Ch. 18: Le Chatelier's Principle

Le Chatelier's Principle

Reversible Reactions

• Reactions can bounce back and forth, forming products and reforming reactants; these are known as reversible reactions.
• Reversible reactions occur simultaneously in both directions.
• Example: A + B \rightleftharpoons C

Chemical Equilibrium

• At chemical equilibrium, there is no net change in the actual amounts of the components of the system.
• The rates of the forward and reverse reactions are equal at chemical equilibrium.
• The concentrations of components on both sides of the chemical equation are not necessarily the same and can be dramatically different.
• Analogy: Escalators where the number of people using the up escalator is the same as the number of people using the down escalator.
• The number of people upstairs does not have to equal the number of people downstairs; only the transfer between floors must be consistent.

Which Direction is Favored?

• The equilibrium position of a reaction is given by the concentrations of the system’s components at equilibrium.
• The arrow direction indicates whether the components on the left or right side of a reversible reaction are at a higher concentration.
• If A reacts to give B and the mixture at equilibrium contains more of B (e.g., 1% of A vs. 99% of B), the formation of B is favored.
• If the mixture contains 99% of A and 1% of B at equilibrium, then the formation of A is favored.
• Forward direction favored: A \rightleftharpoons B (99% A, 1% B)
• Reverse direction favored: A \rightleftharpoons B (1% A, 99% B)

Reversibility vs. Reality

• In principle, almost all reactions are reversible to some extent under the right conditions.
• In practice, one set of components is often so favored at equilibrium that the other set cannot be detected.
• If one set of components has established equilibrium by converting mostly into products, the reaction has gone to completion.
• When no products can be detected, one can say there is no reaction.
• Reversible reactions occupy a middle ground between the theoretical extremes of irreversibility and no reaction.
• The addition of a catalyst will speed up forward and reverse reactions equally by reducing the energy needed to activate the reaction in both forward and reverse directions.
• Catalysts do not affect the amount of reactants and products present at equilibrium; they simply decrease the time it takes to establish equilibrium.

Equilibrium Expression

• Chemists can express the equilibrium position in terms of a numerical constant.
• The equilibrium constant shows the relationship between the amount of product and reactant at equilibrium.
• For a hypothetical reaction: aA + bB \rightleftharpoons cC + dD

Mass Action Expression

• An expression to show the ratio of product concentrations to reactant concentrations.
• \frac{[C]^c [D]^d}{[A]^a [B]^b}
• The concentration of each substance is raised to a power equal to the number of moles of that substance in the balanced reaction equation.
• Square brackets indicate concentration in Molarity (mol/L).

Equilibrium Constant

• The constant is dependent on the temperature; if the temperature changes, so does the constant.
• The resulting ratio of the equilibrium is called the equilibrium constant or K_{eq}.
• K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Examples of Mass Action Expressions

• 2SO2(g) + O2(g) \rightleftharpoons 2SO_3(g)
• Bi2S3(s) \rightleftharpoons 2Bi^{+3}(aq) + 3S^{-2}(aq)

Equilibrium Constant—Who Cares?

• Equilibrium constants provide valuable chemical information.
• They show whether the products or the reactants are favored in a reaction (spontaneous or non-spontaneous).
• Always written as a ratio of products over reactants.
• A value of K_{eq} > 1 means that products are favored.
• K_{eq} < 1 means that reactants are favored.
• K_{eq} > 1, products favored at equilibrium
• K_{eq} \leq 1, reactants favored at equilibrium

Sample Problem 1

Dinitrogen tetroxide (N2O4), a colorless gas, and nitrogen dioxide (NO2), a brown gas, exist in equilibrium with each other according to the following equation: N2O4(g) \rightleftharpoons 2NO2(g)
A 1.0-liter gas mixture at 10°C at equilibrium contains 0.0045 mol N2O4 & 0.030 mol NO2. Write the mass action expression and calculate K{eq} for the reaction.

• Known:
  • [N2O4] = 0.0045 \frac{mol}{1.0 L}
  • [NO_2] = 0.030 \frac{mol}{1.0 L}
• Unknown:
  • Mass action expression = ?
  • K_{eq} = ?
• At equilibrium, there is no net change in the amount of N2O4 or NO_2 at any given instant.
• The only product of the reaction is NO_2, which has a coefficient of 2 in the balanced equation.
• The only reactant N2O4 has a coefficient of 1 in the balanced equation.
• The mass action expression is:
  \frac{[NO2]^2}{[N2O_4]^1}
• K_{eq} = \frac{[0.030M]^2}{[0.0045M]^1} = 0.20
• K_{eq} < 1, therefore the reaction doesn’t favor products.

Reaction Quotient

• We can also determine if a reaction has reached equilibrium by calculating a reaction quotient (Q).
• It’s like taking a snapshot of a reaction at a given time and interpreting how far along the reaction is.
• Once the reaction quotient is solved, it is compared to the equilibrium constant.
  • If Q < K, movement toward equilibrium is towards the products.
  • If Q > K, movement toward equilibrium is towards the reactants.
  • If Q = K, reactants and products are at equilibrium.

Sample Problem

At a certain temperature, K{eq} = 55, and a reaction vessel contains a mixture with the following concentrations: [SO3] = 0.85 M, [NO] = 1.2 M, [SO2] = 1.5 M, [NO2] = 2.0 M.
SO3(g) + NO(g) \rightleftharpoons SO2(g) + NO_2(g)
Is the reaction at equilibrium, and if not, which direction will the reaction proceed?

• Solve just as if you were solving for the equilibrium constant.
• Then analyze the resulting quotient with the given K_{eq}.
• Q = \frac{[SO2][NO2]}{[SO_3][NO]} = \frac{[1.5][2.0]}{[0.85][1.2]} = 2.94
• Q = 2.94, which is < K_{eq} (55).
• If Q < K{eq}, then the numerator of our quotient must increase; therefore, the reaction continues in order to increase [products] until it reaches K{eq}.

Le Chatelier’s Principle

• There is a principle that can be studied to govern changes in equilibrium: Le Chatelier’s Principle.
• Le Chatelier’s Principle states: “If a stress is applied to a system in dynamic equilibrium, the system changes to relieve the stress.”
• Stresses are changes in temperature, pressure, concentration of reactants, or concentration of products.

Concentration & Equilibrium

• Adjusting the concentrations of either reactants or products can have a dramatic impact on the equilibrium.
• If we add more of reactant A to a system at equilibrium, the system will strive to reestablish equilibrium at a new equilibrium position.
• The reaction will push to use up the extra A and generate more C.
• [A] \uparrow, reaction will shift toward products.
• Adjusting the concentrations of either reactants or products can have a dramatic impact on the equilibrium.
• If we add more of product C to a system at equilibrium, the system will strive to reestablish equilibrium at a new equilibrium position.
• The reaction will push to use up the extra C and generate more A and B.
• [C] \uparrow, reaction will shift toward reactants.

Temperature Effects on Equilibrium

• The impact of temperature changes on an equilibrium is dependent on if the process is endothermic or exothermic.
• Endothermic processes use energy as a reactant, while exothermic processes produce energy.
• K_{eq} is temperature-dependent.
• Exothermic: If T↑, the equilibrium shifts left (250 kJ is a product).
• Endothermic: If T↑, the equilibrium shifts right (energy is a reactant).

Pressure & Equilibrium

• If A, B, and C are all gases, then the equilibrium they establish is pressure-dependent.
• When the pressure is increased, the system relieves the pressure by favoring the direction that produces fewer gas molecules.
• Pressure is # of particles dependent; the more particles, the higher the pressure.
• Fewer gas molecules will exert less pressure, so more product is formed, which overall reduces the pressure; this is a shift right.
• Conversely, a decrease in pressure will favor the reaction that produces the most molecules, so we have a shift to the left.
• P↑, this equilibrium shifts right
• If P↓, this equilibrium shifts left