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For equilibrium reaction: \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+14.6 \mathrm{~J}\), increase in tempera- ture would (a) Favour the formation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (b) Stop reaction (c) Favour the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (d) No alter the equilibrium

Short Answer

Expert verified
Increasing temperature favors the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\).

Step by step solution

01

Determine the Nature of the Reaction

The reaction \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+14.6 \mathrm{~J}\) releases energy (14.6 J), indicating that it is an exothermic reaction. Therefore, increasing temperature should shift the equilibrium towards the endothermic direction to absorb the extra heat.
02

Identify the Endothermic Direction

In an exothermic reaction, the forward reaction (formation of \(\mathrm{N}_{2} \mathrm{O}_{4}\)) releases heat. Thus, the reverse reaction, which is the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) into \(\mathrm{NO}_{2}\), is endothermic. Increasing the temperature will favor this endothermic reverse reaction.
03

Conclude the Effect of Temperature Increase

Therefore, increasing the temperature will favor the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) into \(\mathrm{NO}_{2}\) as the equilibrium shifts in the endothermic direction.

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Key Concepts

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

Exothermic Reaction
An exothermic reaction is a process where energy is released, usually in the form of heat, to the surrounding environment. This means that the products of the reaction have less energy than the reactants, and the difference is expelled as heat. For example, in the reaction \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+14.6 \mathrm{~J}\), the formation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) releases 14.6 J of energy. Some features of exothermic reactions include:
  • Energy is released, which often makes the surroundings warmer.
  • Exothermic reactions are usually spontaneous once initiated.
  • The change in enthalpy (∆H) is negative, indicating a release of energy.
If you think about the reaction in terms of temperature, increasing the temperature will generally favor the reverse reaction (decomposition) in an exothermic system, as adding heat supports the endothermic response.
Endothermic Reaction
Endothermic reactions require energy input. This means that heat is absorbed from the surroundings, making the reaction products have more energy than the reactants. In the case of the reaction \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+14.6 \mathrm{~J}\), the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) into \(\mathrm{NO}_{2}\) is the endothermic part of the equilibrium.Important characteristics of endothermic reactions include:
  • Energy is absorbed, often making the surroundings cooler.
  • The reaction requires a constant input of energy to be sustained.
  • The change in enthalpy (∆H) is positive, indicating an absorption of energy.
Upon increasing the temperature in this reaction, the equilibrium state favors the conversion of \(\mathrm{N}_{2} \mathrm{O}_{4}\) back to \(\mathrm{NO}_{2}\), which absorbs more energy, fitting the profile of an endothermic process.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that predicts how a system in equilibrium will react to external changes. When an equilibrium system is subjected to a change in concentration, temperature, or pressure, the system will adjust in such a way as to counteract that change and establish a new equilibrium.Key aspects of Le Chatelier's Principle:
  • If the concentration of a reactant or product is changed, the system will shift to reestablish equilibrium by favoring the forward or backward reaction.
  • Increasing the temperature in an exothermic reaction shifts equilibrium towards the endothermic direction (backward reaction) to absorb the extra heat.
  • Similarly, in an endothermic reaction, a decrease in temperature shifts equilibrium towards the exothermic direction to release heat.
In our reaction example, increasing the temperature will push the equilibrium towards the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) into \(\mathrm{NO}_{2}\), aligning with Le Chatelier's Principle that aims to absorb the added heat.

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Most popular questions from this chapter

Consider an endothermic reaction \(\mathrm{X} \longrightarrow \mathrm{Y}\) with the activation energies \(E_{b}\) and \(E_{f}\) for the backward and forward reactions, respectively. In general: (a) \(\mathrm{E}_{\mathrm{b}}<\mathrm{E}_{\mathrm{f}}\) (b) \(\mathrm{E}_{\mathrm{b}}>\mathrm{E}_{\mathrm{f}}\) (c) \(\mathrm{E}_{\mathrm{b}}=\mathrm{E}_{\mathrm{f}}\) (d) There is no definite relation between \(\mathrm{E}_{\mathrm{b}}\) and \(\mathrm{E}_{\mathrm{f}}\)

In which of the following reactions, equilibrium is independent of pressure: (a) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) ; \Delta \mathrm{H}=+\mathrm{ve}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (c) \(3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-\mathrm{ve}\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}=+\mathrm{ve}\)

One of the following equilibria is not affected by change in volume of the flask: (a) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g}) \mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) (d) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\)

The equilibrium constant for the following reaction will be \(3 \mathrm{~A}+2 \mathrm{~B} \rightleftharpoons \mathrm{C}\) : (a) \(\frac{[3 \mathrm{~A}][2 \mathrm{~B}]}{[\mathrm{C}]}\) (b) \(\frac{[\mathrm{C}]}{[3 \mathrm{~A}][2 \mathrm{~B}]}\) (c) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{2}[\mathrm{~B}]^{2}}\) (d) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)

At \(550 \mathrm{~K}\), the \(\mathrm{K}\) for the following reaction is \(10^{4} \mathrm{~mol}^{-1}\) lit \(\mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{Z}(\mathrm{g})\) At equilibrium, it was observed that \([\mathrm{X}]=\frac{1}{2}[\mathrm{Y}]=\frac{1}{2}[\mathrm{Z}]\) What is the value of \([\mathrm{Z}]\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium: (a) \(2 \times 10^{-4}\) (b) \(10^{-4}\) (c) \(2 \times 10^{4}\) (d) \(10^{4}\)

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