Question

For which reaction high pressure and high temperature is helpful in obtaining a high cquilibrium yicld? (1) \(2 \mathrm{NF}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \quad 54.40 \mathrm{kcal}\) (2) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{II}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NI}_{3}(\mathrm{~g})+22.08 \mathrm{kcal}\) (3) \(\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClO}_{3}(\mathrm{~g}) \quad 49.4 \mathrm{kcal}\) (4) \(2 \mathrm{Cl}_{2} \mathrm{O}_{7}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Cl}_{2}(\mathrm{~g}) \times 7 \mathrm{O}_{2}(\mathrm{~g})+126.8 \mathrm{kcal}\)

Short answer

The reactions that are helped by high temperature and high pressure are: (1) and (4).

Step-by-step solution

Analyze the Reactions

Each reaction should be evaluated based on the changes in pressure and temperature that will favor a high equilibrium yield. Use Le Chatelier's principle, which states that the system will adjust to counteract changes, to guide this analysis.

Identify Endothermic and Exothermic Reactions

Determine whether the reaction is endothermic (absorbs heat) or exothermic (releases heat). Endothermic reactions are favored by high temperatures, while exothermic reactions are favored by low temperatures.

Reaction 1 Analysis

The first reaction: \[2 \text{NF}_{3}(\text{g}) \rightleftharpoons \text{N}_{2}(\text{g}) + 3 \text{F}_{2}(\text{g}) \text{ (Δ}H = 54.40 \text{ kcal, endothermic)}\] This reaction is endothermic and will be favored by high temperature. Analyzing the number of gas molecules, we see it goes from 2 to 4, so it will also be favored by high pressure due to an increase in the number of moles of gas on the product side.

Reaction 2 Analysis

The second reaction: \[\text{N}_{2}(\text{g}) + 3 \text{H}_{2}(\text{g}) \rightleftharpoons 2 \text{NH}_{3}(\text{g}) + 22.08 \text{ kcal}\] This reaction is exothermic, and thus favored by low temperature. The number of gas molecules goes from 4 to 2, favoring high pressure to shift equilibrium to the right.

Reaction 3 Analysis

The third reaction:\[\text{Cl}_{2}(\text{g}) + 2 \text{O}_{2}(\text{g}) \rightleftharpoons 2 \text{ClO}_{3}(\text{g}) \text{ (Δ}H = 49.4 \text{ kcal, endothermic)}\] This reaction is endothermic and favored by high temperature. It goes from 3 moles of gas to 2 moles, so low pressure would favor the forward reaction.

Reaction 4 Analysis

The fourth reaction: \[2 \text{Cl}_{2} \text{O}_{7}(\text{g}) \rightleftharpoons 2 \text{Cl}_{2}(\text{g}) + 7 \text{O}_{2}(\text{g})\] This reaction is endothermic and will be favored by high temperature. The gas molecules increase from 2 to 9, so high pressure will favor the forward reaction.

Conclusion

Based on the analysis, reactions that are favored by both high temperature and high pressure include reactions 1 and 4, as both conditions align with Le Chatelier's principle for those specific reactions.

Key concepts

Le Chatelier's Principle

One of the most essential concepts in chemical equilibrium is Le Chatelier's principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. In other words, the system tries to restore balance.
When a change like pressure, temperature, or concentration is applied to a reaction, the system adjusts itself to minimize the disturbance. For example:

• Increased pressure will shift the equilibrium toward the side with fewer gas molecules.
• Decreased pressure will shift the equilibrium toward the side with more gas molecules.
• Increased temperature will favor the endothermic reaction (absorbs heat).
• Decreased temperature will favor the exothermic reaction (releases heat).

Understanding Le Chatelier's principle helps in predicting the direction in which the equilibrium will shift when the reaction conditions are changed.

Endothermic and Exothermic Reactions

Chemical reactions can be classified as either endothermic or exothermic based on whether they absorb or release heat.
1. Endothermic Reactions: These reactions absorb heat from the surroundings. The system requires energy input to proceed. So, such reactions are favored by high temperatures.

• If the reaction becomes colder (temperature decreases), the system will try to produce more heat by shifting the equilibrium toward the reactants.
• Example: The decomposition of calcium carbonate \(\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)\) absorbs heat.

2. Exothermic Reactions: These reactions release heat into the surroundings. In this case, the system releases energy, and so lower temperatures favor them.

• If the reaction becomes warmer (temperature increases), the system will try to absorb the excess heat by shifting the equilibrium toward the reactants.
• Example: The synthesis of ammonia\(\text{N}_2(\text{g}) + 3 \text{H}_2(\text{g}) \rightarrow 2 \text{NH}_3(\text{g})\) releases heat.

Reaction 1 and Reaction 4 from the exercise are endothermic. They will be favored by high temperatures.

Reaction Pressure and Temperature

In chemical equilibrium, the effects of pressure and temperature are vital in determining the yield of products.
Pressure:

• Increasing pressure favors the side of the reaction with fewer gas molecules. It shifts equilibrium toward the side that takes up less volume.
• Decreasing pressure favors the side with more gas molecules.

Temperature:

• For endothermic reactions, high temperature shifts the equilibrium to the right, favoring products.
• For exothermic reactions, low temperature shifts the equilibrium to the right, favoring products.

Applying these principles:

• Reaction 1: High pressure and high temperature favor the products, as it is endothermic, spreading from 2 moles to 4 moles of gas.
• Reaction 4: Also endothermic and favored by high temperature and pressure, increasing from 2 to 9 moles of gas.

Thus, understanding the combination of pressure and temperature effects allows us to predict equilibrium yields accurately.