Problem 66
Question
Which of the following equilibria will shift toward formation of more products if the volume of a reaction mixture at equilibrium increases by a factor of \(2 ?\) a. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) b. \(\mathrm{NO}(g)+\mathrm{O}_{3}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) c. \(2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) d. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\)
Step-by-Step Solution
Verified Answer
Answer: Reactions c and d will shift towards the formation of more products.
1Step 1: Analyze the first reaction
Consider reaction a: \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\). Compare the moles of gaseous reactants and products:
- Reactants: \(2 \, \text{moles of} \, \mathrm{SO}_{2} + 1 \, \text{mole of} \, \mathrm{O}_{2} = 3 \, \text{moles}\)
- Products: \(2 \, \text{moles of} \, \mathrm{SO}_{3} = 2 \, \text{moles}\)
Here, the reactants have more moles of gaseous substances than the products.
2Step 2: Analyze the second reaction
Consider reaction b: \(\mathrm{NO}(g)+\mathrm{O}_{3}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\). Compare the moles of gaseous reactants and products:
- Reactants: \(1 \, \text{mole of} \, \mathrm{NO} + 1 \, \text{mole of} \, \mathrm{O}_{3} = 2 \, \text{moles}\)
- Products: \(1 \, \text{mole of} \, \mathrm{NO}_{2} + 1 \, \text{mole of} \, \mathrm{O}_{2} = 2 \, \text{moles}\)
Both sides have equal moles of gaseous substances.
3Step 3: Analyze the third reaction
Consider reaction c: \(2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\). Compare the moles of gaseous reactants and products:
- Reactants: \(2 \, \text{moles of} \, \mathrm{N}_{2}\mathrm{O}_{5} = 2 \, \text{moles}\)
- Products: \(4 \, \text{moles of} \, \mathrm{NO}_{2} + 1 \, \text{mole of} \, \mathrm{O}_{2} = 5 \, \text{moles}\)
Here, the products have more moles of gaseous substances than the reactants.
4Step 4: Analyze the fourth reaction
Consider reaction d: \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\). Compare the moles of gaseous reactants and products:
- Reactants: \(1 \, \text{mole of} \, \mathrm{N}_{2}\mathrm{O}_{4} = 1 \, \text{mole}\)
- Products: \(2 \, \text{moles of} \, \mathrm{NO}_{2} = 2 \, \text{moles}\)
Here, the products have more moles of gaseous substances than the reactants.
5Step 5: Determine the reaction with a shift towards more products
According to Le Châtelier's principle, the reaction will shift towards the direction with more moles of gaseous substances when increasing the volume. In this case, reactions c and d have more moles of gaseous products than reactants, so they will shift towards the formation of more products. Thus, the answer is c. \(2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) and d. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\).
Key Concepts
Chemical EquilibriumReaction QuotientGaseous Equilibrium Reactions
Chemical Equilibrium
Chemical equilibrium is the state of a chemical reaction where the concentrations of reactants and products remain constant over time. This occurs when the rate of the forward reaction, the conversion of reactants to products, equals the rate of the reverse reaction, the formation of reactants from products.
Understanding chemical equilibrium is crucial for predicting how a reaction will respond to changes in conditions. It forms the core of many industrial processes and can explain the behavior of gases in closed systems. An interesting aspect of equilibrium is that even though the net changes of reactants and products are zero, the individual molecules are constantly moving and reacting. It's like a bustling city where the number of people arriving is equal to the number of people leaving.
To visualize this, imagine a bottle of carbonated soda: Initially, carbon dioxide escapes from the liquid to the gas phase. Over time, the rate of escape slows down as pressure builds up. Eventually, the rate of carbon dioxide escaping the liquid equals the rate of it dissolving back into the soda, achieving equilibrium.
Understanding chemical equilibrium is crucial for predicting how a reaction will respond to changes in conditions. It forms the core of many industrial processes and can explain the behavior of gases in closed systems. An interesting aspect of equilibrium is that even though the net changes of reactants and products are zero, the individual molecules are constantly moving and reacting. It's like a bustling city where the number of people arriving is equal to the number of people leaving.
To visualize this, imagine a bottle of carbonated soda: Initially, carbon dioxide escapes from the liquid to the gas phase. Over time, the rate of escape slows down as pressure builds up. Eventually, the rate of carbon dioxide escaping the liquid equals the rate of it dissolving back into the soda, achieving equilibrium.
Reaction Quotient
The reaction quotient (Q) portrays the ratio of product concentrations to reactant concentrations at any point in time during a reaction. It is defined by the same expression as the equilibrium constant (K), but it's not limited to equilibrium states.
You can think of Q like a snapshot of a reaction at a specific moment. By comparing Q to the equilibrium constant K, we can predict the direction in which the reaction will proceed to reach equilibrium:
Analyzing the reaction quotient helps chemists adjust conditions to optimize reactions for different outcomes, such as increasing yields in industrial synthesis or controlling the direction of metabolic pathways in biological systems.
You can think of Q like a snapshot of a reaction at a specific moment. By comparing Q to the equilibrium constant K, we can predict the direction in which the reaction will proceed to reach equilibrium:
- If Q < K, the forward reaction is favored, and more products will form until equilibrium is reached.
- If Q > K, the reverse reaction is favored, and more reactants will form until equilibrium is reached.
- If Q = K, the system is already at equilibrium, and no net change will occur.
Analyzing the reaction quotient helps chemists adjust conditions to optimize reactions for different outcomes, such as increasing yields in industrial synthesis or controlling the direction of metabolic pathways in biological systems.
Gaseous Equilibrium Reactions
Gaseous equilibrium reactions involve substances in the gas phase and are especially sensitive to changes in pressure and volume due to the nature of gases. According to Le Châtelier's principle, if an external stress such as a change in volume or pressure is applied to such a system at equilibrium, the system will adjust to counteract that change.
When the volume of a reaction vessel increases, the pressure decreases, and the equilibrium will shift toward the side with more moles of gas to increase pressure. Conversely, if the volume decreases, the pressure increases, and the equilibrium will shift toward the side with fewer moles of gas to reduce pressure. This principle is not only essential for predicting how equilibria shift but also for designing processes where control over reaction conditions can lead to the desired product.
By understanding these nuances, students can grasp why certain reactions are designed with specific volume constraints in industrial applications and how altering these conditions affect the yields of the products involved.
When the volume of a reaction vessel increases, the pressure decreases, and the equilibrium will shift toward the side with more moles of gas to increase pressure. Conversely, if the volume decreases, the pressure increases, and the equilibrium will shift toward the side with fewer moles of gas to reduce pressure. This principle is not only essential for predicting how equilibria shift but also for designing processes where control over reaction conditions can lead to the desired product.
By understanding these nuances, students can grasp why certain reactions are designed with specific volume constraints in industrial applications and how altering these conditions affect the yields of the products involved.
Other exercises in this chapter
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