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

What will be the effect on the equilibrium constant on increasing temperature, if the reaction neither absorbs heat nor releases heat? (a) Equilibrium constant will remain constant. (b) Equilibrium constant will decrease. (c) Equilibrium constant will increase. (d) Can not be predicted.

Short Answer

Expert verified
The equilibrium constant will remain constant.

Step by step solution

01

Understand the Nature of the Reaction

Identify whether the reaction is exothermic or endothermic. In this exercise, it is given that the reaction neither absorbs heat nor releases heat, indicating that it is a thermally neutral reaction.
02

Apply Le Chatelier's Principle

Le Chatelier's Principle suggests that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In this case, since there is no heat absorbed or released, changing temperature would not affect the equilibrium.
03

Analyze the Effect on the Equilibrium Constant

Since the reaction is thermally neutral, and the equilibrium position is unaffected by changes in temperature, the equilibrium constant (K) will remain unchanged.

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.

Le Chatelier's Principle
Le Chatelier's Principle is a foundational concept in chemistry that describes how a chemical system at dynamic equilibrium responds to a change in conditions. If there's an alteration in concentration, temperature, or pressure, the system will adjust to minimize that change. For instance, if you increase the concentration of a reactant in a balanced reaction, the equilibrium will shift to produce more products. This principle helps us predict the direction in which a reaction will shift when it is subjected to external stress. It's essential for students, especially those preparing for competitive exams like JEE Physical Chemistry, to understand this principle because it applies to a vast range of chemical processes. Consequentially, in thermally neutral reactions, Le Chatelier's Principle implies no shift in equilibrium with temperature changes, as temperature is neither a reactant nor a product in these scenarios.
Thermally Neutral Reaction
A thermally neutral reaction is one in which the heat exchange with the surroundings is zero. In simpler terms, this type of reaction neither absorbs heat from nor releases heat to its environment. As a result, thermodynamically, there is no enthalpy change, symbolized as \( \Delta H = 0 \). Students of JEE Physical Chemistry must grasp that for such reactions, temperature has no effect on the equilibrium constant according to Le Chatelier's Principle. So when they encounter a problem like the exercise provided, they can confidently conclude that the equilibrium constant will remain unchanged when temperature changes because there is no heat absorbed or released to drive the shift in equilibrium.
Dynamic Equilibrium
Dynamic equilibrium in chemistry is the condition where the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentrations of reactants and products remain constant over time, despite the continuous exchange, and contributes to a state of no net change. When teaching this concept for JEE Physical Chemistry, it is emphasized that at dynamic equilibrium, any changes in conditions (such as temperature and pressure) have the potential to alter the position of equilibrium. Yet, this does not apply to thermally neutral reactions, reflecting back on the exercise question, where temperature changes do not affect the reactants and products because the process does not involve heat transfer.
JEE Physical Chemistry
The Joint Entrance Examination (JEE) for admission into various engineering colleges across India covers a wide range of subjects, including Physical Chemistry. This discipline of chemistry involves understanding the principles and concepts of chemistry through a quantitative lens. Topics like dynamic equilibrium, thermodynamic principles, and reaction kinetics are central to JEE Physical Chemistry, and require a deep understanding to solve related problems efficiently. Such topics are frequently tested in JEE through conceptual questions like the exercise example on equilibrium constants and temperature effects. Students must not only memorize the facts but also be able to apply concepts like Le Chatelier's Principle to solve complex problems and predict the behavior of chemical systems under various conditions.

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

At a certain temperature, only \(50 \% \mathrm{HI}\) is dissociated at equilibrium in the following reaction: $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ The equilibrium constant for this reaction is : (a) \(0.25\) (b) \(1.0\) (c) \(3.0\) (d) \(0.5\)

Some inert gas is added at constant volume to the following reaction at equilibrium $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ Predict the effect of adding the inert gas: (a) the equilibrium shifts in the forward direction (b) the equilibrium shifts in the backward direction (c) the equilibrium remains unaffected (d) the value of \(K_{p}\) is increased

Given \(\left[\mathrm{CS}_{2}\right]=0.120 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.10,\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.20\) and \(\left[\mathrm{CH}_{4}\right]=8.40 \times 10^{-5} M\) for the following reaction at \(900^{\circ} \mathrm{C}\), at eq. Calculate the equilibrium constant \(\left(K_{c}\right)\). $$ \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) $$ (a) \(0.0120\) (b) \(0.0980\) (c) \(0.280\) (d) \(0.120\)

Consider the following reactions. In which cases is the product formation favoured by decreased pressure? (1) \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) ; \quad \Delta H^{\circ}=+172.5 \mathrm{~kJ}\) (2) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; \quad \Delta H^{\circ}=-91.8 \mathrm{~kJ}\) (3) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) ; \quad \Delta H^{\circ}=181 \mathrm{~kJ}\) (4) \(2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) ; \quad \Delta H^{\circ}=484.6 \mathrm{~kJ}\) (a) 2,3 (b) 3,4 (c) 2,4 (d) 1,4

An equilibrium mixture of the reaction \(2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)\) had \(0.5\) mole \(\mathrm{H}_{2} \mathrm{~S}, 0.10\) mole \(\mathrm{H}_{2}\) and \(0.4\) mole \(\mathrm{S}_{2}\) in one litre vessel. The value of equilibrium constant \((\mathrm{K})\) in \(\mathrm{mol}\) litre \(^{-1}\) is : (a) \(0.004\) (b) \(0.008\) (c) \(0.016\) (d) \(0.160\)

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