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Chapter 15: Problem 39

Explain Le Châtelier's principle. How can this principle help us maximize the yields of reactions?

Short Answer

Expert verified
Le Châtelier's principle states that if a system at equilibrium is disturbed, the system will adjust to minimize that disturbance. The principle is applied to maximize the yields of reactions by manipulating the conditions of the reaction, such as temperature, pressure, and concentration to 'force' the reaction towards producing more of the desired product.

Step by step solution

01

Understanding of Le Châtelier's Principle

Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing conditions, the system adjusts to restore the equilibrium. In the context of chemical reactions, it means if any change is made to a system in equilibrium, such as temperature, pressure, or concentration changes, the system will react to relieve the change.
02

Working Mechanism of Le Châtelier's Principle

To illustrate this, consider a system in chemical equilibrium, and then one of the reactants' concentration is increased. According to Le Châtelier's principle, the system will react by consuming the added reactant, shifting the equilibrium towards the 'product' side of the reaction. Similarly, if a product is removed from the system, the equilibrium will shift to replace the lost product.
03

Application of Le Châtelier's Principle in Maximizing Yields

The principle is widely used in industry to increase the yield of reactions. By manipulating the factors under control, like temperature, pressure, and concentration, it's possible to 'force' the reaction to produce more of the desired results. For instance, if a reaction releases heat (exothermic), it can be shifted toward the products by cooling the system. Similarly, in a reaction where reducing pressure favors the formation of the product, using a large vessel can increase the yield. Hence, understanding and manipulating the equilibrium conditions allows us to maximize yield.

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Most popular questions from this chapter

Consider the heterogeneous equilibrium process: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ At \(700^{\circ} \mathrm{C}\), the total pressure of the system is found to be 4.50 atm. If the equilibrium constant \(K_{P}\) is 1.52 , calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\)

List four factors that can shift the position of an equilibrium. Which one can alter the value of the equilibrium constant?

A sealed glass bulb contains a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases. When the bulb is heated from \(20^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\), what happens to these properties of the gases: (a) color, (b) pressure, (c) average molar mass, (d) degree of dissociation (from \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) ), (e) density? Assume that volume remains constant. (Hint: \(\mathrm{NO}_{2}\) is a brown gas; \(\mathrm{N}_{2} \mathrm{O}_{4}\) is colorless.)

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ is found to be \(2 \times 10^{-42}\) at \(25^{\circ} \mathrm{C}\). (a) What is \(K_{\mathrm{c}}\) for the reaction at the same temperature? (b) The very small value of \(K_{P}\left(\right.\) and \(\left.K_{\mathrm{c}}\right)\) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ is \(3.8 \times 10^{-5}\) at \(727^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the equilibrium $$ 2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g) $$ at the same temperature.
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