Problem 25
Question
Dinitrogen trioxide decomposes to \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in an endothermic process \(\left(\Delta_{\mathrm{r}} H^{\circ}=40.5 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\right)\) $$ \mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) $$ Predict the effect of the following changes on the position of the equilibrium; that is, state which way the equilibrium will shift (left, right, or no change) when each of the following changes is made. (a) adding more \(\mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{g})\) (b) adding more \(\mathrm{NO}_{2}(\mathrm{g})\) (c) increasing the volume of the reaction flask (d) lowering the temperature
Step-by-Step Solution
Verified Answer
(a) Right; (b) Left; (c) Right; (d) Left.
1Step 1: Understanding the Reaction
The reaction given is an endothermic equilibrium: \( \mathrm{N}_2 \mathrm{O}_3(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g})+\mathrm{NO}_2(\mathrm{g}) \). Because it is endothermic, \( \Delta_{\mathrm{r}} H^{\circ} = 40.5 \text{ kJ/mol-rxn} \), heat is absorbed when the equilibrium shifts to the right.
2Step 2: Effect of Adding \(\mathrm{N}_2 \mathrm{O}_3\)
According to Le Chatelier's principle, adding more \( \mathrm{N}_2 \mathrm{O}_3(\mathrm{g}) \) increases the concentration of reactants, thus shifting the equilibrium to the right to produce more products \( \mathrm{NO}(\mathrm{g}) \) and \( \mathrm{NO}_2(\mathrm{g}) \).
3Step 3: Effect of Adding More \(\mathrm{NO}_2\)
Adding more \( \mathrm{NO}_2(\mathrm{g}) \) increases the concentration of products, causing the equilibrium to shift to the left to produce more \(\mathrm{N}_2 \mathrm{O}_3 \) to offset the added \(\mathrm{NO}_2\).
4Step 4: Effect of Increasing Volume
Increasing the volume of the reaction flask decreases the pressure. The equilibrium will shift towards the side with more moles of gas to increase the pressure, which is to the right (2 moles on the right vs 1 mole on the left).
5Step 5: Effect of Lowering Temperature
Lowering the temperature of an endothermic reaction shifts the equilibrium to the left since the system will release heat by forming more \( \mathrm{N}_2 \mathrm{O}_3 \) (reactant side), opposing the heat removal.
Key Concepts
Le Chatelier's PrincipleEndothermic ReactionsEquilibrium ShiftsReaction Dynamics
Le Chatelier's Principle
Le Chatelier's Principle helps predict the direction of an equilibrium shift when a change occurs in a system. Imagine a reaction at equilibrium as a teeter-totter, balanced in the middle. If something new is added or a condition is changed, the teeter-totter tips to reestablish balance. In simple terms, Le Chatelier's Principle states that if a change is imposed on a system at equilibrium, the system will adjust itself to counteract the effect of the change. It is essential for identifying how reactants and products behave when disturbed.
For example:
For example:
- Adding a reactant or product will shift the equilibrium in the opposite direction to restore balance.
- If heat is added to an exothermic reaction, the equilibrium will shift to absorb the extra heat.
- Changes in pressure and volume will cause shifts favoring the side with more or fewer gas moles.
Endothermic Reactions
Endothermic reactions play a significant role in chemical equilibrium. These reactions absorb heat from their surroundings. This means when a reaction shifts to the right (towards products) in an endothermic process, it requires energy in the form of heat.
Some characteristics of endothermic reactions include:
Some characteristics of endothermic reactions include:
- Absorbing heat, indicated by a positive \(\Delta_r H^{\circ}\).
- Reactions feel cold to the touch as they draw in heat.
- Increased temperature favors the formation of products in an endothermic reaction.
Equilibrium Shifts
Equilibrium shifts are the system's response to restore balance when a condition changes. When a disturbance occurs, the equilibrium position will move to reestablish equilibrium.
Consider these common causes of equilibrium shifts:
Consider these common causes of equilibrium shifts:
- Change in concentration: Adding or removing a reactant or product directly shifts the balance.
- Change in volume and pressure: Increasing the volume reduces the pressure, favoring the side with more gas molecules. Decreasing volume does the opposite, favoring the side with fewer gas molecules.
- Temperature changes: In endothermic reactions, increased temperature shifts the equilibrium toward the products.
Reaction Dynamics
Reaction dynamics delve into the complexities of how chemical reactions transition from reactants to products. In every chemical reaction, there is a dynamic process governing how molecules collide, react, and balance. Understanding these dynamics is crucial for predicting the behavior of reactions under various conditions.
In the context of the decomposition of dinitrogen trioxide:
In the context of the decomposition of dinitrogen trioxide:
- The given reaction is reversible, meaning it can go forward and backward, constantly shifting around the equilibrium position.
- The reaction's direction and speed are influenced by changes in conditions like temperature, pressure, and concentration.
- Insight into reaction dynamics aids in controlling reaction rates to favor desired products and optimize efficiency.
Other exercises in this chapter
Problem 23
Calculate \(K\) for the reaction $$ \operatorname{SnO}_{2}(\mathrm{s})+2 \mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Sn}(\mathrm{s})+2 \mathrm{CO}_{2}(\mat
View solution Problem 24
Calculate \(K\) for the reaction $$ \mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftarrows \mathrm{FeO}(\mathrm{s})+\mathrm{H}_{2}(\math
View solution Problem 26
\(K_{\mathrm{p}}\) for the following reaction is 0.16 at \(25^{\circ} \mathrm{C}\) $$ 2 \mathrm{NOBr}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}(\mathrm{g})+\ma
View solution Problem 27
Consider the isomerization of butane with an equilibrium constant of \(K=2.5 .\) (See Study Question 13.) The system is originally at equilibrium with [butane]
View solution