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Chemical Equilibrium

Chapter 15 Homogeneous equilibrium:

When all reacting species are in the same phase. Heterogeneous equilibrium:

When all reacting species are NOT in the same phase. Heterogeneous Equilibrium:

When reacting species are in DIFFERENT phases. SOLID and LIQUID phases are EXCLUDED from the equilibrium expression because they do not have “

A state achieved when the RATES of the forward and reverse reactions are EQUAL & the concentrations of the reactants and products remain constant.

The equilibrium constant (Keq) is independent of changes, but dependent on the . Writing Equilibrium Constant Expressions

For the general reaction α A + β B ↔ c C + d D α β Forward Rate = kF [A] [B] c d Reverse Rate = kR [C] [D] At Equilibrium Forward Rate = Reverse Rate Forward Rate = Reverse Rate Writing Equilibrium Constant Expressions

For the general reaction α A + β B ↔ c C + d D At Equilibrium α β c d kF [A] [B] = kR [C] [D] Put CONSTANTS on one side of equation And Concentrations on other side of equation Equilibrium Constant

• The equilibrium expression compares reactant & concentrations. [Products]p K = [Reactants]r Size of Equilibrium Constant

3 If Kc > 10 , products predominate

–3 If Kc < 10 , reactants predominate

–3 3 If Kc is in the range 10 to 10 , appreciable concentrations of both reactants and products are present. Writing Equilibrium Constant Expressions

α A + β B ↔ c C + d D At Equilibrium α β c d kF [A] [B] = kR [C] [D]

» ÿ c d k F … C ⁄Ÿ [D ] K eq = = β » ÿ k R α B … A ⁄Ÿ [ ]

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The point at which the rate of decomposition

N2O4(g) → 2NO2(g) equals the rate of dimerization

2NO2(g) → N2O4(g). is a .

The equilibrium is dynamic because the reaction has not stopped: the opposing rates are equal. N 2 O 4 ( g ) 2 N O 2 ( g )

At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4 The double arrow implies the process is dynamic

Forward reaction: Rate = kf[N2O4]

Reverse reaction: B → A Rate = kr[NO2]

At equilibrium kf[N2O4] = kr[NO2] The Direction of the Chemical

Equation and Keq

An equilibrium can be approached from any direction.

For Example: N2O4(g) 2NO2(g) has P2 K = NO2 = 6.46 eq P N2O4 In the reverse direction:

2NO2(g) N2O4(g)

PN O 1 K = 2 4 = 0.155 = eq P2 6.46 NO2 Other Ways to Manipulate Chemical Equations and Keq Values

The reaction 2N2O4(g) 4NO2(g) P4 NO2 Has Keq = P2 N2O4 • which is the square of the equilibrium constant for N2O4(g) 2NO2(g) Equilibrium Concentration

Amounts of components are given as molarity { moles solute / liters of or of a gas

2 2 [NO ] P = 2 NO2 Kc [ ] Kp = N2O4 PN2O4

Equilibrium Constant

Write the Kp expressions for:

2 N2O5(g) ↔ 4 NO2(g) + O2(g)

4 2 Kp = (P NO2 ) (P O2) / (P N2O5) Equilibrium Constant

Write the Kc expressions for:

2 N2O5(g) ↔ 4 NO2(g) + O2(g)

4 2 KC = [NO2 ] [ O2] / [N2O5] 2 N2O5(g) ↔ 4 NO2(g) + O2(g)

The Kp and Kc expressions :

4 2 Kp = (P NO2 ) (P O2) / (P N2O5)

4 2 KC = [NO2 ] [ O2] / [N2O5]

Is Kp = KC ? No ! Relation between Gas Pressure and Concentration

• P V = n R T Ideal Gas Equation • Concentration (M) = moles / Liter • M = n / V • Rearranging P V = n R T n P M = = V RT When are KP and KC EQUAL ?

For which of the following is KP = KC

(a) CO2(g) + C(s) ↔ 2 CO(g)

2+ 2+ (b) Hg(l) + Hg (aq) ↔ Hg2 (aq)

(c)2Fe(s) + 3H2O(g) ↔ Fe2O3(s) + 3H2(g)

(d) 2 H2O(l) ↔ 2 H2(g) + O2(g) Homogeneous And Heterogeneous Equilibria

• Homogeneous Equilibrium When all reacting species are in the same phase.

• Heterogeneous Equilibrium: When all reacting species are NOT in the same phase. Heterogeneous Equilibrium:

When reacting species are in DIFFERENT phases. SOLID and LIQUID phases are EXCLUDED from the equilibrium expression because they do not have “concentrations” Equilibrium Constant

The equilibrium concentrations at 1000 K for the reaction

CO(g) + Cl2(g) ↔ COCl2(g) are

–2 [CO] = 1.2x10 M, [Cl2]=0.054 M,

[COCl2] = 0.14 M

Calculate Kc and Kp. CO(g) + Cl2(g) ↔ COCl2(g)

–2 [CO] = 1.2x10 M, [Cl2]=0.054 M, [COCl2] = 0.14 M

Kc = [COCl2] / [CO][Cl2]

–2 Kc = 0.14 /(1.2x10 )(0.054 ) = 21578.2984 and

KC = KP (RT) Why ?

Kp = KC / RT = KC / (0.0821)(1000) = Sulfur dioxide + Oxygen = Sulfur trioxide

SO2 (g) + O2 (g) = SO3 (g)

2 SO2 (g) + O2 (g) = 2 SO3 (g)

2 [SO3 ] KC = 2 [SO2 ] [O2 ] The Initial Change Equilibrium method Is used to calculate equilibrium constant

2 SO2 (g) + O2 (g) = 2 SO3 (g) Initial Moles # # #

Change in Moles –2x -x + 2x Equilibrium # - 2x # - x # + 2x Finally Concentrations at Equilibrium The Initial Change Equilibrium method Is used to calculate equilibrium constant

1.00 mole of SO2 and 1.00 mole of O2 are put into a 1.00 liter flask. At equilibrium, 0.0925

mole of SO3 is formed. Calculate Kc at 1000 K for the reaction

2 SO2 (g) + O2 (g) = 2 SO3 (g) The Initial Change Equilibrium method Is used to calculate equilibrium constant

2 SO2 (g) + O2 (g) = 2 SO3 (g) Initial Moles 1.00 1.00 0

Change in Moles -0.925 -0.925/2 +0.925 Equilibrium 0.075 0.537 0.925 Equil. Conc 0.075 M 0.537 M 0.925 M The Initial Change Equilibrium method Is used to calculate equilibrium constant

2 [SO3 ] KC = 2 [SO2 ] [O2 ]

≈ ’2 ∆ 0.925÷ K = « ◊ = 2.8x102 C ∆≈ ÷’2 ∆≈ ÷’ « 0.075◊ « 0.537◊ Methane (CH4) reacts with sulfide to yield H2 and carbon disulfide

What is the value of Kp if the partial pressures in an equilibrium mixture

are 0.20 atm of CH4, 0.25 atm of H2S,

0.52 atm of CS2, and 0.10 atm of H2? How do you solve the problem?

• 1st Write Reaction • 2nd Balance Reaction • 3rd Write Equilibrium Constant Equation • 4th Put Equilibrium Pressures in Equation • Do the Arithmetic Using Equilibrium Constants

The equilibrium constant (K c ) for the formation of nitrosyl chloride,

2 NO(g) + Cl2(g) ↔ 2 NOCl(g) from & chlorine gas is 6.5 x 10 4 at 35 °C. 4 2NO(g) + Cl2(g) ↔ 2NOCl(g) Kc= 6.5x10

2.0 x 10–2 moles of NO, –3 8.3 x 10 moles of Cl2, & 6.8 moles of NOCl are mixed in a 2.0–L flask.

Is the system at equilibrium? If not, what will happen? 4 2NO(g) + Cl2(g) ↔ 2NOCl(g) Kc= 6.5x10

2 4 [ NOCl ] K = 6 .5 x10 = 2 [ NO ] [ Cl 2 ]

2.0 x 10–2 moles of NO, 8.3 x 10–3 moles of

Cl2, 6.8 moles of NOCl in a 2.0–L flask.

[3.4]2 = 2.8x107 [1.0x10−2 ]2[4.15x10−3 ] The (Qc)

Predicts reaction direction.

Qc > Kc System proceeds to form reactants.

Qc = Kc System is at equilibrium.

Qc < Kc System proceeds to form products. Using Equilibrium Constants

•A mixture of 0.500mol H2 and 0.500mol I2 was placed in a 1.00–L stainless steel flask at

700°C. The equilibrium constant Kc is 57

H2(g) + I2(g) ↔ 2HI(g) •Calculate the equilibrium concentrations. Le Chatelier’s principle:

If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset. Le Chatelier’s principle:

Stress may be changes in • concentration, • pressure, volume, or • temperature that removes the system from equilibrium. Concentration Changes: The concentration stress of an added reactant or product is relieved by reaction in the direction that consumes the added substance. Concentration Changes:

The concentration stress of a removed reactant or product is relieved by reaction in the direction that replenishes the removed substance. Use the synthesis of as an example

N2(g) + 3H2(g) ↔ 2NH3(g)

At equilibrium [N2 ] = 0.50M [H2 ] = 3.00M

and [NH3 ] = 1.98 M at 700K N2(g) + 3H2(g) ↔ 2NH3(g)

At equilibrium [N2 ] = 0.50M [H2 ] = 3.00M

and [NH3 ] = 1.98 M at 700K what happens when the concentration

of N2 is increased to 1.50M? N2(g) + 3H2(g) ↔ 2NH3(g)

when the concentration of N2 is increased Le Châtelier’s principle says the reaction will relieve the stress by

converting the N2 to NH3 N2(g) + 3H2(g) ↔ 2NH3(g) Kc = 0.291 at 700K

If H2 is increased

Le Châtelier’s principle tells us the reaction will relieve the stress by

converting the extra H2 to NH3 N2(g) + 3H2(g) ↔ 2NH3(g) Kc = 0.291 at 700K

If NH3 is removed

Le Châtelier’s principle tells us the reaction will relieve the stress by

producing more NH3 Volume and Pressure Changes P V = n R T Pressure is inversely proportional to volume Increasing pressure = Decreasing volume From P V = n R T P = M (RT) increasing pressure or decreasing volume increases concentration. N2(g) + 3H2(g) ↔ 2NH3(g)

Use the synthesis of ammonia as an example There are Four (4) units of gas reactants but only Two (2) units of gas products

Therefore, an increase in pressure would shift the equilibrium to the right Le Châtelier’s Principle

Volume and Pressure Changes Only reactions containing gases are affected by changes in volume and pressure. The reaction of iron (III) oxide with carbon monoxide occurs in a blast furnace when iron ore is reduced to iron metal:

Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g) Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)

• Use Le Châtelier’s principle to predict the direction of reaction when an equilibrium mixture is disturbed by: (a) Adding CO SHIFT TO RIGHT Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)

• Use Le Châtelier’s principle to predict the direction of reaction when an equilibrium mixture is disturbed by:

(b) Removing CO2 SHIFT TO RIGHT Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)

• Use Le Châtelier’s principle to predict the direction of reaction when an equilibrium mixture is disturbed by: (c) Increasing Pressure NO EFFECT Why ???? Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)

• Use Le Châtelier’s principle to predict the direction of reaction when an equilibrium mixture is disturbed by:

(d) Adding Fe2O3 NO EFFECT Why ???? Temperature Changes:

1. Changes the equilibrium constant

2. Endothermic processes are favored when temperature increases.

3. Exothermic processes are favored when temperature decreases. Le Châtelier’s Principle & Temperature In the first step of the Ostwald process for synthesis of nitric , ammonia is oxidized to monoxide

4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g) ∆H° = –905.6 kJ 4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g) ∆H° = –905.6 kJ

What happens as temperature is increased ? Reaction is EXO thermic therefore an increase in temperature will shift the equilibrium toward the reactants Le Châtelier’s Principle : No effect.