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Chapter 15: Problem 55

Continuous removal of one of the products of a chemical reaction has the effect of causing the reaction to go to completion. Explain this fact in terms of Le Châtelier's principle.

Short Answer

Expert verified
When a product of a chemical reaction is continuously removed, it disturbs the equilibrium of the system. To counteract this change, in accordance with Le Châtelier's principle, the system shifts towards the side of the products, driving the reaction towards completion.

Step by step solution

01

Explaining Le Châtelier's Principle

Le Châtelier's Principle can be simplified as a law of 'cause and effect' in chemistry. If there's a change in the conditions of a chemical system at equilibrium, the system will react in such a way as to counteract that change.
02

Relation to Removal of a Product

When one of the products in a chemical reaction is continuously removed from the mixture, it disturbs the equilibrium condition. According to Le Châtelier’s principle, the equilibrium will shift in the direction where it can counteract this change.
03

Explaining the Effect on Completion of the Reaction

If a product is constantly removed, the system will respond by trying to produce more of that product to restore the equilibrium. Thus, the reaction is pushed towards the 'product' side until all reactants have been used up and the reaction goes to completion.

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Most popular questions from this chapter

Determine \(K_{c}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})+\) \(\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g}),\) given the following data at \(298 \mathrm{K}\) $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.0 \times 10^{-9}$$ $$\operatorname{NOCl}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=1.1 \times 10^{2}$$ $$\mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=0.3$$

Can a mixture of \(2.2 \mathrm{mol} \mathrm{O}_{2}, 3.6 \mathrm{mol} \mathrm{SO}_{2},\) and \(1.8 \mathrm{mol}\) \(\mathrm{SO}_{3}\) be maintained indefinitely in a \(7.2 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=100\) in this reaction? Explain. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) $$

Starting with \(0.280 \mathrm{mol} \mathrm{SbCl}_{3}\) and \(0.160 \mathrm{mol} \mathrm{Cl}_{2},\) how many moles of \(\mathrm{SbCl}_{5}, \mathrm{SbCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present when equilibrium is established at \(248^{\circ} \mathrm{C}\) in a 2.50 L flask? $$\begin{aligned} \mathrm{SbCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 2.5 \times 10^{-2} \mathrm{at} \ 248^{\circ} \mathrm{C} \end{aligned}$$

Starting with \(\mathrm{SO}_{3}(\mathrm{g})\) at \(1.00 \mathrm{atm},\) what will be the total pressure when equilibrium is reached in the following reaction at \(700 \mathrm{K} ?\) \(2 \mathrm{SO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.6 \times 10^{-5}\)

\(1.00 \mathrm{g}\) each of \(\mathrm{CO}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{H}_{2}\) are sealed in a \(1.41 \mathrm{L}\) vessel and brought to equilibrium at 600 K. How many grams of \(\mathrm{CO}_{2}\) will be present in the equilibrium mixture? $$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=23.2$$
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