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Dinitrogen trioxide decomposes to \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in an endothermic process \(\left(\Delta_{r} H^{\circ}=40.5 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\right)\) $$ \mathrm{N}, \mathrm{O}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) $$ 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

Short Answer

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(a) Right (b) Left (c) Right (d) Left

Step by step solution

01

Analyze the Reaction

The given reaction is an equilibrium reaction described as \(\mathrm{N}_2\mathrm{O}_3(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) + \mathrm{NO}_2(\mathrm{g})\). It is endothermic, indicating heat is absorbed during the reaction (\(\Delta_{r} H^{\circ}=40.5 \mathrm{kJ}/\mathrm{mol}-\mathrm{rxn}\)). According to Le Chatelier's principle, an endothermic reaction will be treated as having heat on the reactants' side.
02

Change (a) - Adding More \(\mathrm{N}_2\mathrm{O}_3(\mathrm{g})\)

When more \(\mathrm{N}_2\mathrm{O}_3(\mathrm{g})\) is added, the reaction will shift to the right to reduce this stress by producing more products \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{NO}_2(\mathrm{g})\).
03

Change (b) - Adding More \(\mathrm{NO}_2(\mathrm{g})\)

Adding more \(\mathrm{NO}_2(\mathrm{g})\) will cause the equilibrium to shift to the left to re-establish equilibrium by converting \(\mathrm{NO}_2(\mathrm{g})\) back to \(\mathrm{N}_2\mathrm{O}_3(\mathrm{g})\).
04

Change (c) - Increasing the Volume of the Reaction Flask

Increasing the volume decreases pressure. For gases, the equilibrium shifts towards the side with more moles of gas to reduce this change. Here, the right side has 2 moles and left side has 1 mole, so the equilibrium will shift to the right.
05

Change (d) - Lowering the Temperature

Lowering the temperature in an endothermic reaction will result in the equilibrium shifting to the left, towards the exothermic direction, as it seeks to absorb less heat.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemical equilibrium. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and re-establish balance. This principle helps predict how a system at equilibrium reacts to changes like concentration, temperature, or pressure alterations. - **Concentration Changes**: Adding or removing substances shifts the equilibrium to either increase the concentration of reactants or products to restore balance. - **Pressure and Volume Changes**: For reactions involving gases, an increase in pressure by decreasing volume will shift the equilibrium towards the side with fewer gas molecules. - **Temperature Changes**: A rise in temperature shifts the equilibrium in the direction that absorbs heat, favoring endothermic processes, whereas a decrease favors exothermic ones. Understanding this principle provides crucial insight into manipulating reactions to achieve desired outcomes.
Endothermic Reaction
An endothermic reaction is one that absorbs heat from its surroundings. In the equilibrium reaction \( ext{N}_2 ext{O}_3( ext{g}) ightleftharpoons ext{NO}( ext{g}) + ext{NO}_2( ext{g}) \), endothermic nature is signified by the positive enthalpy change (\( riangle_r H^{ ext{°}} = 40.5 ext{kJ/mol} ext{-rxn} \)). - **Heat as a Reactant**: Here, heat can be considered as a reactant. When the temperature increases, more heat is available, pushing the reaction towards the products side, promoting more \( ext{NO} ext{ and } ext{NO}_2 \).- **Energy Requirement**: These reactions require continuous energy input to proceed, unlike exothermic reactions that release energy.Endothermic reactions are often sensitive to temperature changes. This characteristic is crucial in industrial reactions where temperature control can optimize product yield.
Equilibrium Shift
The equilibrium shift refers to the direction in which a reaction's equilibrium moves in response to changes in concentration, pressure, or temperature. When a system is out of equilibrium, it shifts to minimize disruptions and return to a balanced state.- **Right Shift**: Indicates an increased formation of products. This is seen when adding reactants or removing products.- **Left Shift**: Indicates an increased formation of reactants. Happens when adding products or removing reactants.In the given endothermic reaction, adding \( ext{N}_2 ext{O}_3 \) will cause a rightward shift, producing more \( ext{NO} ext{ and } ext{NO}_2 \). Conversely, adding \( ext{NO}_2 \) or reducing temperature causes a leftward shift, favoring \( ext{N}_2 ext{O}_3 \) formation. Shifts are vital for understanding and controlling chemical reactions to favor specific products or reactants.
Reaction Dynamics
Reaction dynamics explores the factors influencing the speed and progression of chemical reactions. It delves into the particular conditions under which reactions occur and how these influence reaction rates and equilibria. - **Factors Affecting Reaction Rates**: Include temperature, concentration, surface area, catalysts, and the nature of reactants. - **Reaction Paths**: Indicate the sequential steps or pathways a reaction follows to convert reactants into products. - **Pollution and Emission Control**: Managing these dynamics is essential in processes like pollution control, where optimizing reaction conditions can minimize harmful emissions. Understanding reaction dynamics in the context of equilibrium helps to predict the course and speed of reactions, enabling better control and efficiency in chemical processes. It’s about understanding not just where and why the equilibrium shifts, but also how fast changes occur when a new equilibrium is achieved.

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Most popular questions from this chapter

Heating a metal carbonate leads to decomposition. $$ \mathrm{BaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{BaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ Predict the effect on the equilibrium of each change listed below. Answer by choosing (i) no change, (ii) shifts left, or (iii) shifts right. (a) add \(\mathrm{BaCO}_{3}\) (c) add BaO (b) add \(\mathrm{CO}_{2}\) (d) raise the temperature (e) increase the volume of the flask containing the reaction

The equilibrium constant for the reaction $$ \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) $$ is \(1.7 \times 10^{-3}\) at \(2300 K\) (a) What is \(K\) for the reaction when written as follows? $$ 1 / 2 \mathrm{N}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) $$ (b) What is \(K\) for the following reaction? $$ 2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$

You place 0.0300 mol of pure \(\mathrm{SO}_{3}\) in an \(8.00-\mathrm{L}\) flask at \(1150 \mathrm{K}\). At equilibrium, 0.0058 mol of \(\mathrm{O}_{2}\) has been formed. Calculate \(K_{c}\) for the reaction at \(1150 \mathrm{K}\) $$ 2 \mathrm{sO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{sO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$

A \(\mathrm{COCl}_{2}\) decomposes to \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) at high temperatures. \(K_{c}\) at \(600 \mathrm{K}\) for the reaction is \(0.0071 .\) $$ \mathrm{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ If 0.050 mol of \(\mathrm{COCl}_{2}\) is placed in a 12.5 -L. flask, what is the total pressure at equilibrium at \(600 \mathrm{K} ?\)

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas with a pressure of 1.00 atm is placed in a flask. When equilibrium is achieved, \(20.0 \%\) of the \(\mathrm{N}_{2} \mathrm{O}_{4}\) has been converted to \(\mathrm{NO}_{2}\) gas. (a) Calculate \(K_{\mathrm{p}}\) (b) If the original pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is 0.10 atm, what is the percent dissociation of the gas? Is the result in agreement with Le Chatelier's principle?

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