Problem 79 - Chemistry The Science in Context

Question

The value of \(K_{\mathrm{p}}\) for the reaction $$ \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g) $$ is \(1.5 \times 10^{6}\) at \(25^{\circ} \mathrm{C} .\) At equilibrium, what is the ratio of \(P_{\mathrm{NO}_{2}}\) to \(P_{\mathrm{NO}}\) in air at \(25^{\circ} \mathrm{C} ?\) Assume that \(P_{\mathrm{O}_{2}}=0.21 \mathrm{atm}\) and does not change.

Step-by-Step Solution

Answer

Answer: The ratio of partial pressures $P_{\mathrm{NO_2}}/P_{\mathrm{NO}}$ at equilibrium is approximately $1.03 \times 10^{6}$.

Step 1: Write down the balanced chemical equation and the expression for Kp

Write down the balanced chemical equation for the given reaction: $$ \mathrm{NO}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g) $$ Write down the expression for the equilibrium constant, Kp, for this reaction: $$ K_p = \frac{P_{\mathrm{NO_2}}}{P_{\mathrm{NO}} \cdot \sqrt{P_{\mathrm{O_2}}}} $$ We are given \(K_p = 1.5 \times 10^6\), and \(P_{\mathrm{O_2}} = 0.21\,\mathrm{atm}\).

Step 2: Replace the given values and solve for the ratio of the partial pressures

Now, substitute the given values, \(K_p\) and \(P_{\mathrm{O_2}}\), into the expression for \(K_p\): $$ 1.5 \times 10^6 = \frac{P_{\mathrm{NO_2}}}{P_{\mathrm{NO}} \cdot \sqrt{0.21\,\mathrm{atm}}} $$ As we need to find the ratio \(\frac{P_{\mathrm{NO_2}}}{P_{\mathrm{NO}}}\), we can rearrange the equation as: $$ \frac{P_{\mathrm{NO_2}}}{P_{\mathrm{NO}}} = 1.5 \times 10^6 \cdot \sqrt{0.21\,\mathrm{atm}} $$

Step 3: Perform the calculation

Now, calculate the value of the ratio \(\frac{P_{\mathrm{NO_2}}}{P_{\mathrm{NO}}}\) using the given values: $$ \frac{P_{\mathrm{NO_2}}}{P_{\mathrm{NO}}} = 1.5 \times 10^{6} \cdot \sqrt{0.21\,\mathrm{atm}} $$ $$ \frac{P_{\mathrm{NO_2}}}{P_{\mathrm{NO}}} \approx 1.03 \times 10^{6} $$ So, the ratio of \(P_{\mathrm{NO_2}}\) to \(P_{\mathrm{NO}}\) in air at \(25^{\circ}\mathrm{C}\) at equilibrium is approximately \(1.03 \times 10^{6}\).

Key Concepts

Chemical Equilibrium

Imagine a busy city intersection where cars are constantly coming from all directions but manage to flow smoothly without ever causing a complete standstill. This dynamic balance is akin to chemical equilibrium, a state in which the rate of the forward chemical reaction is equal to the rate of the reverse reaction. At this point, there is no net change in the concentrations of reactants and products.

Take the reaction in the exercise, for instance, where nitrogen monoxide (NO) and oxygen (O₂) react to form nitrogen dioxide (NO₂). When this system reaches equilibrium, it means the creation of NO₂ from NO and O₂ happens at the same rate as NO₂ decomposing back into NO and O₂. However, it's crucial to understand that while the macroscopic properties remain constant, the microscopic processes of reaction and decomposition continue unabated.

Partial Pressure

Walking into a room filled with balloons of various colors, each balloon can be thought of as a different gas occupying part of the room's air. In this analogy, the space that each colored balloon occupies corresponds to the partial pressure of a specific gas in a mixture. Partial pressure, denoted as P, is the individual pressure exerted by a particular gas within a mixture of gases.

In the exercise, the focus is on the equilibrium state involving the gases NO, O₂, and NO₂ at a temperature of 25°C. Each of these gases exerts its partial pressure. For example, the partial pressure of oxygen (O₂), P_O2, has a critical role in the calculation of the equilibrium constant (K_p) and directly influences the ratio of the partial pressures of NO₂ to NO.

Reaction Quotient

Think of a football game, where the scoreboard at any given point gives us a snapshot of the game's progress. The reaction quotient (Q) serves a similar purpose in chemistry, giving us an instant 'score' of a reaction's progress towards equilibrium. It compares the concentrations (or partial pressures in case of gases) of the products and reactants at any point in time before the reaction reaches equilibrium.

Mathematically, Q has the same form as the equilibrium constant (K), but while K values are associated with a reaction at equilibrium, Q can be calculated at any point in time. If Q is less than K, the reaction will proceed forward, increasing the concentration of the products. Conversely, if Q is greater than K, the reaction will shift in the reverse direction, increasing the concentrations of the reactants, until Q equals K, and the system achieves equilibrium.

Le Chatelier's Principle

Imagine sitting on a seesaw, and trying to keep it perfectly balanced. If someone hops on one end, you'd need to adjust to regain balance. This is what Le Chatelier's Principle suggests about chemical reactions. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This principle helps predict how a change in concentration, temperature, or pressure will affect the system.

Referring back to our exercise, if the partial pressure of NO₂ were to increase, the principle indicates that the equilibrium would shift to reduce this pressure, favoring the reverse reaction. By understanding Le Chatelier's principle, it becomes easier to predict how changes in conditions like the fixed partial pressure of O₂ will affect the equilibrium ratio of NO₂ to NO and the overall system.