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AP Chemistry of 13Mr. Markic

Chapter 14 - Chemical Equilibrium

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 Chemical equilibrium H2O (l) H2O (g) N2O4 (g) 2NO2 (g)

N2O4 (g) 2NO2 (g)

Figure 14.2Change in the concentrations of NO2 and N2O4 with time, in three situations.

(a) Initially only NO2 is present.(b) Initially N2O4 is present(c) Initially NO2 and N2O4 are present.Note, that even though equilibrium is reached in all cases, the equilibrium concentrations of NO2 and N2O4 are not the same

N2O4 (g) 2NO2 (g)

K = ¿¿ = 4.63 x 10-3

aA + bB cC + dD

K = ¿¿


Equilibrium Will

K >> 1 Lie to the right Favor productsK << 1 Lie to the left Favor reactants

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.

N2O4 (g) 2NO2 (g)

Kc = ¿¿ Kp = PNO 2

2PN 2O4

In most cases Kc Kp

aA (g) + bB (g) cC (g) + dD (g)

Kp = Kc(RT)n

∆n = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

Homogeneous EquilibriumCH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)

Kc’ = ¿¿ [H2O] = constant Kc = ¿¿ = Kc’ [H2O]

General practice not to include units for the equilibrium constant.

Sample ExerciseWrite the expressions for KC, and KP if applicable, for the following reversible reactions at equilibrium:(a) HF(aq) + H2O(l) H3O+(aq) + F-(aq) (b) 2NO(g) + O2(g) 2NO2(g)

Law of Mass Action


(c) CH3COOH(aq) + C2H5OH(aq) CH3COOC2H5(aq) + H2O(l)

Practice ExerciseWrite Kc and Kp for the decomposition of dinitrogen pentoxide. 2NO(g) ↔ 4NO2(g) + O2(g)

Sample ExerciseThe following equilibrium process has been studied at 230°C.

2NO(g) + O2(g) 2NO2(g)In one experiment the concentrations of the reacting species at equilibrium are found to be [NO] = 0.0542 M, [O 2] = 0.127 M, and [NO2] = 15.5 M. Calculate the equilibrium constant (Kc) of the reaction at this temperature

Practice ExerciseCarbonyl chloride (COCl2), also called phosgene, was used in World War I as a poisonous gas. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form carbonyl chloride

CO(g) + Cl2 ↔ COCl2(g)

At 74°C are [CO] = 1.2 x 10-2 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constant (Kc).

The equilibrium constant KP for the decomposition of phosphorous pentachloride to phosphorous trichloride and molecular chlorine

PCl5(g) PCl3(g) + Cl2(g)

Is found to be 1.05 at 250°C. If the equilibrium of the partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250°C?

The equilibrium constant Kp for the reaction: 2NO2(g) ↔ 2NO(g) + O2(g)


Is 158 at 1000 K. Calculate PO2 if PNO2 = 0.400 atm and PNO = 0.270 atm.

Methanol (CH3OH) is manufactured industrially by the reaction: CO(g) + 2H2(g) CH3OH(g)The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of KP at this temperature?

For the reaction: N2 (g) + 3H2 (g) ↔ 2NH3(g)KP is 4.3 x 10-4 at 375°C. Calculate Kc for the reaction.

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.

CaCO3 (s) CaO (s) + CO2 (g)

Kc’ = [CaO ] [CO2][CaCO3]

Kc = [CO2] = Kc’ x [CaCO3][CaO]

Kp = PCO2

[CaCO3] = constant

[CaO] = constant

CaCO3 (s) CaO (s) + CO2 (g)

PCO2 = Kp

PCO 2 does not depend on the amount of CaCO3 or CaO

Sample ExerciseWrite the equilibrium constant expression Kc, and Kp if applicable, for each of the following heterogeneous systems(a) (NH4)2Se(s) 2NH3(g) + H2Se(g) (b) AgCl(s) Ag+(aq) + Cl-(aq) (c) P4(s) + 6Cl2(g) 4PCl3(l)

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.


Write the equilibrium constant expression for Kc and Kp for the formation of nickel tetracarbonyl, which is used to separate nickel from other impurities: Ni(s) + 4CO(g) ↔ Ni(CO)4(g)

Consider the following heterogeneous equilibrium:CaCO3(s) CaO(s) + CO2(g)

At 800°C, the pressure of CO2 is 0.236 atm. Calculate (a)Kp and (b) Kc for the reaction at this temperature

Consider the following equilibrium at 395 K: NH4HS(s) ↔ NH3(g) + H2S(g)

The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction.

Review of ConceptsFor which of the following reactions is Kc equal to Kp?

(a) 4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g)

(b) 2H2O2(aq) ↔ 2H2O(l) + O2(g)

(c) PCl3(g) +3NH3(g) ↔ 3HCl(g) + P(NH2)3(g)

A + B C + D Kc’ Kc’ = [C ] [D ][A ][B]

Kc’’ = [E ] [F ][C ] [D ]

C + D E + F Kc’’

A + B E + F Kc Kc = [E ][F][A ][B]

so Kc = Kc’ x Kc’’

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.


N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g)

K = ¿¿ = 4.63 x 10-3 K’ = ¿¿ = 1K = 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.

Sample ExerciseThe reaction for the production of ammonia can be written in a number of ways:

(a) N2(g) + 3H2(g) 2NH3(g)(b)

12 N2(g) +

32 H2(g)

NH3(g)

(c) 13 N2(g) + H2(g)

23

H3(g)

Write the equilibrium constant expression for each formulation. (Express the concentrations of the reacting species in mol/L.)

(d) How are the equilibrium constants related to one another?

Practice ExerciseWrite the equilibrium expression (Kc) for each of the following reactions and show how they are related to each other:

(a) 3O2(g) ↔ 2O3(g)(b) O2(g) ↔

23 O3(g)

Review of ConceptsFrom the following equilibrium constant expression, write a balanced chemical equation for the gas-phase reaction.

Kc = ¿¿

Writing Equilibrium Constant Expressions• 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.

• The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.

• The equilibrium constant is a dimensionless quantity.

• In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.

• 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.


Chemical Kinetics and Chemical Equilibrium

A + 2B K f↔

K r AB2 Rate = kf [A][B]2 ratef = rater

k fkr

= Kc = [A B2][A ]¿¿

Rate = kr [AB2] kf [A][B]2 = kr [AB2]

Review of ConceptsThe equilibrium constant (Kc) for the reaction A ↔ B + C is 4.8 x 10-2 at 80°C. If the forward rate constant is 3.2 x 102 s-1, calculate the reverse rate constant.

The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.

IF• Qc > Kc system proceeds from right to left to reach equilibrium• Qc = Kc the system is at equilibrium• Qc < Kc system proceeds from left to right to reach equilibrium

Sample ExerciseAt the start of the reaction, there are 0.249 mol N2, 3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a 3.50-L reaction vessel at 375°C. If the equilibrium constant (Kc) for the reaction

N2(g) + 3H2(g) 2NH3(g)Is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed.

The equilibrium constant (Kc) for the formation of nitrosyl chloride, and orange-yellow compound, from nitric oxide and molecular chlorine 2NO(g) + Cl2(g) 2NOClis 6.5 x 104 at 35C. In a certain experiment, 2.0 x 10-2 mole of NO, 8.3 x 10-3 mole of Cl2. and 6.8 moles of NOCl are mixed in a 2.0-L flask. In which direction will the system proceed to reach equilibrium?

Review of ConceptsThe equilibrium constant (Kc) for the A2 + B2 ↔ 2AB reaction is 3 at a certain temperature. Which of the diagrams shown here corresponds to the reaction at equilibrium? For those mixtures that are not at equilibrium, will the net reaction move in the forward or reverse direction to reach equilibrium?


Calculating Equilibrium Concentrations1. 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.

Sample ExerciseA mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00-L stainless-steel flask at 430°C. The equilibrium constant Kc for the reaction H2(g) + I2(g) 2HI(g) is 54.3 at this temperature. Calculate the concentrations of H2, I2, and HI at equilibrium.

For the same reaction and temperature as in the prior example, suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium

Le Châtelier’s PrincipleIf 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.

Changes in Concentration


N2 (g) + 3H2 (g) ↔ 2NH3 (g)

Change Shifts the EquilibriumIncrease concentration of product(s) leftDecrease concentration of product(s) rightIncrease concentration of reactant(s) rightDecrease concentration of reactant(s) left

aA (g) + bB (g) cC (g) + dD (g)

Change Shifts the EquilibriumIncrease concentration of product(s)Decrease concentration of product(s)Increase concentration of reactant(s)Decrease concentration of reactant(s)

Sample ExerciseAt 720°C, the equilibrium constant Kc for the reaction: N2(g) + 3H2(g) 2NH3(g) is 2.37 x 10-3. In a certain experiment the equilibrium concentrations are [N2] = 0.683 M, [H2] = 8.80 M, and [NH3] = 1.05 M. Suppose some NH3 is added to mixture so that its concentration is increased to 3.65 M

(a) Use Le Chatelier’s principle to predict the shift in direction of the net reaction to reach a new equilibrium

(b) Confirm your prediction by calculating the reaction quotient QC and comparing its value with Kc

Practice ExerciseAt 430°C, the equilibrium constant (KP) for the reaction: 2NO(g) + O2(g) 2NO2(g)Is 1.5 x 105. In one experiment, the initial pressures of NO, O2, and NO2 are 2.1 x 10-3 atm, 1.1 x 10-2 atm, and 0.14 atm, respectively. Calculate QP, and predict the direction that the net reaction will shift to reach equilibrium

AddNH3

Equilibrium shifts left to offset stress


Changes in Volume and Pressure

A(g) + B(g) ↔ C(g)

Change Shifts the EquilibriumIncrease pressure Side with fewest moles of gasDecrease pressure Side with most moles of gasIncrease volume Side with most moles of gasDecrease volume Side with fewest moles of gasConsider the following equilibrium systems, predict the direction of the net reaction in each case as a result of increasing the pressure (decreasing the volume) on the system at constant temperature):

(a) 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) (b) PCl5(g) PCl3(g) + Cl2(g) (c) H2(g) + CO2(g) H2O(g) + CO(g)

Consider the equilibrium reaction involving nitrosyl chloride, nitric oxide, and molecular chlorine

2NOCl(g) ↔ 2NO(g) + Cl2(g)

Predict the direction of the net reaction as a result of decreasing the pressure (increasing the volume) on the system at constant temperature.

Review of ConceptsThe diagram here shows the gaseous reaction 2A ↔ A2 at equilibrium. If the pressure is decreased by increasing the volume at constant temperature, how would the concentrations of A and A2 change when a new equilibrium is established?

Changes in TemperatureChange Exothermic Rx Endothermic RxIncrease temperature K decreases K increasesDecrease temperature K increases K decreases

Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner

Catalyst lowers Ea for both forward and reverse reactions.

catalyzeduncatalyzed


Catalyst does not change equilibrium constantor shift equilibrium.

Le Châtelier’s PrincipleChange Equilibrium

Change Shift Equilibrium Constant Concentration yes no

Pressure yes* no

Volume yes* no

Temperature yes yes

Catalyst no no

*Dependent on relative moles of gaseous reactants and productsReview of ConceptsThe diagram shown here represents the reaction X2 + Y2 ↔ 2XY at equilibrium at two temperatures (T2 > T1). Is the reaction endothermic or exothermic?

Sample ExerciseConsider the following equilibrium process between dinitrogen tetrafluoride (N2F4) and nitrogen difluoride (NF2):

N2F4(g) 2NF2 (g) DH° = 38.5 kJ/mol

Predict the changes in the equilibrium if:

(a) The reacting mixture is heated at constant volume

(b) Some N2F4 gas is removed from the reacting mixture at constant temperature and volume

(c) The pressure of the reacting mixture is decreased at constant temperature

(d) A catalyst is added to the reacting mixture

Practice ExerciseConsider the equilibrium between molecular oxygen and ozone

3O2(g) ↔ 2O3(g) ΔH° = 284 kJ/mol

What would be the effect of:

(a) Increasing the pressure of the system by decreasing the volume

(b) Adding O2 to the system at constant volume

(c) Decreasing the temperature

(d) Adding a catalyst


Big Idea 6: Any bond or intermolecular attraction that can be formed can be broken. These two processes are in dynamic competition, sensitive to initial conditions and external perturbations.Duration: Early MarchTextbook Chapter: 14Enduring Understanding Essential Knowledge6.A: Chemical equilibrium is a dynamic, reversible state in which rates of opposing processes are equal.

6.A.1: In many classes of reactions, it is important to consider both the forward and reverse reaction.6.A.2: The current state of a system undergoing a reversible reaction can be characterized by the extent to which reactants have been converted to products. The relative quantities of reaction components are quantitatively described by the reaction quotient, Q.6.A.3: When a system is at equilibrium, all macroscopic variables, such as concentrations, partial pressures, and temperature, do not change over time. Equilibrium results from an equality between the rates of the forward and reverse reactions, at which point Q = K.6.A.4: The magnitude of the equilibrium constant, K, can be used to determine whether the equilibrium lies toward the reactant side or product side.

6.B: Systems at equilibrium are responsive to external perturbations, with the response leading to a change in the composition of the system.

6.B.1: Systems at equilibrium respond to disturbances by partially countering the effect of the disturbance (Le Chatelier’s principle).6.B.2: A disturbance to a system at equilibrium causes Q to differ from K, thereby taking the system out of the original equilibrium state. The system responds by bringing Q back into agreement with K, thereby establishing a new equilibrium state.

Learning Objective5.16 The student can use Le Chatelier’s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product.5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction.5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g., reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K.6.3 The student can connect kinetics to equilibrium by using reasoning about equilibrium, such as Le Chatelier’s principle, to infer the relative rates of the forward and reverse reactions.6.4 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use the tendency of Q to approach K to predict and justify the prediction as to whether the reaction will proceed toward products or reactants as equilibrium is approached.


6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained, calculate the equilibrium constant, K.6.6 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the conditions at equilibrium for a system involving a single reversible reaction.6.7 The student is able, for a reversible reaction that has a large or small K, to determine which chemical species will have very large versus very small concentrations at equilibrium.6.8 The student is able to use Le Chatelier’s principle to predict the direction of the shift resulting from various possible stresses on a system at chemical equilibrium.6.9 The student is able to use Le Chatelier’s principle to design a set of conditions that will optimize a desired outcome, such as product yield.6.10 The student is able to connect Le Chatelier’s principle to the comparison of Q to K by explaining the effects of the stress on Q and K.

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