Chemical Equilibrium Chapter 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Physical equilibrium H 2 O (l) Chemical equilibrium N2O4 (g)N2O4 (g) 14.1 H 2 O (g) 2NO 2 (g)

N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium 14.1

constant

Chemical reactions often seem to stop before they are complete. Actually, such reactions are reversible. That is, the original reactants form products, but then the products react with themselves to give back the original reactants. When these two reactions—forward and reverse— occur at the same rate, a chemical equilibrium exists.

The graph shows how the amounts of reactants and products change as the reaction approaches equilibrium. CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g)

This graph shows how the rates of the forward reaction and the reverse reaction change as the reaction approaches equilibrium. CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g)

Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal. We can apply stoichiometry to compute the content of the reaction mixture at equilibrium.

When heated PCl 5, phosphorus pentachloride, forms PCl 3 and Cl 2 as follows: PCl 5 (g) PCl 3 (g) + Cl 2 (g) When 1.00 mol PCl 5 in a 1.00-L container is allowed to come to equilibrium at a given temperature, the mixture is found to contain mol PCl 3. What is the molar composition of the mixture?

We will organize this problem by using the chemical reaction to set up a table of initial, change, and equilibrium amounts. Initially we had 1.00 mol PCl 5 and no PCl 3 or Cl 2. The change in each is stoichiometric: If x moles of PCl 5 react, then x moles of PCl 3 and x moles of Cl 2 are produced. For reactants, this amount is subtracted from the original amount; for products, it is added to the original amount.

We were told that the equilibrium amount of PCl 3 is mol. That means x = mol. We can now find the amounts of the other substances. PCl 5 (g)PCl 3 (g) +Cl 2 (g) Initial1.00 mol00 Change –x–x+x+x+x+x Equilibrium1.00 – xxx

Moles PCl 5 = 1.00 – = 0.87 mol (2 decimal places) Moles PCl 3 = mol (given with 3 significant figures) Moles Cl 2 = mol

The Equilibrium Constant, K c The equilibrium constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration term to a power equal to its coefficient in the balanced chemical equation. The equilibrium constant, K c, is the value obtained for the K c expression when equilibrium concentrations are substituted.

N 2 O 4 (g) 2NO 2 (g) = 4.63 x K = [NO 2 ] 2 [N 2 O 4 ] aA + bB cC + dD K = [C] c [D] d [A] a [B] b 14.1

K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Equilibrium Will K = [C] c [D] d [A] a [B] b aA + bB cC + dD 14.1

Methanol (also called wood alcohol) is made commercially by hydrogenation of carbon monoxide at elevated temperature and pressure in the presence of a catalyst: 2H 2 (g) + CO(g) CH 3 OH(g) What is the K c expression for this reaction?

When K c is very large (>10 2 ), the equilibrium mixture is mostly products. When K c is very small (<10 -2 ), the equilibrium mixture is mostly reactants. When K c approaches 1, the equilibrium mixture contains appreciable amounts of both reactants and products.

Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen: 2CO 2 (g) 2CO(g) + O 2 (g) At 3000 K, 2.00 mol CO 2 is placed into a 1.00-L container and allowed to come to equilibrium. At equilibrium, 0.90 mol CO 2 remains. What is the value of K c at this temperature?

We can find the value of x – 2x = = 2x x = 0.55 mol 0.55 mol1.10 mol0.90 mol x2x2x2.00 – 2xEquilibrium +x+x+2x –2x–2xChange molInitial O2(g)O2(g)2CO(g) +2CO 2 (g) 2x = 2(0.55) = 1.10 mol

2CO 2 (g) 2CO(g) + O 2 (g)

In a heterogeneous equilibrium, in the K c expression, the concentrations of solids and pure liquids are constant (due to these substances’ constant density). As a result, we incorporate those constants into the value of K c, thereby making a new constant, K c. In other words, equilibrium is not affected by solids and pure liquids as long as some of each is present. More simply, we write the K c expression by replacing the concentration of a solid or pure liquid with 1.

Write the K c expression for the following reaction: H 2 O(g) + C(s) CO(g) + H 2 (g)

Given: aA + bB cC + dD; K 1 When the reaction is reversed: cC + dD aA + bB; K 2 The equilibrium constant expression is inverted: K 2 =

Given: aA + bB cC + dD; K 1 When the reaction is doubled: 2aA + 2bB 2cC + 2dD; K 2 The equilibrium constant expression, K 2, is the square of the equilibrium constant expression, K 1 : K 2 =

For the reaction aA(g) + bB(g) cC(g) + dD(g) The equilibrium constant expressions are K c = and K p =

How are these related? We know From the ideal gas law, we know that So,

When you express an equilibrium constant for a gaseous reaction in terms of partial pressures, you call it the equilibrium constant, K p. In general, the value of K p is different from that of K c. We will explore this relationship on the next slides. Recall the ideal gas law and the relationship between pressure and molarity of a gas:

K p = K c (RT)  n

For catalytic methanation, CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) the equilibrium expression in terms of partial pressures becomes and

The value of K c at 227°C is for the following reaction: CH 3 OH(g) CO(g) + 2H 2 (g) What is K p at this temperature? K p = (RT)  n where T = = 500. K R = L  atm/(mol  K)  n = 2 K p = 1.60 × 10 2

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] K p = NO 2 P2P2 N2O4N2O4 P aA (g) + bB (g) cC (g) + dD (g) 14.2 K p = K c (RT)  n  n = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b) In most cases K c  K p

Homogeneous Equilibrium CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) K c = ‘ [CH 3 COO - ][H 3 O + ] [CH 3 COOH][H 2 O] [H 2 O] = constant K c = [CH 3 COO - ][H 3 O + ] [CH 3 COOH] =K c [H 2 O] ‘ General practice not to include units for the equilibrium constant. 14.2

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = M, [Cl 2 ] = M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = x = 220 K p = K c (RT)  n  n = 1 – 2 = -1 R = T = = 347 K K p = 220 x ( x 347) -1 =

The equilibrium constant K p for the reaction is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = atm and P NO = atm? 2 2NO 2 (g) 2NO (g) + O 2 (g) 14.2 K p = 2 P NO P O 2 P NO 2 2 POPO 2 = K p P NO POPO 2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) K c = ‘ [CaO][CO 2 ] [CaCO 3 ] [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K c x ‘ [CaCO 3 ] [CaO] K p = P CO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. 14.2

P CO 2 = K p CaCO 3 (s) CaO (s) + CO 2 (g) P CO 2 does not depend on the amount of CaCO 3 or CaO 14.2

Consider the following equilibrium at 295 K: The partial pressure of each gas is atm. Calculate K p and K c for the reaction? NH 4 HS (s) NH 3 (g) + H 2 S (g) K p = P NH 3 H2SH2S P= x = K p = K c (RT)  n K c = K p (RT) -  n  n = 2 – 0 = 2 T = 295 K K c = x ( x 295) -2 = 1.20 x

A + B C + D C + D E + F A + B E + F K c = ‘ [C][D] [A][B] K c = ‘ ‘ [E][F] [C][D] [E][F] [A][B] K c = KcKc ‘ KcKc ‘‘ KcKc KcKc ‘‘ KcKc ‘ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

N 2 O 4 (g) 2NO 2 (g) = 4.63 x K = [NO 2 ] 2 [N 2 O 4 ] 2NO 2 (g) N 2 O 4 (g) K = [N 2 O 4 ] [NO 2 ] 2 ‘ = 1 K = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 14.2

Writing Equilibrium Constant Expressions 1.The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. 2.The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. 3.The equilibrium constant is a dimensionless quantity. 4.In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. 5.If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

14.3 Chemical Kinetics and Chemical Equilibrium A + 2B AB 2 kfkf krkr rate f = k f [A][B] 2 rate r = k r [AB 2 ] Equilibrium rate f = rate r k f [A][B] 2 = k r [AB 2 ] kfkf krkr [AB 2 ] [A][B] 2 =K c =

The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium 14.4

Nickel(II) oxide can be reduced to the metal by treatment with carbon monoxide. CO(g) + NiO(s) ⇌ CO 2 (g) + Ni(s) If the partial pressure of CO is 100. mmHg and the total pressure of CO and CO 2 does not exceed 1.0 atm, will this reaction occur at 1500 K at equilibrium? (K p = 700. at 1500 K.)

Nitrogen and oxygen form nitric oxide. N 2 (g) + O 2 (g) 2NO(g) If an equilibrium mixture at 25°C contains M N 2 and M O 2, what is the concentration of NO in this mixture? The equilibrium constant at 25°C is 1.0 ×

N 2 (g) + O 2 (g) 2NO(g)

Given: H 2 (g) + F 2 (g) 2HF(g); K c = 1.15 × mol of each species is put in a L vessel. What is the equilibrium concentration of each species? First calculate the initial concentrations:

H 2 (g) +F2(g)F2(g)2HF(g) Initial2.000 M Change –x–x –x–x+2x Equilibrium2.000 – x x

Now compute the equilibrium concentrations: Double-check by substituting these equilibrium concentrations into the K c expression and solving. The answer should be the value of K c. This is within round-off error.

Calculating Equilibrium Concentrations 1.Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 2.Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 3.Having solved for x, calculate the equilibrium concentrations of all species. 14.4

At C the equilibrium constant (K c ) for the reaction Is 1.1 x If the initial concentrations are [Br 2 ] = M and [Br] = M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) x-x+2x x x [Br] 2 [Br 2 ] K c = ( x) x = 1.1 x Solve for x 14.4

K c = ( x) x = 1.1 x x x = – x 4x x = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) x-x+2x x x x = x = At equilibrium, [Br] = x = Mor M At equilibrium, [Br 2 ] = – x = M 14.4

If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Le Châtelier’s Principle Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress 14.5

Le Châtelier’s Principle When a system in chemical equilibrium is disturbed by a change in temperature, pressure, or concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable.

When a substance that is part of the equilibrium is added to the mixture, the equilibrium shifts to use it (in a direction that makes the substance a reactant). When a substance that is part of the equilibrium is removed from the mixture, the equilibrium shifts to produce it (in a direction that makes the substance a product).

Le Châtelier’s Principle Changes in Concentration continued ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left 14.5 aA + bB cC + dD Add Remove

Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas 14.5

Le Châtelier’s Principle Changes in Temperature ChangeExothermic Rx Increase temperatureK decreases Decrease temperatureK increases Endothermic Rx K increases K decreases 14.5 colder hotter

uncatalyzedcatalyzed 14.5 Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle

Chemistry In Action Life at High Altitudes and Hemoglobin Production K c = [HbO 2 ] [Hb][O 2 ] Hb (aq) + O 2 (aq) HbO 2 (aq)

Chemistry In Action: The Haber Process N 2 (g) + 3H 2 (g) 2NH 3 (g)  H 0 = kJ/mol

Le Châtelier’s Principle ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressureyesno Volumeyesno Temperatureyes Catalystno 14.5