Problem 25
Question
Disturbing a Chemical Equilibrium Dinitrogen trioxide decomposes to NO and \(\mathrm{NO}_{2}\) in an endothermic process \((\Delta H=40.5 \mathrm{kJ} / \mathrm{mol})\) $$ \mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) $$ 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) No change, (d) Left.
1Step 1: Analyzing the Reaction
The chemical equation for the decomposition of dinitrogen trioxide is \( \mathrm{N}_2 \mathrm{O}_3(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) + \mathrm{NO}_2(\mathrm{g}) \). This is an endothermic reaction, meaning heat is absorbed during the process.
2Step 2: Applying Le Chatelier's Principle
Le Chatelier's Principle states that if a system at equilibrium is subjected to a change in concentration, pressure, volume, or temperature, the system will adjust to counteract that change and re-establish equilibrium.
3Step 3: Change (a) - Adding \( \mathrm{N}_2 \mathrm{O}_3(\mathrm{g}) \)
Increasing the concentration of a reactant, \( \mathrm{N}_2 \mathrm{O}_3(\mathrm{g}) \), will shift the equilibrium to the right to produce more products and restore equilibrium.
4Step 4: Change (b) - Adding \( \mathrm{NO}_2(\mathrm{g}) \)
Increasing the concentration of a product, \( \mathrm{NO}_2(\mathrm{g}) \), will shift the equilibrium to the left to produce more reactants and restore equilibrium.
5Step 5: Change (c) - Increasing the Volume of the Reaction Flask
Increasing the volume of the container will decrease the pressure. Since there is an equal number of moles of gas on both sides of the equation (1 mol \( \mathrm{N}_2 \mathrm{O}_3 \) and 1 mol of \( \mathrm{NO} + \mathrm{NO}_2 \)), the change in pressure will not affect the equilibrium position.
6Step 6: Change (d) - Lowering the Temperature
Lowering the temperature of an endothermic reaction, where the process absorbs heat, will shift the equilibrium to the left as the system compensates by producing more reactants to generate heat.
Key Concepts
Le Chatelier's PrincipleEndothermic ReactionEquilibrium ShiftReaction Flask VolumeTemperature 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 various changes. Imagine a balance scale that tips when you add weight to one side. Similarly, when a change is introduced to a chemical system at equilibrium, such as changing the concentration, pressure, volume, or temperature, the system will adjust itself to minimize that change and re-establish a balance, or equilibrium. This can lead to a shift in the position of the equilibrium either to the left (towards the reactants) or to the right (towards the products).
- If the concentration of a reactant or product is changed, the system will shift to oppose this change, favoring either the forward or reverse reaction.
- Adjustments in temperature or volume will also cause the system to shift, depending on whether the reaction absorbs heat or involves a change in gas volume.
Endothermic Reaction
An endothermic reaction is one where heat is absorbed from the surroundings. This type of reaction requires energy input to proceed, essentially using heat as a reactant. In the context of our exercise, the decomposition of dinitrogen trioxide (N\(_2\)O\(_3\)) into nitric oxide (NO) and nitrogen dioxide (NO\(_2\)) is endothermic, with an enthalpy change, \(\Delta H = 40.5 \, \text{kJ/mol}\), which reflects the energy required for the reaction.Every endothermic reaction benefits from additional heat:
- Increasing the temperature provides more energy, favoring the product side of the reaction.
- Conversely, decreasing the temperature removes energy from the system, pushing the equilibrium towards the reactants as the system tries to "generate" heat.
Equilibrium Shift
The equilibrium shift refers to the directional change the reacting system undergoes in response to an external modification, such as concentration, temperature, or pressure changes. This shift aims to counterbalance the imposed change and restore equilibrium.
- Addition of Reactants or Products: When extra reactants are added, the equilibrium shifts right towards products, whilst adding products shifts it left towards reactants.
- Volume and Pressure: Changes in volume can affect pressure. If pressure increases, the system shifts towards the side with fewer moles of gas. Conversely, if pressure decreases, it shifts towards the side with more moles.
Reaction Flask Volume
The volume of the reaction flask plays a significant role in determining how the equilibrium position adjusts when volume changes occur. For gaseous reactions, altering the volume alters the pressure inside the flask as well, owing to the relationship between pressure, volume, and moles of gas outlined by the ideal gas law (PV=nRT).
- Increasing the volume of a flask reduces the pressure, which could shift the equilibrium towards the side with more moles of gas, provided there is a different number of moles on each side. However, when the number of gas moles is equivalent on both sides of the equilibrium (as seen in this reaction), the equilibrium position remains largely unaffected by volume changes.
Temperature Effects on Equilibrium
Temperature is another crucial factor that can significantly influence the position of a chemical equilibrium, depending on whether the reaction is endothermic or exothermic. In our context of an endothermic reaction, temperature plays a pivotal role in determining the reaction direction:
- Raising the temperature inserts additional heat into the system, which shifts the equilibrium right, towards the product side. This happens because the system compensates by favoring the endothermic forward reaction, which absorbs the excess heat.
- Conversely, lowering the temperature pulls heat out of the system. This decreases the rate of the forward reaction and shifts the equilibrium to the left, towards the reactant side, because the system tries to generate heat to compensate for the loss.
Other exercises in this chapter
Problem 23
Calculate \(K\) for the reaction $$ \mathrm{SnO}_{2}(\mathrm{s})+2 \mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Sn}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}
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