Problem 74 - Chemistry

Question

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

Step-by-Step Solution

Answer

a. The equilibrium position will shift to the left. b. The equilibrium position will shift to the right. c. There will be no shift in the equilibrium position. d. The equilibrium position will shift to the right. e. The equilibrium position will shift to the right.

Step 1: Reaction a: Analyzing the equilibrium condition

For the equilibrium reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\), notice that there are 4 moles of gas on the left-hand side (1 mole of \(\mathrm{N}_{2}\) and 3 moles of \(\mathrm{H}_{2}\)) and 2 moles of \(\mathrm{NH}_{3}\) gas on the right-hand side, for a total of 4 moles on the left and 2 moles on the right.

Step 2: Reaction a: Predicting the shift

When the volume of the reaction container is increased, the pressure decreases. According to Le Chatelier's principle, the equilibrium will shift in the direction that counteracts this pressure change. In this case, the reaction will shift in the direction that has more moles of gas, which is the left-hand side (the reactant side). Therefore, the equilibrium position will shift to the left.

Step 3: Reaction b: Analyzing the equilibrium condition

For the equilibrium reaction \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\), there is 1 mole of \(\mathrm{PCl}_{5}\) gas on the left-hand side and 1 mole of \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) each on the right-hand side (2 moles in total).

Step 4: Reaction b: Predicting the shift

Upon increasing the volume of the reaction container and decreasing the pressure, the equilibrium will shift in the direction with more moles of gas molecules, which is the right-hand side (the product side) in this case. Therefore, the equilibrium position will shift to the right.

Step 5: Reaction c: Analyzing the equilibrium condition

For the equilibrium reaction \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\), there are 2 moles of gas on both the left-hand and right-hand sides (1 mole each of \(\mathrm{H}_{2}\) and \(\mathrm{F}_{2}\) on the left, and 2 moles of \(\mathrm{HF}\) on the right).

Step 6: Reaction c: Predicting the shift

Since there are equal moles of gas molecules on both sides of the equilibrium, there will be no shift in the equilibrium position when the volume of the reaction container is increased.

Step 7: Reaction d: Analyzing the equilibrium condition

For the equilibrium reaction \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\), there is 1 mole of \(\mathrm{COCl}_{2}\) gas on the left-hand side and 1 mole of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) each on the right-hand side (2 moles in total).

Step 8: Reaction d: Predicting the shift

Upon increasing the volume of the reaction container and decreasing the pressure, the equilibrium will shift to the side with more moles of gas molecules, which is the right-hand side (the product side), in this case. Therefore, the equilibrium position will shift to the right.

Step 9: Reaction e: Analyzing the equilibrium condition

For the equilibrium reaction \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\), there are no moles of gas on the left-hand side and 1 mole of \(\mathrm{CO}_{2}\) gas on the right-hand side.

Step 10: Reaction e: Predicting the shift

When the volume of the reaction container is increased, the pressure decreases. The equilibrium will shift to the side with more moles of gas molecules, which is the right-hand side (the product side) in this case. Therefore, the equilibrium position will shift to the right.

Key Concepts

Equilibrium Shift

Le Chatelier's principle explains how chemical equilibria respond to changes in external conditions, such as concentration, temperature, or pressure. When a change happens, the equilibrium will shift to try to restore balance. For example, if the volume of a gas reaction container decreases, the pressure increases, and the system will adjust by shifting in the direction that has fewer gas molecules, to lower the pressure.
In contrast, if the volume increases, leading to decreased pressure, the equilibrium will prefer the side with more gas molecules in an attempt to increase the pressure. This reactive adjustment helps in recognizing and predicting the direction of the equilibrium shift in response to any disturbances.

Reaction Container Volume

The volume of the reaction container plays a crucial role in determining the equilibrium position of gas-phase reactions. When the reaction container is expanded, the pressure of the gases inside decreases.
Since the system seeks to counteract the reduction in pressure, it favors the side of the reaction with more moles of gas. For reactions with the same number of moles on both sides, an increase in volume has no effect. Conversely, decreasing the volume results in increased pressure and the system will shift towards the side with fewer moles of gas to compensate.

• Increasing volume: shifts equilibrium towards more moles of gas.
• Decreasing volume: shifts equilibrium towards fewer moles of gas.

Mole Comparison

Understanding the number of moles of gases on each side of a chemical equation is essential for predicting how equilibrium might shift. When comparing the moles, think of each mole as a contributor to the overall pressure within the system.
In reactions like a. o. \[ N_{2}(g) + 3 \, H_{2}(g) \rightleftharpoons 2 \, NH_{3}(g) \]The left side has 4 moles compared to the right side's 2 moles. Therefore, if volume increases, equilibrium shifts to the left. Meanwhile, reactions with equal moles on both sides, such as Reaction c: \[ H_{2}(g) + F_{2}(g) \rightleftharpoons 2 \, HF(g) \]will remain unchanged when volume changes. Recognizing this can make predicting equilibrium shifts more intuitive.
Looking at moles helps in not just predicting shifts, but also understanding chemical kinetics and behavior under different conditions.

Gas Pressure Change

The pressure of a gas is directly related to both the volume of the container it is in and the amount of gaseous molecules present. If a container's volume is increased, gas molecules have more space and thus the overall gas pressure decreases.
Under Le Chatelier's principle, when pressure changes, the system attempts to minimize this change. Increasing the volume, which decreases pressure, will make the equilibrium shift toward producing more gas molecules to re-increase pressure.

• More gas molecules = higher pressure.
• Less volume = higher pressure, favoring side with fewer molecules.
• More volume = lower pressure, favoring side with more molecules.

These correlations can be crucial for controlling reactions, especially in industrial settings where the balance of pressure and volume affects yield and efficiency.