Problem 4
Question
In the reaction \(\mathrm{BaO}_{2}(\mathrm{~s}) \longrightarrow \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})\) \(\Delta \mathrm{H}=\) tve. In equilibrium condition, pressure of \(\mathrm{O}_{2}\) depends on (a) increase mass of \(\mathrm{BaO}_{2}\) (b) increase mass of \(\mathrm{BaO}\) (c) temperature of equilibrium (d) mass of \(\mathrm{BaO}_{2}\) and \(\mathrm{BaO}\) both
Step-by-Step Solution
Verified Answer
The pressure of \(\mathrm{O}_{2}\) depends on (c) temperature of equilibrium.
1Step 1: Analyze Reaction
The given reaction is \(\mathrm{BaO}_{2}(\mathrm{~s}) \rightarrow \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})\). \(\Delta \mathrm{H}=\) positive implies it is endothermic. In such reactions, increasing temperature shifts equilibrium towards more product formation.
2Step 2: Understand Le Chatelier's Principle
Using Le Chatelier's Principle, at equilibrium, a system will adjust to counteract changes. Since adding solid either \(\mathrm{BaO}_{2}\) or \(\mathrm{BaO}\) will not affect equilibrium pressure, only changes to gas or temperature typically do.
3Step 3: Evaluate Effect of Temperature
Since the reaction is endothermic, increasing temperature adds more heat. System will shift right, increasing \(\mathrm{O}_{2}\) production. Therefore, equilibrium pressure of \(\mathrm{O}_{2}\) increases with temperature.
4Step 4: Determine Effect of Mass
Changing the mass of pure solids \(\mathrm{BaO}_{2}\) or \(\mathrm{BaO}\) will not affect the equilibrium pressure of \(\mathrm{O}_{2}\) because they do not alter concentration or volume of available reactants or products in gaseous form.
Key Concepts
Le Chatelier's PrincipleEndothermic ReactionEffect of Temperature
Le Chatelier's Principle
Le Chatelier's Principle is a key concept in understanding chemical equilibrium. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore a new balance. Imagine your reaction is like a group of friends on a see-saw. If one friend moves, the rest adjust until it's balanced again.
If any change such as pressure, concentration, or temperature is applied to the system, the equilibrium will shift its position to reduce the effect of the change.
In the case of the reaction \( \mathrm{BaO}_{2}(\mathrm{~s}) \rightarrow \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \), the addition of solids like \( \mathrm{BaO}_{2} \) or \( \mathrm{BaO} \) will not disturb the equilibrium significantly. That's because solids don't influence the gaseous concentration as gases in equilibrium processes often do. Instead, attention should be paid to changes in gas or temperature to see the principle in full swing.
If any change such as pressure, concentration, or temperature is applied to the system, the equilibrium will shift its position to reduce the effect of the change.
In the case of the reaction \( \mathrm{BaO}_{2}(\mathrm{~s}) \rightarrow \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \), the addition of solids like \( \mathrm{BaO}_{2} \) or \( \mathrm{BaO} \) will not disturb the equilibrium significantly. That's because solids don't influence the gaseous concentration as gases in equilibrium processes often do. Instead, attention should be paid to changes in gas or temperature to see the principle in full swing.
Endothermic Reaction
An endothermic reaction is one that absorbs heat from its surroundings. The reaction \( \mathrm{BaO}_{2}(\mathrm{~s}) \rightarrow \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \) has a positive \( \Delta \mathrm{H} \), indicating it is endothermic.
This means it requires heat to proceed and favors higher temperatures to maintain product formation. Think of the reaction as a plant that needs sunlight (heat) to grow (produce \( \mathrm{O}_{2} \)).
So when the temperature increases, more heat is available for the reaction, thereby pushing the equilibrium towards the products. This happens because the system seeks to absorb the added heat by producing more \( \mathrm{O}_{2} \), which generally increases the concentration of the product in an endothermic reaction.
Understanding these reactions is vital for processes like photosynthesis, where plants use sunlight to create the conditions for equilibrium in nature.
This means it requires heat to proceed and favors higher temperatures to maintain product formation. Think of the reaction as a plant that needs sunlight (heat) to grow (produce \( \mathrm{O}_{2} \)).
So when the temperature increases, more heat is available for the reaction, thereby pushing the equilibrium towards the products. This happens because the system seeks to absorb the added heat by producing more \( \mathrm{O}_{2} \), which generally increases the concentration of the product in an endothermic reaction.
Understanding these reactions is vital for processes like photosynthesis, where plants use sunlight to create the conditions for equilibrium in nature.
Effect of Temperature
Temperature is a powerful lever for shifting chemical equilibrium, especially in endothermic reactions. In the given reaction, increasing the temperature provides more energy, causing the equilibrium to shift towards the products, in this case generating more \( \mathrm{O}_{2} \). Thus, an increase in temperature leads to higher pressure of oxygen gas at equilibrium.
Let's think of the reaction as a snowball rolling downhill. The warmer the slopes (temperature), the more energy the snowball has to roll further (product formation). Conversely, lowering the temperature in endothermic reactions will shift the equilibrium towards the reactants, decreasing the production of \( \mathrm{O}_{2} \).
Whenever you need to predict the effect of temperature changes on any equilibrium, start by identifying whether your reaction is endothermic or exothermic, then apply the concept of heat as a reactant or product. For endothermic reactions, increasing temperature is similar to adding more reactant, encouraging product formation.
Let's think of the reaction as a snowball rolling downhill. The warmer the slopes (temperature), the more energy the snowball has to roll further (product formation). Conversely, lowering the temperature in endothermic reactions will shift the equilibrium towards the reactants, decreasing the production of \( \mathrm{O}_{2} \).
Whenever you need to predict the effect of temperature changes on any equilibrium, start by identifying whether your reaction is endothermic or exothermic, then apply the concept of heat as a reactant or product. For endothermic reactions, increasing temperature is similar to adding more reactant, encouraging product formation.
Other exercises in this chapter
Problem 2
For a reversible reaction, if the concentrations of the reactants are doubled, at constant temperature the equilibrium constant will be (a) one-fourth (b) halve
View solution Problem 3
The value of \(K_{p}\) in the reaction \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(\mat
View solution Problem 6
If equilibrium constant for the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is \(\mathrm{K}_{e}\), then the equilibrium co
View solution Problem 8
For a reversible reaction, the concentration of the reactants are doubled, then the equilibrium constant (a) becomes one-fourth (b) is doubled (c) is halved (d)
View solution