Chapter 14 Chemical Equilibrium

Chapter 14 Chemical Equilibrium Fu-Yin Hsu

Chemical reaction The speed of a chemical reaction is determined by kinetics. The extent of a chemical reaction is determined by thermodynamics.

14.1 Fetal Hemoglobin and Equilibrium A protein called hemoglobin (Hb) reacts with oxygen according to the chemical equation The double arrows in this equation indicate that the reaction can occur in both the forward and reverse directions and can reach chemical equilibrium.

Equilibrium constant The concentrations of the reactants and products in a reaction at equilibrium are described by the equilibrium constant, K. A large value of K means that the reaction lies far to the right at equilibrium a high concentration of products and a low concentration of reactants. Small value of K means that the reaction lies far to the left at equilibrium a high concentration of reactants and a low concentration of products

O 2 Transport In the lungs: High concentration of O 2 The equilibrium shifts to the right Hb and O 2 combine to make more HbO 2 In the muscles: Low concentration of O 2, The equilibrium shifts to the right HbO 2 breaks down (dissociates) increasing the amount of free O 2.

Fetal Hemoglobin HbF + O 2 HbFO 2 Fetal hemoglobin s equilibrium constant is larger than adult hemoglobin s constant. Fetal hemoglobin is more efficient at binding O 2. O 2 is transferred to the fetal hemoglobin from the mother s hemoglobin in the placenta.

FIGURE 14.1 Oxygen Exchange between the Maternal and Fetal Circulation

14.2 The Concept of Dynamic Equilibrium H 2 and I 2 react to form 2HI molecules, but the 2 HI molecules can also react to re-form H 2 and I 2. A reaction that can proceed in both the forward and reverse directions is said to be reversible as H 2 and I 2 react, their concentrations decrease, which in turn decreases the rate of the forward reaction. At the same time, HI begins to form, and the reverse reaction begins to occur at a faster and faster rate. Eventually the rate of the reverse reaction equals the rate of the forward reaction. At that point, dynamic equilibrium is reached

Dynamic equilibrium Dynamic equilibrium for a chemical reaction is the condition in which the rate of the forward reaction equals the rate of the reverse reaction. Once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant constant because the chemicals are being consumed and made at the same rate. Note: 反應非停止正反應速率 = 逆反應速率反應物濃度 = 產物濃度

Dynamic equilibrium H 2 (g) + I 2 (g) 2 HI(g) As the concentration of product increases and the concentrations of reactants decrease, the rate of the forward reaction slows down, and the rate of the reverse reaction speeds up.

14.3 The Equilibrium Constant ( K ) For the general reaction: aa + bb gg + hh The equilibrium expression is: [G] g [H] h K c = [A]a [B] b The relationship between the chemical equation and the concentrations of reactants and products is called the law of mass action.

The Equilibrium Constant ( K ) The equilibrium constant is constant regardless of the initial concentrations of reactants and products. This constant is denoted by the symbol Kc and is called the concentration equilibrium constant. Concentrations of the products appear in the numerator( 分子 ) and concentrations of the reactants appear in the denominator( 分母 ). The exponents of the concentrations are identical to the stoichiometric coefficients in the chemical equation. The concentrations within Kc should always be written in moles per liter (M);. Kc is unitless. [G] g [H] h K c = [A]a [B] b

EXAMPLE 14.1 Expressing Equilibrium Constants for Chemical Equations Express the equilibrium constant for the chemical equation: Solution :

Writing Equilibrium Constant Expressions 2 N 2 O 5(g) 4 NO 2(g) + O 2(g) the equilibrium constant expression is as follows:

The Significance of the Equilibrium Constant

Relationships between the Equilibrium Constant and the Chemical Equation Consider the reaction: 2 NO(g) + O 2 (g) 2 NO 2 (g) [NO 2 ] 2 K c = = 4.67 x 10 13 (at 298 K) [NO] 2 [O 2 ] Now consider the reaction: 2 NO 2 (g) 2 NO(g) + O 2 (g) What will be the equilibrium constant K' c for the new reaction? [NO] 2 [O 2 ] 1 K' c = = = [NO 2 ] 2 [NO 2 ] 2 [NO] 2 [O 2 ] 1 1 = = 2.14 x 10 14 K c 4.67 x 10 13

Relationships between the Equilibrium Constant and the Chemical Equation Consider the reaction: 2 NO(g) + O 2 (g) 2 NO 2 (g) [NO 2 ] 2 K c = = 4.67 x 10 13 (at 298 K) [NO] 2 [O 2 ] Now consider the reaction: NO 2 (g) NO(g) + ½ O 2 (g) What will be the equilibrium constant K" c for the new reaction? [NO] [O 2 ] 1/2 1 1/2 K" c = = [NO 2 ] 2 K c = 2.14 x 10 14 = 1.46 x 10 7

Relationships between the Equilibrium Constant and the Chemical Equation Suppose we need: N 2 O(g) + 3/2 O 2 (g) 2 NO 2 (g) K c (1) =?? and we re given: N 2 O(g) + ½ O 2 (g) 2 NO(g) K c (2) = 1.7 x 10 13 2 NO(g) + O 2 (g) 2 NO 2 (g) K c (3) = 4.67 x 10 13 Overall N 2 O(g) + 3/2 O 2 (g) 2 NO 2 (g) K c (1) =?? [NO] 2 [N 2 O][O 2 ] [NO 2 ] 2 [NO 2 ] 2 1/2 [NO] 2 [O 2 ] = [N O][O 2 2 ]3/2 Kc(2) Kc(3) = Kc(1)

Summary For the reverse reaction, K is the reciprocal ( 倒數 ) of K for the forward reaction. When an equation is divided by two, K for the new reaction is the square root of K for the original reaction. When the coefficients of an equation are multiplied by a common factor n to produce a new equation, we raise the original K c value to the power n to obtain the new equilibrium constant. When the equations for individual reactions are added to obtain an overall equation, their equilibrium constants are multiplied the overall reaction

EXAMPLE 14.2 Manipulating the Equilibrium Constant to Reflect Changes in the Chemical Equation

Solution:

14.4 Expressing the Equilibrium Constant in Terms of Pressure In reactions involving gases, it is often convenient to measure partial pressures rather than molarities. a A(g) +b B(g) + ------ g G(g) + h H(g) + --- In these cases, a partial pressure equilibrium constant, K p, is used. K g h ( PG) ( PH)... p a b ( PA) ( PB)...

14.4 Expressing the Equilibrium Constant in Terms of Pressure a A(g) +b B(g) + ------ g G(g) + h H(g) + --- K g h ( PG) ( PH)... p a b ( PA) ( PB)... = ([G]RT)g ([H]RT) h ([A]RT) a ([B]RT) b = [G]g [H] h [A] a [B] b (RT) (g+h+---)-(a+b+----) = k c (RT) (g+h+---)-(a+b+----) n is the difference between the number of moles of reactants and moles of products (note :PV= nrt) P= (n/v) RT P = MRT

EXAMPLE 14.3 Relating Kp and Kc Nitrogen monoxide, a pollutant in automobile exhaust, is oxidized to nitrogen dioxide in the atmosphere according to the equation:

Solution

Heterogeneous Equilibria Many chemical reactions involve pure solids or pure liquids as reactants or products. EX: The concentrations of pure solids and pure liquids do not change during the course of a reaction. Because their concentration doesn t change, solids and liquids are not included in the equilibrium constant expression.

Example Since H 2 O(/) is pure liquid, it is omitted from the equilibrium expression: CaCO 3 (s) CaO(s) + CO 2 (g) [CaO] [CO 2 ] K c = [CaCO 3 ] K c = [CO 2 ]

Calculating the Equilibrium Constant from Measured Equilibrium Concentrations The most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibrium. The equilibrium mixture may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same, as long as the temperature is kept constant. The value of the equilibrium constant is independent of the initial amounts of reactants and products.

Calculating the Equilibrium Constant from Measured Equilibrium Concentrations

Calculating the Equilibrium Constant from Measured Ex: Equilibrium Concentrations consider the simple reaction: Suppose that we have a reaction mixture in which the initial concentration of A is 1.00 M and the initial concentration of B is 0.00 M. When equilibrium is reached, the concentration of A is 0.75 M.

Solution:

EXAMPLE 14.5 Finding Equilibrium Constants from Experimental Concentration Measurements Consider the following reaction: A reaction mixture at 780 initially contains [CO] = 0.500 M and [H 2 ] = 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of the equilibrium constant?

Solution: Step1 Step3 Step2 Step4

14.7 The Reaction Quotient: Predicting the Direction of Change Question: If a reaction mixture containing both reactants and products is not at equilibrium, how can we determine in which direction it will proceed? Answer: For nonequilibrium conditions, the expression having the same form as K c or K p is called the reaction quotient ( 反應商數 ), Q c or Q p. The reaction quotient is not constant for a reaction, but is useful for predicting the direction in which a net change must occur to establish equilibrium.

The Reaction Quotient For the gas phase reaction the reaction quotient is as follows: Qc < Kc, the reaction moves in the forward direction Qc > Kc, the reaction moves in the reverse direction Qc = Kc, the reaction reaches equilibrium

FIGURE 14.7 Q, K, and the Direction of a Reaction

EXAMPLE 14.7 Predicting the Direction of a Reaction by Comparing Q and K Solution:

14.8 Finding Equilibrium Concentrations

Solution:

Finding Equilibrium Concentrations from the Equilibrium Constant and Initial Concentrations or Pressures Example 14-9 Consider the reaction: A reaction mixture at 2000 C initially contains [N 2 ] = 0.200 M and [O 2 ] = 0.200 M. Find the equilibrium concentrations of the reactants and product at this temperature.

Solution Step 1 Step 2

Solution Step3

EXAMPLE 14.11 Finding Equilibrium Partial Pressures When You Are Given the Equilibrium Constant and Initial Partial Pressures

Solution: Step1

Step2:

Step 3

EXAMPLE 14.12 Finding Equilibrium Concentrations from Initial Concentrations in Cases with a Small Equilibrium Constant Consider the reaction for the decomposition of hydrogen disulfide: A 0.500 L reaction vessel initially contains 0.0125 mol of H 2 S at 800. Find the equilibrium concentrations of H 2 and S 2.

Solution: Step1 Q = 0; the reaction will proceed to the right

Step 2

Step 3

Step 3 cont Note: ( 計算時可忽略 )

14.9 Le Chatelier s Principle: How a System at Equilibrium Responds to Disturbances Once a reaction is at equilibrium, the concentrations of all the reactants and products remain constant. However, if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is restored. The new concentrations will be different, but the equilibrium constant will be the same, unless you change the temperature.

Le Châtelier s Principle Le Chatelier s principle: When a chemical system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance. Ex: If the concentration of one of the reactants is increased, the denominator of the reaction quotient increases and Q is, therefore, now less than Kc To counteract this, the concentrations of the products increase and the equilibrium is pushed to the right

FIGURE 14.9 Le Chatelier s Principle: The Effect of a Concentration Change Le Châtelier s Principle: Changing Concentration

Le Châtelier s Principle: Graphical Representation

Le Châtelier s Principle Addition or removal of pure solids or pure liquids from a system at equilibrium does not affect the equilibrium. EXAMPLE 14.14 The Effect of Concentration Change on Equilibrium Sol: 1. Adding additional CO 2 increases the concentration of CO 2 and causes the reaction to shift to the left. 2. Adding additional CaCO 3 has no effect on the position of the equilibrium

The Effect of a Volume (or Pressure) Change on Equilibrium The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium: Adding a gaseous reactant increases its partial pressure, causing the equilibrium to shift to the right. Increasing its partial pressure increases its concentration. It does not increase the partial pressure of the other gases in the mixture. Adding an inert gas to the mixture has no effect on the position of equilibrium. It does not affect the partial pressures of the gases in the reaction.

Effect of Volume Change (or External Pressure) on Equilibrium Decreasing the volume of the container increases the concentration of all the gases in the container. It increases their partial pressures. If their partial pressures increase, then the total pressure in the container will increase. According to Le Châtelier s Principle, the equilibrium should shift to remove that pressure. The way the system reduces the pressure is to reduce the number of gas molecules in the container When the volume decreases, the equilibrium shifts to the side with fewer gas molecules.

Effect of Volume Change (or External Pressure) on Equilibrium

Effect of Volume Change (or External Pressure) on Equilibrium Ex: H 2 (g) + I 2 (g) 2 HI (g) equilibrium is unaffected by External pressure changes EX: EX: N 2 O 4 (g) 2 NO 2 (g) when external pressure is increased, equilibrium is pushed to the left N 2 O 4 (g) 2 NO 2 (g) when H 2 (gas)is added at a constant external pressure equilibrium is pushed to the right ( volume )

EXAMPLE 14.15 The Effect of a Volume Change on Equilibrium Sol: 1. Decreasing the volume of the reaction mixture increases the pressure and causes the reaction to shift to the left. 2. Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to the right. 3. Adding an inert gas has no effect.

The Effect of a Temperature Change on Equilibrium Exothermic reactions release energy and endothermic reactions absorb energy Writing heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, helps us use Le Châtelier s principle to predict the effect of temperature changes

The Effect of Temperature Changes on Equilibrium for Exothermic Reactions Increasing the temperature is like adding heat. Raising the temperature of an equilibrium mixture shifts the equilibrium condition in the direction of the endothermic reaction. Lowering the temperature causes a shift in the direction of the exothermic reaction.

Temperature and equilibrium constant Changing the temperature does change the value of the equilibrium constant.

EXAMPLE 14.16 The Effect of a Temperature Change on Equilibrium The following reaction is endothermic: What is the effect of increasing the temperature of the reaction mixture? Decreasing the temperature? 1. Raising the temperature is equivalent to adding a reactant, causing the reaction to shift to the right. 2. Lowering the temperature is equivalent to removing a reactant, causing the reaction to shift to the left.