State Le Châtelier's principle using the words undo and partially.

Short Answer

Expert verified
In conclusion, Le Châtelier's principle can be stated as follows: When a chemical system in equilibrium is disturbed, the system will adjust itself to partially undo the effect of the disturbance, thereby re-establishing a new equilibrium state.

Step by step solution

01

Introducing Le Châtelier's principle

Le Châtelier's principle is a useful tool in understanding the behavior of chemical reactions at equilibrium when an external change is applied to the system. The principle states that if an equilibrium system is subjected to a disturbance, such as a change in temperature, pressure, or concentration, the system will adjust its position to partially counteract or undo the effect of the disturbance, in an effort to re-establish a new equilibrium state.
02

Understand the meaning of 'undo'

In the context of Le Châtelier's principle, the term 'undo' refers to the chemical system's response to counteract or minimize the imposed change upon it. By adjusting the equilibrium position of the reaction, the system tries to revert to a state closer to its original equilibrium, partially undoing the effect of the disturbance that caused the shift.
03

Understand the meaning of 'partially'

When discussing Le Châtelier's principle, 'partially' means that the system's adjustment will not completely cancel out the imposed disturbance. Instead, it will minimize the impact of the disturbance to a certain extent, hence only partially undoing the effect. This is because the system will establish a new equilibrium state that reflects a balance between the original state and the imposed change.
04

Apply the principle to various conditions

Now, let's consider the application of Le Châtelier's principle to changes in the following conditions: 1. Concentration: If the concentration of a reactant or product is increased, the system will partially undo the change by shifting the equilibrium position to consume the added substance. Conversely, if the concentration is decreased, the system will partially undo the change by shifting the equilibrium position to produce more of that substance. 2. Temperature: If the temperature of an exothermic reaction (one that releases heat) is increased, the system will partially undo the change by shifting the equilibrium position towards the endothermic (heat-absorbing) direction, in order to consume the added heat. Conversely, if the temperature is decreased, the system will partially undo the change by shifting the equilibrium position towards the exothermic direction to release more heat. 3. Pressure: If the pressure on a gaseous reaction system involving a change in the number of moles is increased, the system will partially undo the change by shifting the equilibrium position in the direction with fewer moles of gas. Conversely, if the pressure is decreased, the system will partially undo the change by shifting the equilibrium position in the direction with more moles of gas. Remember, in each case, the adjustments made by the system are not absolute but partial in nature, and a new equilibrium position will be established.
05

State Le Châtelier's principle in the required form

In conclusion, Le Châtelier's principle can be stated as follows: When a chemical system in equilibrium is disturbed, the system will adjust itself to partially undo the effect of the disturbance, therebŷ re-establishing a new equilibrium state.

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

Indicate with an arrow the direction of the equilibrium shift and predict what will happen to the amount of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (increases, decreases, unchanged, need more information) when the following stresses are applied to the following exothermic reaction: \(2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g)\) The reaction is cooled. \(\mathrm{H}_{2}\) gas is added. \(\mathrm{H}_{2} \mathrm{O}\) is removed. Volume is reduced. A catalyst is added. Fe is added while the reaction temperature is increased.

One way of preparing hydrogen is by decomposition of water: \(2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta E_{\mathrm{rxn}}=484 \mathrm{~kJ}\) (a) Would you expect the decomposition to be more complete at equilibrium when it is run at high temperature or when it is run at low temperature? Explain. (b) According to your answer to (a), would the reaction speed up, slow down, or occur at the same rate as before the temperature was changed? Explain.

Suppose you are making ammonia \(\left(\mathrm{NH}_{3}\right)\) by the Haber reaction, at \(472{ }^{\circ} \mathrm{C}\) : \(3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g) K_{\mathrm{eq}}=0.105\) (a) Describe qualitatively where the equilibrium lies for this reaction. (b) On the face of it, would this reaction be a good one for isolating pure ammonia? (c) What would happen if you could keep feeding \(\mathrm{H}_{2}\) and \(\mathrm{N}_{2}\) into the reaction vessel while at the same time removing \(\mathrm{NH}_{3}\) ?

Suppose you have a reaction with many reactants. When you write the equilibrium expression for the reaction, do the reactant concentrations all go in the numerator or in the denominator? What mathematical operation(s) should be used for these concentrations?

What allows us to incorporate the concentrations of pure solids and liquids into \(K_{\text {eq }}\) instead of writing these concentrations explicitly in the equilibrium constant expression?

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