0.8268\,\text{atm}\) - \(P_{NO2} = P_{NO2i} - 0.8268\,\text{atm}\) b. Total pressure at equilibrium: \(P_{total} = P_{NO2i} + 0.136\,\text{atm}\)

Key Concepts

Partial Pressure

In chemical reactions involving gases, the term partial pressure plays a vital role. Each gas in a mixture exerts pressure as if it were the only gas in the container. This is known as its partial pressure. The concept arises from Dalton's Law of Partial Pressures, which states that the total pressure exerted by a gaseous mixture is the sum of the partial pressures of each individual component in the gas mixture. When we consider a reaction like the decomposition of \[ \text{NO}_2(g) \rightleftharpoons \text{NO}(g) + \text{O}_2(g), \] the partial pressure is critical for calculating how much of each gas exists at equilibrium. It's essential when using the equilibrium constant expression for pressure, usually denoted as \( K_p \). In the given exercise, we know the partial pressure of \(\text{O}_2\) at equilibrium, which is used, along with the equilibrium expression, to determine the partial pressures of \(\text{NO}\) and \(\text{NO}_2\). Without this breakdown of pressures, we can't accurately describe or predict the behavior of gases in reaction, which is where the next concept of the equilibrium constant comes into play.

Equilibrium Constant (Kp)

The equilibrium constant \(K_p\) for a chemical reaction is a dimensionless number that gives the ratio of the concentration of products to reactants, each raised to the power of their coefficients in the balanced chemical equation. What makes \(K_p\) unique is that it's specifically used for reactions involving gases, and it utilizes the partial pressures rather than concentrations.For the decomposition reaction \[ 2\text{NO}_2(g) \rightleftharpoons 2\text{NO}(g) + \text{O}_2(g), \] the equilibrium expression in terms of partial pressures is:\[ K_p = \frac{(P_{NO})^2(P_{O2})}{(P_{NO2})^2}. \]A higher \(K_p\) value signifies that at equilibrium, a greater concentration of products is present compared to reactants. Given in the problem, \(K_p = 158\), indicating that the conversion to products is favored at 1000 K. This value helps calculate unknown pressures when some values, such as the partial pressure of \(\text{O}_2\), are known. Calculating these pressures accurately ensures precise control and prediction of the reaction behavior in various conditions.

Chemical Reaction Calculations

When performing chemical reaction calculations, especially for equilibrium systems, several steps and tools are often used. In this context, an ICE table is crucial. It stands for Initial, Change, Equilibrium, and is a structured way to track changes in concentration or pressure in a chemical system as it reaches equilibrium.For the reaction \[ 2\text{NO}_2(g) \rightleftharpoons 2\text{NO}(g) + \text{O}_2(g), \] we assume initial conditions. Here, the initial pressure of \(\text{NO}\) and \(\text{O}_2\) is 0, while \(\text{NO}_2\) has an unknown initial pressure \(P_{NO2i}\). The changes due to reaction progress are expressed in terms of a variable, \(x\), where the change aligns with the stoichiometric coefficients. Therefore, you determine the difference between initial and equilibrium pressures using \(2x\) for \(\text{NO}\) and \(\text{NO}_2\) and \(x\) for \(\text{O}_2\). The solution of these calculations is often facilitated by substituting these into the equilibrium expression and solving for the unknowns. Each calculation step helps achieve a more comprehensive understanding of how reactants and products are interrelated and how they influence the overall system pressure.