Suppose human’s memory decays exponentially as a first order

1 / 78

# Suppose human’s memory decays exponentially as a first order - PowerPoint PPT Presentation

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?

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Suppose human’s memory decays exponentially as a first order' - gavin-valentine

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
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

Chemical Equilibrium

Equilibrium

Two opposite processes have reached the same rate.

Note: equilibrium is dynamic!

Reactants  Products

Reactants  Products

Reversible Reaction

Reactants ⇌ Products

kf

N2O4(g) ⇌ 2NO2(g)

kr

Both directions are elementary reactions

What are the differential rate laws?

rf = kf[N2O4]

rr = kr[NO2]2

kf

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

kf

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.

kf

N2O4(g) ⇌ 2NO2(g)

kr

rf = kf[N2O4]

rr = kr[NO2]2

kf[N2O4] = kr[NO2]2

K =

a A + b B ⇌ c C + d D

K: equilibrium constant.

[ ]: concentration at equilibrium!

K: independent of concentration,

dependent upon temperature.

recall d=m/V

K: no unit.

(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 ⇌ 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.
Haber process: N2(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

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

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

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) ⇌ 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) + O2(g) ⇌ 2NO2(g) K = 5.0 x 1012

What is the equilibrium constant for

NO2(g) ⇌NO(g) + ½ O2(g)

N2(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

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

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

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)

equilibrium constant expression for heterogeneous equilibria.

2H2O(l) ⇌ 2H2(g) + O2(g)

2H2O(g) ⇌ 2H2(g) + O2(g)

• PCl5(s) ⇌ PCl3(l) + Cl2(g)
• CuSO4 • 5H2O (s) ⇌ CuSO4(s) + 5H2O(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 ⇌ c C + d D

[ ]: concentration at equilibrium!

reaction quotient

[ ]: concentration at a particular moment.

a A + b B ⇌ 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: N2(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 Kc or Kp

equilibrium concentrations or pressures

N2O4(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

CO(g) + H2O(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

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.

H2(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

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.

H2(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

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 ⇌ 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.
value of Kc or Kp

equilibrium concentrations or pressures

Example 14.5, page 628

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

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?

a A + b B ⇌ 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

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

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

As4O6(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
Effect of a change in pressure

1) Add or remove a gaseous reactant or product.

Effect of a change in pressure

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 NH3(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

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

• each of the following processes when the volume is reduced.
• P4(s) + 6Cl2(g) ⇌ 4PCl3(l)
• PCl3(g) + Cl2(g) ⇌ PCl5(g)
• PCl3(g) + 3NH3(g) ⇌ P(NH2)3(g) + 3HCl(g)
Effect of a change in temperature

K is a function of temperature

Do experiments

(brown) 2NO2⇌ 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

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

• changes as the temperature is increased.
• N2(g) + O2(g) ⇌ 2NO(g) ∆H = 181 kJ
• 2SO2(g) + O2(g) ⇌ 2SO3(g) ∆H = −198 kJ
Le Châtelier's Principle

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 ⇌ B

∆Ea

Ef’

Er’

Catalyst does not shift equilibrium

For the reaction
• 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?