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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

The vapor pressure of mercury is \(0.0020 \mathrm{mmHg}\) at \(26^{\circ} \mathrm{C}\). (a) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the process \(\mathrm{Hg}(l) \rightleftharpoons \mathrm{Hg}(g) .\) (b) A chemist breaks a thermometer and spills mercury onto the floor of a laboratory measuring \(6.1 \mathrm{~m}\) long, \(5.3 \mathrm{~m}\) wide, and \(3.1 \mathrm{~m}\) high. Calculate the mass of mercury (in grams) vaporized at equilibrium and the concentration of mercury vapor in \(\mathrm{mg} / \mathrm{m}^{3}\). Does this concentration exceed the safety limit of \(0.050 \mathrm{mg} / \mathrm{m}^{3} ?\) (Ignore the volume of furniture and other objects in the laboratory.)

A mixture of 0.47 mole of \(H_{2}\) and 3.59 moles of \(H C l\) is heated to \(2800^{\circ} \mathrm{C}\). Calculate the equilibrium partial pressures of \(\mathrm{H}_{2}, \mathrm{Cl}_{2}\), and \(\mathrm{HCl}\) if the total pressure is 2.00 atm. The \(K_{P}\) for the reaction \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)\) is 193 at \(2800^{\circ} \mathrm{C}.\)

In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it speeds up both the forward and reverse rates to the same extent. To test this statement, consider a situation in which an equilibrium of the type $$ 2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g) $$ is established inside a cylinder fitted with a weightless piston. The piston is attached by a string to the cover of a box containing a catalyst. When the piston moves upward (expanding against atmospheric pressure), the cover is lifted and the catalyst is exposed to the gases. When the piston moves downward, the box is closed. Assume that the catalyst speeds up the forward reaction \((2 \mathrm{~A} \longrightarrow \mathrm{B})\) but does not affect the reverse process \((\mathrm{B} \longrightarrow 2 \mathrm{~A})\). Suppose the catalyst is suddenly exposed to the equilibrium system as shown below. Describe what would happen subsequently. How does this "thought" experiment convince you that no such catalyst can exist?

The equilibrium constant \(K_{\mathrm{c}}\) for the following reaction is 0.65 at \(395^{\circ} \mathrm{C}\). $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ (a) What is the value of \(K_{P}\) for this reaction? (b) What is the value of the equilibrium constant \(K_{\mathrm{c}}\) for \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) ?\) (c) What is \(K_{\mathrm{c}}\) for \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g) ?\) (d) What are the values of \(K_{P}\) for the reactions described in (b) and (c)?

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 \(\mathrm{CO}\)

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