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When hydrogen iodide is heated, the degree of dissociation increases. Is the dissociation reaction exothermic or endothermic? Explain.

Short Answer

Expert verified
The dissociation of hydrogen iodide is an endothermic reaction.

Step by step solution

01

Identify The Meaning Of Different Reactions

An exothermic reaction is a chemical reaction that releases energy by light or heat. It is the opposite of an endothermic reaction. Expressed in a chemical equation, reactants lead to products + energy. On the other hand, an endothermic reaction is a process or reaction in which the system absorbs energy from its surroundings in the form of heat.
02

Understand The Effect Of Heat On The Reactions

According to Le Chatelier's Principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. For an exothermic reaction, increasing the temperature shifts the equilibrium towards the reactants. Conversely, for an endothermic reaction, increasing the temperature shifts the equilibrium towards the products (increases the degree of dissociation).
03

Apply The Principle To The Given Exercise

Since the degree of dissociation for hydrogen iodide increases with an increase in temperature, we can conclude that the dissociation of hydrogen iodide is an endothermic reaction.

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Key Concepts

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

Endothermic Reactions
In chemical reactions, energy plays a crucial role. Endothermic reactions are those where energy is absorbed from the surroundings. This means that the reaction requires heat to proceed. The temperature of the surroundings often decreases as the reaction absorbs heat. This is opposite to exothermic reactions, which release energy.

For example, when ice melts to water, it absorbs heat from the environment, making it an endothermic process. Similarly, in chemical equations, you'll notice reactants plus heat lead to products. A common sign of an endothermic reaction is an increase in the degree of dissociation when heated, as seen in the dissociation of hydrogen iodide.
Understanding Le Chatelier's Principle
Le Chatelier's Principle helps us understand how a system at equilibrium responds to external changes. It states that if a system at equilibrium experiences a change in temperature, pressure, or concentration, the equilibrium will shift to counteract that change. This principle is incredibly useful in predicting the behavior of reactions.

For endothermic reactions, such as the dissociation of hydrogen iodide, an increase in temperature shifts the equilibrium towards the products.
  • This means more reactants convert to products as temperature rises.
  • It favors the side of the reaction that absorbs heat.

This shift results in an increased degree of dissociation, further confirming the reaction's endothermic nature.
Dissociation of Hydrogen Iodide
The dissociation of hydrogen iodide ( HI ) into hydrogen ( H_2 ) and iodine ( I_2 ) is a fascinating reaction. As temperature increases, hydrogen iodide molecules break apart, leading to more products.

This behavior aligns with an endothermic reaction, as the system absorbs heat to dissociate the molecules. The equilibrium shifts to produce more H_2 and I_2 , demonstrating Le Chatelier's Principle in action.
  • Heat acts as a driving force for the dissociation.
  • The process results in a higher degree of dissociation with rising temperature.

Understanding this helps in predicting and manipulating reactions in industrial and laboratory settings.

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Most popular questions from this chapter

Write an equilibrium constant, \(K_{c},\) for the formation from its gaseous elements of \((a) 1\) mol \(\mathrm{HF}(\mathrm{g})\) (b) \(2 \mathrm{mol} \mathrm{NH}_{3}(\mathrm{g}) ;(\mathrm{c}) 2 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) ;(\mathrm{d}) 1 \mathrm{mol} \mathrm{ClF}_{3}(1)\)

Determine \(K_{c}\) for the reaction $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NOBr}(\mathrm{g})$$ from the following information (at \(298 \mathrm{K}\) ). $$\begin{aligned} 2 \mathrm{NO}(\mathrm{g}) & \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) K_{\mathrm{c}}=2.1 \times 10^{30} \\ \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{NOBr}(\mathrm{g}) \quad K_{\mathrm{c}}=1.4 \end{aligned}$$

Starting with \(0.3500 \mathrm{mol} \mathrm{CO}(\mathrm{g})\) and \(0.05500 \mathrm{mol}\) \(\mathrm{COCl}_{2}(\mathrm{g})\) in a \(3.050 \mathrm{L}\) flask at \(668 \mathrm{K},\) how many moles of \(\mathrm{Cl}_{2}(\mathrm{g})\) will be present at equilibrium? $$\begin{aligned} \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g}) & \\ K_{\mathrm{c}} &=1.2 \times 10^{3} \mathrm{at} \ 668 \mathrm{K} \end{aligned}$$

In the equilibrium described in Example \(15-12,\) the percent dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) can be expressed as $$\frac{3.00 \times 10^{-3} \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4}}{0.0240 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4} \text { initially }} \times 100 \%=12.5 \%$$ What must be the total pressure of the gaseous mixture if \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) is to be \(10.0 \%\) dissociated at \(298 \mathrm{K} ?\) $$ \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.113 \text { at } 298 \mathrm{K} $$

Equilibrium is established at \(1000 \mathrm{K},\) where \(K_{\mathrm{c}}=281\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{O}_{2}(\mathrm{g})\) in a \(0.185 \mathrm{L}\) flask is 0.00247 mol. What is the ratio of \(\left[\mathrm{SO}_{2}\right]\) to \(\left[\mathrm{SO}_{3}\right]\) in this equilibrium mixture?

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