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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write the equilibrium constant expression for the following reaction, $$\begin{array}{r} \mathrm{Fe}(\mathrm{OH})_{3}+3 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ K=9.1 \times 10^{3} \end{array}$$ and compute the equilibrium concentration for \(\left[\mathrm{Fe}^{3+}\right]\) at \(\left.\mathrm{pH}=7 \text { (i.e., }\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7}\right)\)

Lead metal is added to \(0.100 \mathrm{M} \mathrm{Cr}^{3+}(\mathrm{aq}) .\) What are \(\left[\mathrm{Pb}^{2+}\right],\left[\mathrm{Cr}^{2+}\right],\) and \(\left[\mathrm{Cr}^{3+}\right]\) when equilibrium is established in the reaction? $$\begin{aligned} \mathrm{Pb}(\mathrm{s})+2 \mathrm{Cr}^{3+}(\mathrm{aq}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cr}^{2+}(\mathrm{aq}) & \\ K_{\mathrm{c}}=3.2 \times 10^{-10} & \end{aligned}$$

In one of Fritz Haber's experiments to establish the conditions required for the ammonia synthesis reaction, pure \(\mathrm{NH}_{3}(\mathrm{g})\) was passed over an iron catalyst at \(901^{\circ} \mathrm{C}\) and 30.0 atm. The gas leaving the reactor was bubbled through 20.00 mL of a HCl(aq) solution. In this way, the \(\mathrm{NH}_{3}(\mathrm{g})\) present was removed by reaction with HCl. The remaining gas occupied a volume of 1.82 L at STP. The \(20.00 \mathrm{mL}\) of \(\mathrm{HCl}(\mathrm{aq})\) through which the gas had been bubbled required \(15.42 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Another \(20.00 \mathrm{mL}\) sample of the same HCl(aq) through which no gas had been bubbled required \(18.72 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Use these data to obtain a value of \(K_{\mathrm{p}}\) at \(901^{\circ} \mathrm{C}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})\)

Write equilibrium constant expressions, \(K_{\mathrm{p}},\) for the reactions (a) \(\mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) (b) \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

In the human body, the enzyme carbonic anahydrase catalyzes the interconversion of \(\mathrm{CO}_{2}\) and \(\mathrm{HCO}_{3}^{-}\) by either adding or removing the hydroxide anion. The overall reaction is endothermic. Explain how the following affect the amount of carbon dioxide: (a) increasing the amount of bicarbonate anion; (b) increasing the pressure of carbon dioxide; (c) increasing the amount of carbonic anhydrase; (d) decreasing the temperature.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free