Problem 36
Question
For the chemical reaction \(3 \mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{X}_{3} \mathrm{Y}(\mathrm{g})\), the amount of \(\mathrm{X}_{3} \mathrm{Y}\) at equilibrium is affected by (a) temperature and pressure (b) temperature only (c) pressure only (d) temperature, pressure and catalyst
Step-by-Step Solution
Verified Answer
(a) Temperature and pressure
1Step 1: Assess the Reaction Type
The reaction given is an equilibrium reaction, meaning it can proceed in both forward and reverse directions. Such reactions are typically affected by changes in temperature, pressure, and catalysts.
2Step 2: Consider Temperature Effects
For equilibrium reactions, changing the temperature can alter the position of equilibrium based on the enthalpy change of the reaction. If the reaction is exothermic, increasing temperature shifts equilibrium to the left (favoring reactants), and vice versa for endothermic reactions.
3Step 3: Consider Pressure Effects
In reactions involving gases, pressure changes shift equilibrium toward the side with fewer moles of gas, according to Le Chatelier's principle. In this case, the reaction goes from 4 moles (3 moles of X and 1 mole of Y) to 1 mole of X₃Y, so increasing pressure will favor the formation of X₃Y.
4Step 4: Consider Catalyst Effects
Catalysts speed up the rate at which equilibrium is reached, but they do not change the position of equilibrium or the amount of products and reactants at equilibrium. Thus, catalysts do not affect the amount of X₃Y at equilibrium.
Key Concepts
Le Chatelier's PrincipleTemperature Effects on EquilibriumPressure Effects on Equilibrium
Le Chatelier's Principle
Le Chatelier's principle is a fundamental concept in chemistry that helps us understand how a system at equilibrium responds to external changes. When a system at equilibrium experiences a change in concentration, temperature, or pressure, it adjusts itself to counteract the change and restore a new equilibrium.
For chemical reactions like the one given, where 3 moles of X react with one mole of Y to form X₃Y, Le Chatelier's principle allows us to predict how the equilibrium will shift.
For chemical reactions like the one given, where 3 moles of X react with one mole of Y to form X₃Y, Le Chatelier's principle allows us to predict how the equilibrium will shift.
- If there's an increase in the concentration of reactants (X or Y), the system will shift to the right, producing more X₃Y.
- A decrease in reactants will cause the equilibrium to shift to the left, reducing X₃Y production.
Temperature Effects on Equilibrium
Temperature changes can significantly affect equilibrium reactions, predominantly through the principle of endothermic and exothermic reactions. An exothermic reaction is one that releases heat, while an endothermic reaction absorbs heat.
In the given reaction, knowing whether it's endothermic or exothermic would allow us to predict the direction of the equilibrium shift:
In the given reaction, knowing whether it's endothermic or exothermic would allow us to predict the direction of the equilibrium shift:
- If the reaction is exothermic, increasing the temperature will favor the reverse reaction, shifting the equilibrium to the left, towards the reactants.
- Conversely, if it is endothermic, increasing temperature will favor the forward reaction, producing more X₃Y.
Pressure Effects on Equilibrium
Pressure changes primarily affect equilibrium reactions that involve gases, as it exploits the relationship between pressure and volume. Le Chatelier's principle informs us that when a system at equilibrium is subjected to an increase in pressure, the equilibrium will shift toward the side with fewer moles of gas.
In the given reaction, the reactants consist of 3 moles of X and one mole of Y—a total of 4 moles—while the product X₃Y consists of only 1 mole.
In the given reaction, the reactants consist of 3 moles of X and one mole of Y—a total of 4 moles—while the product X₃Y consists of only 1 mole.
- An increase in pressure will therefore shift the equilibrium towards the product side, favoring the formation of X₃Y because it has fewer moles of gas.
- Conversely, a decrease in pressure would shift the equilibrium towards the reactants, as they occupy a larger volume.
Other exercises in this chapter
Problem 33
\(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=\frac{\mathrm{Fe} / \mathrm{Mo}, 500^{\circ} \mathrm{C}}{=\mathrm{V} 200-900 \mathrm{~atm}} 2 \mathr
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Which of the following reaction will be favoured at low pressure? (a) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (b) \(\mathrm{H}_
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At constant temperature, the equilibrium constant \(\left(\mathrm{K}_{\mathrm{p}}\right)\) for the decomposition reaction, \(\mathrm{N}_{2} \mathrm{O}_{4} \long
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For the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) the forward reaction at constan
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