Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The exothermic formation of \(\mathrm{ClF}_{3}\) is represented by the equation: \(\mathrm{Cl}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-329 \mathrm{~kJ}\). Which of the following will increase the quantity of \(\mathrm{CIF}_{3}\) in an equilibrium mixture of \(\mathrm{Cl}_{2}, \mathrm{~F}_{2}\) and \(\mathrm{ClF}_{3}\) ? (a) Increasing the temperature (b) Removing \(\mathrm{Cl}_{2}\)

Short Answer

Expert verified
Neither increasing temperature nor removing \(\mathrm{Cl}_{2}\) will increase \(\mathrm{ClF}_{3}\).

Step by step solution

01

Understand Exothermic Reaction

The reaction \(\mathrm{Cl}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}_{3}(\mathrm{~g})\) is exothermic, with \(\Delta \mathrm{H} = -329 \mathrm{~kJ}\). An exothermic reaction releases heat, and therefore, increasing the temperature will shift the equilibrium to favor the endothermic direction (reactants).
02

Apply Le Chatelier's Principle to Temperature Increase

According to Le Chatelier’s Principle, increasing the temperature of an exothermic reaction will decrease the formation of products. Therefore, increasing the temperature will not increase the quantity of \(\mathrm{ClF}_{3}\).
03

Consider Effect of Removing a Reactant

If \(\mathrm{Cl}_{2}\) is removed from the equilibrium mixture, according to Le Chatelier's Principle, the equilibrium will shift towards the reactant side to produce more \(\mathrm{Cl}_{2}\), thus decreasing \(\mathrm{ClF}_{3}\) quantity.
04

Choose the Correct Option

After analyzing both options, neither increasing the temperature nor removing \(\mathrm{Cl}_{2}\) will increase the \(\mathrm{ClF}_{3}\) quantity. Hence, neither option (a) nor (b) will increase \(\mathrm{ClF}_{3}\) in the mixture when considering standard options given.

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!

Key Concepts

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

Exothermic Reaction
An exothermic reaction is a type of chemical reaction that releases energy by light or heat into its surroundings. In the given reaction of chlorine (\(\mathrm{Cl}_2\)) and fluorine (\(\mathrm{F}_2\)) to form chlorine trifluoride (\(\mathrm{ClF}_3\)), the reaction releases 329 kJ of energy. This indicates that the system is losing energy as it progresses, which is characteristic of exothermic reactions.
As the reaction moves forward, the products have less energy than the reactants because some of the reactants' potential energy has been converted into thermal energy.
  • Heat is viewed as a "product" in this context.
  • This relationship is crucial when deciding how changes in conditions affect the position of equilibrium.
Understanding that the reaction is exothermic helps you predict how certain changes, like shifts in temperature, can influence the balance of products and reactants.
Chemical Equilibrium
Chemical equilibrium refers to the state in which both reactants and products are present in concentrations that have no further tendency to change with time. It is a balance achieved when the rates of the forward and reverse reactions are equal.
In the equation representing the formation of \(\mathrm{ClF}_3\), the symbols \(\rightleftharpoons\) show us the reaction is reversible and can reach equilibrium.
  • At equilibrium, the concentrations of chlorine gas (\(\mathrm{Cl}_2\)), fluorine gas (\(\mathrm{F}_2\)), and chlorine trifluoride (\(\mathrm{ClF}_3\)) remain constant.
  • It's important to note that equilibrium does not mean the reactants and products are equal in concentration, but simply that their concentrations stop changing.
At equilibrium, even though the reactants and products are forming at the same rate, external factors can affect this balance, shifting the equilibrium position one way or the other.
Temperature Effects on Equilibrium
Temperature changes can have significant effects on the equilibrium of a chemical reaction, particularly exothermic ones like the \(\mathrm{ClF}_3\) formation.
According to Le Chatelier's Principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
  • In exothermic reactions, increasing the temperature adds heat, "a product," prompting the system to form more reactants to absorb the extra heat.
  • This shifts the equilibrium towards the endothermic direction (reactants), resulting in fewer products.
So, for the exothermic reaction to produce more \(\mathrm{ClF}_3\), the temperature should ideally be decreased, encouraging more product formation to release heat. Understanding these temperature effects helps in controlling and predicting the outcomes of chemical processes, which is crucial in both industrial and laboratory settings.

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

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at 300 is kept in a closed container under one atmosphere. It is heated to 600 when \(20 \%\) by mass of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes to \(\mathrm{NO}_{2}\) (g). The resultant pressure is: (a) \(1.2 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.0 \mathrm{~atm}\)

The equilibrium constant for the reaction: \(\mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is \(\mathrm{K}_{1}\) for reaction \(\mathrm{H}_{2} \mathrm{PO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HPO}_{4}^{2-}\) is \(\mathrm{K}_{2}\) and for reaction \(\mathrm{HPO}_{4}^{2-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-}\) is \(\mathrm{K}_{3}\) The equilibrium constant \((\mathrm{K})\) for \(\mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons 3 \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-}\) will be: (a) \(\mathrm{K}_{1}, \times \mathrm{K}_{2} \times \mathrm{K}_{3}\) (b) \(\mathrm{K}_{1} / \mathrm{K}_{2} \mathrm{~K}_{3}\) (c) \(\mathrm{K}_{2} / \mathrm{K}_{1} \mathrm{~K}_{3}\) (d) \(\mathrm{K}_{1},+\mathrm{K}_{2}+\mathrm{K}_{3}\)

The equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) At temperature \(\mathrm{T}\) is \(4 \times 10^{-4}\). The value of \(\mathrm{K}_{\mathrm{c}}\) for the reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is: (a) \(4 \times 10^{-6}\) (b) \(2.5 \times 10^{2}\) (c) \(0.02\) (d) 50

At \(\mathrm{Kp}\) for the following reaction is 1 atm \(\mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g})\) At equilibrium, \(50 \%\) of \(\mathrm{X}(\mathrm{g})\) is dissociated. The total pressure of the equilibrium system is ' \(\mathrm{P}\) ' atm. what is the partial pressure (in atm) of \(X(g)\) at equilibrium: (a) 1 (b) 4 (c) 2 (d) \(0.5\)

For the reaction: \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) the forward reaction at constant temperature is favoured by 1\. Introducing an inert gas at constant volume 2\. Introducing chlorine gas at constant volume 3\. Introducing an inert gas at constant pressure 4\. Increasing the volume of the container 5\. Introducing \(\mathrm{pc} 1_{5}\) at constant volume (a) \(1,2,3\) (b) 4,5 (c) \(2,3,5\) (d) \(3,4,5\)

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