8.3 LeChatelier's Principle: How We Can Affect Equlibrium

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As we saw in the previous section, reactions proceed in both directions (reactants go to products and products go to reactants). We next address what happens when a system at equilibrium is disturbed so that it is no longer at equilibrium. If a system at equilibrium is subjected to a stress (a change in concentration of one of the species in the law of mass action), the reaction is no longer at equilibrium. Since this stress affects the concentrations of the reactants and the products, the system will either shift toward the products or toward the reactants until $Q$ equilibrium is reattained. This process is described by Le Chatelier's principle. It is important to note that the value of K, the equilibrium constant, does not change if you change the concentration of species in the law of mass action.

Le Chatelier's principle

When a chemical system at equilibrium is stressed, it returns to equilibrium by counteracting the stress.

A chemical system at equilibrium can be temporarily shifted out of equilibrium by adding or removing one or more of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the system to equilibrium. The value of the equilibrium constant does not change.

It is also possible to shift the equilibrium by changing the temperature. However, changing the temperature causes the equilibrium to shift because the value of the equilibrium constant K changes. We will not go into the details of why the value of K changes, you should simply understand that if the temperature is changed,the reaction will shift to reach a new equilibrium as described below in table 1.

Stresses and How the Reaction Shifts

The overview of how different disturbances affect the reaction equilibrium properties is tabulated in Table 1 $7.9. 1$

*Table : Effects of Disturbances of Equilibrium and K*
Disturbance	Direction of Shift	Effect on K
product added	toward reactants	none
reactant added	toward products	none
product removed	towards product	none
reactant removed	towards reactant	none
temperature increase	toward products for endothermic, toward reactants for exothermic	changes
temperature decrease	toward reactants for endothermic, toward products for exothermic	changes
decrease in volume/increase in gas pressure	toward side with fewer moles of gas	none
increase in volume/decrease in gas pressure	toward side with fewer moles of gas	none

Example $\PageIndex{1}$

Write an equilibrium constant expression for each reaction and use this expression to predict in which direction the reaction will shift to reattain equilibrium.

$ 2 H_2O_{(g)} \rightleftharpoons 2H_2_{(l)} +O_{2(g)} $: the amount of H_2O is doubled.
$NH_4HS_{(aq)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)} $: the concentration of H₂S is tripled
$n-butane_{(g)} \rightleftharpoons isobutane_{(g)} $: the concentration of isobutane is halved.

Given: equilibrium systems and changes

Asked for: equilibrium constant expressions and effects of changes

Strategy:

Write the equilibrium constant expression. From this expression, predict the change that must occur to maintain equilibrium when the indicated changes are made.

Solution:

a. For this reaction, K = $\dfrac{[H_2]^2[O_2]}{[H_2O]^2} $. Adding more H₂O means adding more reactant. According to table 1, the reaction will shift to make more products to reattain equilibrium.

b. For this reaction, K = $\dfrac{[NH_3][H_2S]}{[NH_4HS]} $. If the concentration of H₂S triples, you are increasing the amount of product. According to table 1, the reaction must shift left to reattain equilibrium.

c. For this reaction K = $\dfrac{[isobutane]}{[n-butane]} $, so halving the concentration of isobutane means that you have decreased the amount of product. According to table 1, the reaction must shift to the right to reattain equilibrium.

Exercise $\PageIndex{1}$

In which direction will the reaction shift if the stated stress is placed on the reaction at equilibrium?

$ HBr_{(g)} + NaH_{(s)} \rightleftharpoons NaBr_{(aq)} +H_{2(g)} $: the concentration of HBr is decreased by a factor of 3.
$6 Li_{(s)} + N_{2(g)} \rightleftharpoons 2Li_3N_{(s)} $: the amount of N₂ is tripled.
$SO_{2(g)} + Cl_{2(g)} \rightleftharpoons SO_2Cl_{2(l)} $: the concentration of Cl₂ is doubled.

Answer

a. The reaction will shift left to reattain equilibrium because you removed reactant.

b. The reaction will shift right because N₂ is a reactant.

c. The reaction will shift to the right because you added a reactant.

Catalysts Do Not Affect Equilibrium

As we learned during our study of kinetics, a catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations.