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Suppose human’s memory decays exponentially as a first order reaction and the half life is one day. Three days have passed since you came to class last Friday. If you have not reviewed what you learned, what is the percentage of chemistry you remember from last Friday?

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Suppose human’s memory decays exponentially as a first order

reaction and the half life is one day. Three days have passed

since you came to class last Friday. If you have not reviewed

what you learned, what is the percentage of chemistry you

remember from last Friday?

IF you need the equations for first order reaction, here they are

r = k [A]


Chapter 14 order

Chemical Equilibrium


phase equilibrium order

Liquid Gas


Equilibrium order

Two opposite processes have reached the same rate.

Note: equilibrium is dynamic!


Reactants order Products

Reactants  Products

Reversible Reaction

Reactants ⇌ Products


k orderf

N2O4(g) ⇌ 2NO2(g)

kr

Both directions are elementary reactions

What are the differential rate laws?

rf = kf[N2O4]

rr = kr[NO2]2


k orderf

N2O4(g) ⇌ 2NO2(g)

kr

Suppose we start with a flask of pure N2O4 gas

rf = kf[N2O4]

rr = kr[NO2]2

At equilibrium, the forward and reverse reactions are proceeding at the same rate. rf = rr


k orderf

N2O4(g) ⇌ 2NO2(g)

rf = kf[N2O4] = rr = kr[NO2]2

kr

Once equilibrium is achieved, the concentration of each reactant and product remains constant.


k orderf

N2O4(g) ⇌ 2NO2(g)

kr

rf = kf[N2O4]

rr = kr[NO2]2

kf[N2O4] = kr[NO2]2

K =


a A + b B order⇌ c C + d D

K: equilibrium constant.

[ ]: concentration at equilibrium!

K: independent of concentration,

dependent upon temperature.

recall d=m/V

K: no unit.


Write the equilibrium expression for K for the following reactions:

(a) 2O3(g) ⇌ 3O2(g)

(b) 2NO(g) + Cl2(g) ⇌ 2NOCl(g)

(c) Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq)

a A + b B ⇌ c C + d D


At one temperature: one K, infinite number of equilibrium concentrations.

K is independent of concentrations. Recall d = m/V.



a A + b B concentrations.⇌ c C + d D

  • If K>>1, the reaction is product-favored; product predominates at equilibrium.

  • If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.


N concentrations.2(g) + 3H2(g) ⇌ 2NH3(g)

Haber process



Haber process: N Fertilizer2(g) + 3H2(g) ⇌ 2NH3(g)

  • The following equilibrium concentrations were observed for the Haber process

  • at 127 °C: [NH3] = 3.1 x 10−2 mol/L, [N2] = 8.5 x 10−1 mol/L,

  • [H2] = 3.1 x 10−3 mol/L.

  • Calculate the value of K at 127 °C for this reaction.

  • Calculate the value of the equilibrium constant at 127 °C for the reaction

  • 2NH3(g) ⇌ N2(g) + 3H2(g)

c) Calculate the value of the equilibrium constant at 127 °C for the reaction

given by the equation

⇌ NH3(g)


reverse Fertilizer

a A + b B ⇌ c C + d D

c C + d D⇌a A + b B

x n

na A + nb B ⇌ nc C + nd D


For practice 14.2, page 622 Fertilizer

Consider the following chemical equation and equilibrium

constant at 25 °C:

2COF2(g) ⇌ CO2(g) + CF4(g) K = 2.2 x 106

Compute the equilibrium constant for the following reaction

At 25 °C:

2CO2(g) + 2CF4(g) ⇌ 4COF2(g) K’ = ?


Equilibrium constant of a composite reaction Fertilizer

2 NOBr (g) ⇌ 2 NO (g) + Br2(g)

Br2 (g) + Cl2 (g) ⇌ 2 BrCl (g)

+

2NOBr (g) + Cl2 (g) ⇌ 2NO (g) + 2BrCl (g)

Sum of reactions  Product of equilibrium constants


HF (aq) Fertilizer⇌ H+ (aq) + F− (aq)

K1 = 6.8 x 10−4

H2C2O4 (aq) ⇌ 2H+ (aq) + C2O42− (aq)

K2 = 3.8 x 10−6

Determine the equilibrium constant for

2HF (aq) + C2O42− (aq) ⇌ 2F− (aq) + H2C2O4 (aq)

K3 = 0.12

Method: take linear combination of known reactions to construct target reaction.


2NO(g) + O Fertilizer2(g) ⇌ 2NO2(g) K = 5.0 x 1012

What is the equilibrium constant for

NO2(g) ⇌NO(g) + ½ O2(g)


N Fertilizer2(g) + 3H2(g) ⇌ 2NH3(g)

= Kc

a A (g) + b B (g) ⇌ c C (g) + d D (g)

∆n = Sum of the coefficients of gaseous products

− Sum of the coefficients of gaseous reactants


Example 14.3, page 624 Fertilizer

Kp for the following reaction is 2.2 x 1012 at 25 °C, calculate

the value of Kc.

2NO(g) + O2(g) ⇌ 2NO2(g)

Try For Practice 14.3 on the same page


Consider the equilibrium Fertilizer

N2(g) + O2(g) + Br2(g) ⇌ 2NOBr(g)

Calculate the equilibrium constant Kp for this reaction, given

The following information (at 298 K):

2NO(g) + Br2(g) ⇌ 2NOBr(g) Kc, 1 = 2.0

2NO(g) ⇌ N2(g) + O2(g) Kc, 2 = 2.1 x 1030

Answer: Kp = 4.3 x 1028



Homogeneous Equilibrium Fertilizer

All species have the same phase

N2(g) + 3H2(g) ⇌ 2NH3(g)

Heterogeneous Equilibrium

Not all species have the same phase

Kc, Kp

2CO(g) ⇌ CO2(g) + C(s)


The concentration of pure solid or liquid is not included in the

equilibrium constant expression for heterogeneous equilibria.


2H the2O(l) ⇌ 2H2(g) + O2(g)

2H2O(g) ⇌ 2H2(g) + O2(g)


blue

white


Hydrated Copper (II) Sulfate on the Left. Water Applied to Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound


a A + b B Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound⇌ c C + d D

[ ]: concentration at equilibrium!

reaction quotient

[ ]: concentration at a particular moment.


a A + b B Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound⇌ c C + d D

Q = K, system (mixture) is at equilibrium, rr = rf

Q > K → rr > rf → system will shift to left to

reach equilibrium

Q < K → rr < rf → system will shift to right to

reach equilibrium


Haber process: N Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound2(g) + 3H2(g) ⇌ 2NH3(g)

  • At 500 °C, K = 6.01 x 10−2 . Predict the direction in which the

  • system will shift to reach equilibrium in each of the following

  • cases:

  • [NH3]0=1.0 x 10−3 M; [N2]0 = 1.0 x 10−5 M; [H2]0 = 2.0 x10−3 M

  • [NH3]0=2.00 x 10−4 M; [N2]0 = 1.50 x 10−5 M; [H2]0 =3.54 x10−1 M

  • [NH3]0=1.0 x 10−4 M; [N2]0 = 5.0 M; [H2]0 = 1.0 x10−2 M

b) Q = 6.01 x 10−2

c) Q = 2.0 x 10−3

a) Q = 1.3 x 107


value of K Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compoundc or Kp

equilibrium concentrations or pressures


N Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound2O4(g) ⇌ 2NO2(g)

N2O4 was placed in a flask and allowed to reach equilibrium

at a temperature where Kp = 0.133. At equilibrium, the pressure

of N2O4 was found to be 2.71 atm. Calculate the equilibrium

pressure of NO2 (g).

0.600 atm


Try Example 14.8 and For Practice 14.8 Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

on page 632


CO(g) + H Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound2O(g) ⇌ CO2(g) + H2(g)

At 700 K the equilibrium constant is 5.10. Calculate the

equilibrium concentrations of all species if 1.000 mol of

each component is mixed in a 1.000 L flask.

x = 0.387 (M)


Example 14.11, page 636 Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

Consider the following reaction:

I2(g) + Cl2(g) ⇌ 2ICl(g) Kp = 81.9

A reaction mixture at 25 C initially contains PI2 0.100 atm, PC12 0.100 atm, and PICl = 0.100 atm. Find the equilibrium partial pressures of I2, Cl2, and ICl at this temperature.


Try Example 14.9 and For Practice 14.9 Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

on page 634.


H Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound2(g) + F2(g) ⇌ 2HF(g)

3.000 mol H2 and 6.000 mol F2 are mixed in a 3.000 L flask.

K at this temperature is 115. Calculate the equilibrium

concentration of each component.

x = 2.14 (M) or 0.968 (M)


Example 14.10, page 634 Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

N2O4(g) ⇌ 2NO2(g)

Kc = 0.36 at 100 °C. A reaction mixture at 100 °C initially

contains [NO2] = 0.100 M. Find the equilibrium concentration

of NO2 and N2O4 at this temperature.


H Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound2(g) + I2(g) ⇌ 2HI(g)

Kp = 1.00 x 102. Suppose HI at 5.000 x 10−1 atm, H2 at

1.000 x 10−2 atm, and I2 at 5.000 x 10−3 atm are mixed in a

5.000-L flask. Calculate the equilibrium pressure of each

component.

x = 3.55 x 10−2 (atm) or −7.19 x 10−2 (atm)


Example 14.12, page 638 Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

2H2S(g) ⇌ 2H2(g) + S2(g)

Kc = 1.67 x10−7 at 800 °C

A 0.500 L reaction vessel initially contains 0.0125 mol of H2S

at 800 °C. Find the equilibrium concentration of H2 and S2.


a A + b B Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound⇌ c C + d D

  • If K>>1, the reaction is product-favored; product predominates at equilibrium.

  • If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.


Try Example 14.13 on page 638. Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound


value of K Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compoundc or Kp

equilibrium concentrations or pressures


Example 14.5, page 628 Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

Consider the following reaction:

CO(g) + 2H2(g) ⇌ CH3OH(g)

A reaction mixture at 780 C initially contains [CO] = 0.500 M and [H2]= 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of the equilibrium constant?


Example 14.6, page 628 Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

Consider the following reaction:

2CH4(g) ⇌ C2H2(g) + 3H2(g)

A reaction mixture at 1700 C initially contains [CH4] = 0.115 M. At equilibrium, the mixture contains [C2H2] = 0.035 M. What is the value of the equilibrium constant?


Try For Practice 14.5 and 14.6 on page 629. Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound


a A + b B Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound⇌ c C + d D

Q = K, system (mixture) is at equilibrium.

Q > K → rr > rf → system will shift to left to

reach equilibrium

Q < K → rr < rf → system will shift to right to

reach equilibrium


Effect of a change in concentration Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

a A + b B ⇌ c C + d D

add reactants → reactant concentrations ↑ → Q < K

→ system shifts to right

remove reactants → reactant concentrations ↓ → Q > K

→ system shifts to left


Effect of a change in concentration Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

a A + b B ⇌ c C + d D

add products → product concentrations ↑ → Q > K

→ system shifts to left

remove products → product concentrations ↓ → Q < K

→ system shifts to right


Haber process: N Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound2(g) + 3H2(g) ⇌ 2NH3(g)

concentration


As Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound4O6(s) + 6C(s) ⇌ As4(g) + 6CO(g)

  • Predict the direction of the shift of the equilibrium position in

  • response to each of the following changes in conditions.

  • Addition of CO

  • Addition or removal of C or As4O6

  • c) Removal of As4


Try Example 14.14 and For Practice 14.14 Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

on page 644.


Effect of a change in pressure Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

1) Add or remove a gaseous reactant or product.


Haber process: N Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound2(g) + 3H2(g) ⇌ 2NH3(g)


Effect of a change in pressure Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

1) Add or remove a gaseous reactant or product.

2) Add an inert gas (one not involved in the reaction).

Equilibrium does not shift

3) Change the volume of the container.


(a) A Mixture of NH Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound3(g), N2(g), and H2(g) at Equilibrium (b) The Volume is Suddenly Decreased (c) The New Equilibrium Position for the System Containing More NH3 and Less N2 and H2 compared to the old equilibrium (a)

N2(g) + 3H2(g) ⇌ 2NH3(g)


Effect of a change in pressure Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

1) Add or remove a gaseous reactant or product.

2) Add an inert gas (one not involved in the reaction).

Equilibrium does not shift

3) Change the volume of the container.

decrease volume → equilibrium shifts to the direction with less

moles of gases

increase volume → equilibrium shifts to the direction with more

moles of gases




a A + b B for ⇌ c C + d D


Effect of a change in temperature for

K is a function of temperature

Do experiments


(brown) 2NO for 2⇌ N2O4 (colorless)

∆H = − 58 kJ

∆H < 0, exothermic

∆H > 0, endothermic

0 °C

100 °C

decrease T → shifts to right → K increases

increase T → shifts to left → K decreases


Exothermic reactions for

decrease T → shifts to right → K increases

increase T → shifts to left → K decreases

Endothermic reactions

decrease T → shifts to left → K decreases

increase T → shifts to right → K increases




Le Châtelier's Principle of K

If a change is imposed on a system at equilibrium,

the position of the equilibrium will shift in a direction

that tends to reduce that change.



A of K⇌ B


A of K⇌ B

∆Ea

Ef’

Er’

Catalyst does not shift equilibrium


  • For the reaction of K

  • PCl5(g) ⇌ PCl3(g) + Cl2(g) ∆H = 87.9 kJ

  • In which direction will the equilibrium shift when

  • Cl2(g) is removed,

  • The temperature is decreased,

  • The volume of the system is increased,

  • PCl3(g) is added?

  • In b), will the equilibrium constant increase or decrease?

  • In d), will the concentration of Cl2 and PCl5 increase or

  • decrease?


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