Problem 65 - Chemistry The Science in Context

Question

Which of the following equilibria will shift toward formation of more products if an equilibrium mixture is compressed into half its volume? a. \(2 \mathrm{N}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)\) b. \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)\) c. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) d. \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\)

Step-by-Step Solution

Answer

Answer: Choices (b) and (d).

Step 1: Calculate the change in moles of gas in each reaction

Find the difference between the number of moles of reactants and products in each reaction. We will use this information to determine which reaction will shift toward more product formation upon compression. Remember that increasing the pressure will shift the equilibrium towards the side with fewer moles of gas particles. a. \(2 \mathrm{N}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)\) change in moles of gas: \((2 + 1) - 2 = 1\) b. \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)\) change in moles of gas: \(2 - (2 + 1) = -1\) c. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) change in moles of gas: \(2 - (1 + 1) = 0\) d. \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) change in moles of gas: \(2 - (2 + 1) = -1\)

Step 2: Determine which reaction shifts toward more product formation upon compression

Upon compressing the equilibrium mixture (increasing the pressure), the system will shift towards the side with fewer moles of gas particles. We can use the change in moles of gas calculated in step 1 to determine which reaction favors product formation when compressed. a. Since there is an increase of 1 mole of gas, this reaction will shift towards the reactants upon compression. b. Since there is a decrease of 1 mole of gas, this reaction will shift towards the products upon compression. c. Since there is no change in the moles of gas, compression will not affect this reaction's equilibrium. d. Since there is a decrease of 1 mole of gas, this reaction will shift towards the products upon compression. Answer: Choices (b) and (d) will shift toward the formation of more products upon compression because they involve a decrease in moles of gas particles.

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a key concept in understanding how chemical equilibria respond to changes in conditions. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore balance. For gas reactions, when pressure increases by reducing the volume, the equilibrium will shift towards the side with fewer gas molecules. This is because a decrease in volume leads to an increase in pressure, and the system reacts to reduce the pressure. In exercise options "b" and "d," applying Le Chatelier's Principle explains why the reactions shift towards product formation. Both reactions have fewer moles of gas on the product side. Option "b" shows a decrease from 3 to 2 moles, and option "d" also decreases from 3 to 2. When compressed, these equilibria shift to the product side, reducing pressure by forming fewer gas molecules.

Gas Laws

Understanding gas laws helps explain the behavior of gases under different conditions, such as changes in pressure and volume. One fundamental idea is that pressure and volume are inversely related in a closed system of gases, according to Boyle's Law. Boyle's Law states that the pressure of a given mass of gas is inversely proportional to its volume when the temperature remains constant. Mathematically, it is expressed as:\[ P_1 V_1 = P_2 V_2 \]where \( P \) is the pressure and \( V \) is the volume. When the volume of a gas is halved, as in the exercise scenario, its pressure doubles, assuming the same temperature. This increase in pressure encourages the equilibrium to shift toward the side with fewer moles of gas. This principle is directly applied when interpreting which reactions will favor product formation upon the compression of the container.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. Understanding this concept helps predict how changes in conditions, like pressure and volume, affect reactions. In the practice question, you calculate moles of gas to determine how equilibrium shifts under pressure changes. For instance: - **Option (a):** 3 moles of products vs. 2 moles of reactants results in a shift towards reactants upon increased pressure. - **Option (b):** Total reactant moles (3) decrease to 2 product moles, favoring product formation when compressed. - **Option (c):** An equal number of moles for reactants and products means no shift in equilibrium due to pressure change. - **Option (d):** Similarly, opt for 2 moles of products from 3 moles of reactants shifts towards product side under compression. By balancing moles on each side of the equation, you determine which way the equilibrium will move. Understanding stoichiometry is key for predicting how various conditions affect chemical reactions in practice.