Chemistry203 TermTest2 - Learnfaster.calearnfaster.ca/wp-content/uploads/chem203/TT2/C203-TT2...Chemistry203 –TermTest2 1 (C) Pavel Sedach Learnfaster.ca Questions? Suggestions?

Chemistry 203 – Term Test 2 1

(C) Pavel SedachLearnfaster.ca

Questions? Suggestions? Email me and I will try to find time to reply. [email protected]

All the best!

Lecture Slides Booklet Solutions Test Solutions Extra notes and advice Advice on which exams and questions to write Corrections

Available Today at: http://learnfaster.ca/blog/chem203termtest2/

Friday

3:30 to 5:00 ICF, ICE tables and Kinetics

5:00 to 5:30 Break

5:30 to 6:45 Equilibrium

6:45 to 7:00 Break

7:00 to 8:20 Acids and Bases

8:20 to 8:30 Break

8:30 to 9:30 Review

Plotting a stoichiometric reaction

© Pavel SedachLearnfaster.ca

2

all values in moles 3 CaC2(s) + 2 H3PO4 l → 3 H2C2 g +Ca3 PO4 2 s

Initial 0.235 0.288 0 0

Change 3x 2x +3x +x

Final 0 0.0742 0.235 0.0783

time

Concentration (mol/L)

H2C2 g

[CaC2(s)]

Ca3 PO4 2 s

Initial moles Final moles

.235

3= 0.0783 mol = x

CaC2(s) H3PO4 l H2C2 g Ca3 PO4 2 s CaC2(s) H3PO4 l H2C2 g Ca3 PO4 2 s

0.25

0.20

0.10

0.05

0.15

0

[H3PO4 l ]

Kinetics

Rate =Change in Concentrat ion

Change in Time=

ΔΔt

Equ i l ib r ium (a Thermodynamic Qua l i ty ) vs . K inet ics

• Equilibrium deals with the extent of reaction – does it happen? – depends on ΔH

• For instance , do d iamonds last forever? No – they degrade into carbon

• Kinetics answers the question how fast does it happen? –millions of years. – depends on ΔEa


3

Rate = −Δ AaΔt

= −Δ BbΔt

= +Δ CcΔt

= +Δ DdΔt

The rate of consumption of iodide, I aq− in the following elementary reaction is 5.40

mol I aq−

L s:

S2O8(aq)2− + 3 I aq

− → I3 aq + 2 SO4(aq)2−

What is the rate of formation of SO4(aq)2− ?

Method 1 Method 2

a) 1.80 mol SO4(aq)

2−

L s

b) 3.60 mol SO4(aq)

2−

L s

c) 5.40 mol SO4(aq)

2−

L s

d) 8.10 mol SO4(aq)

2−

L s

Kinetics – how fast does a reaction go? aA + bB → cC + dD


4

Which of the following are differential rate laws?


5

A differential rate law is either rate = k A m B n C o or d X

dt= k A m B n C o

By definition, a differential is a change with respect to time or another variable, so if we have rate =d A

dt, that is not a differential

because the time cancels – remember, rate is M/s.

a)d X

dt= k A 2

b) rate = k A 2 B

c) rate = −1

2∙d A

dt

d) −d A

dt= k A 2

e) rate = k

f) rate = −1

2∙d A

dt

Measuring rate:


Spectrometric –Monitor change in color due to appearance/disappearance of species this was one of your labs and has beenon previous exams!

A = 𝜺𝒍𝑐

𝑦 = 𝒎𝑥 + 𝑏

6

slope

Absorbance = absorptioncoefficient

× length × concentration

what is the yintercept?

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 0.25 0.5 0.75 1 1.25 1.5 1.75

A (arbitrary units)


What is the concentration when A is 0.23?

Only 20% was transmitted so absorbance is 80% or 0.8!

Measuring Reaction Rates


7

As [reactant] decreases, rate also decreases. With fewer reactants, there is less opportunity for collisions.

H2O2 aq → H2O l +½ O2 g

H2O2 aq

time (minutes)*can be seconds to trick

Instantaneous Rate

Initial Rate

Average Rate

0 1 2 3 4 5 6 7 8 9

What was the average rate of rxn. from 0 to 5 seconds? Provide units.

What was the instantaneous rate of formation of oxygen at 1.5 seconds?

Slope = 2.30

Slope = 2.10

aA + bB → cC + dD

Rate = k A m B n

If m =1 , the reaction is said be first order with respect to A. If n = 2, the reaction is said to be second-order with respect to B.

The sum of all the exponents (m+n) gives the overall order of the reaction.

Orders of reaction can be (but usually aren’t) integers, fractions, and even negative.

• m, n DO NOT equal a, b UNLESS it’s an ELEMENTARY step.

• An elementary step is like a single physical action.

• The above reaction may occur in five (5) steps! (so it is NOT elementary)

Rate Laws


Based on Reactants ONLY

m, n DO NOT equal a, b

8

(UNLESS it’s an ELEMENTARY step)

Rate Laws


For the reaction: C3H8 + 5 O2 ⟶ 3 CO2 + 4 H2O, if the reaction is first order with respect to C3H8 and second order with

respect to O2, the rate law of the reaction would be:

If the concentration of O2 is doubled, how would the rate change?

Tripled?

Halved?

If the concentration of C3H8 is doubled, how would the rate change?

Tripled?

9

Method of Initial Rates


For the reaction: 2 I aq− + S2O8

2−aq → I2 aq + 2 SO4(aq)

2− the initial rates of reaction were determined at a fixed temperature of

298K:

Calculate the order with respect to S2O82−

aq

Method 1 (actual math – works 100% of the time) Method 2 (fast method – works most of the time)

Rate = k I aq− 𝑚

S2O82−

aq

𝑛

Rate1Rate3

=

# I aq−

0(mol/L) S2O8

2−aq 0

(mol/L) Initial Rate (mol/L s)

1 0.080 0.040 1.25 × 10−4

2 0.040 0.040 6.25 × 10−5

3 0.080 0.020 6.25 × 10−5

10

Method of Initial Rates


For the reaction: 2 l aq− + S2O8

2−aq → I2 aq + 2 SO4(aq)

2− the initial rates of reaction were determined at a fixed temperature of

298K:

Calculate the rate law and the rate constant of the reaction.

What is the overall reaction order? What does the rate constant tell you?

11

# I aq−

0(mol/L) S2O8

2−aq 0

(mol/L) Initial Rate (mol/L s)

1 0.080 0.040 1.25 × 10−4

2 0.040 0.040 6.25 × 10−5

3 0.080 0.020 6.25 × 10−5

Units of the Rate Constant


The units for the rate constant (k) are dependent upon the overall order of the reaction.

Rate = k A m B n

Rate = k A 0 Rate = k A 1 Rate = k A 2 Rate = k A 5

12

0 Order

1st Order

2nd Order

Recognizing 0, 1st and 2nd Order Kinetics 1 A → 1B+ 1 C © Pavel SedachLearnfaster.ca

[A]

time

13

A 0 order means the reaction does NOT depend on this reactant

rate = k A 0

1st order can mean the reactant decays into a product

rate = k A 1

If a reagent is 2nd order, this means it likely collides with itself

All the reactions in thisgraph have the same k (normally impossible)

rate = k A 2

0

5

10

15

20

0 25 50 75 100

0 Order Kinetics 1st Order Kinetics 2nd Order Kinetics© Pavel SedachLearnfaster.ca

[A]

ln[A]

1A

ln A t = −kt + ln A 0

1A t

= kt +1A 0time (s)

14

0

0.2

0.4

0.6

0.8

1

0 25 50 75 100

2.5

1.5

0.5 0 25 50 75 100

time (s)

time (s)

A t = −kt + A 0y = mx + b

time (s)

time (s)

time (s)

[A]

ln[A]

1A

0

0.5

1

1.5

0 50 100 150 200

3

2

1

0

0 50 100 150 200

0

5

10

15

20

0 50 100 150 200

0

0.5

1

1.5

0 150 300 450 600 750

2.8

2.1

1.4

0.7

0

0 150 300 450 600 750

0

5

10

15

20

0 150 300 450 600 750time (s)

time (s)

time (s)

[A]

ln[A]

1A

1st order (fast takeaways)


15


time (s)

time (s)

time (s)

[A]

1A

0

0.2

0.4

0.6

0.8

1

1.2

0 50 100 150 200

4

2

0

0 50 100 150 200

0

10

20

0 50 100 150 200


ln[A]

Time (s) Concentration [A](mol/L)

0 order (fast takeaways)


16A t = −kt + A 0

0

10

20

0 25 50 75 100

[A]

ln[A]

1A

time (s)

0

0.2

0.4

0.6

0.8

1

0 25 50 75 100

2.5

0.5 0 25 50 75 100

time (s)

time (s)

A t = −kt + A 0


2nd order (fast takeaways)


17

1A t

= kt +1A 0

1A t

= kt +1A 0

0

0.2

0.4

0.6

0.8

1

1.2

0 150 300 450 600 750

2.8

2.1

1.4

0.7

0

0 250 500 750

0

10

20

0 150 300 450 600 750time (s)

time (s)

time (s)

[A]

ln[A]

1A


Rates – F2016 Q 15 – Practically Guaranteed to be on Term Test 2


18

1. What is the order of this reaction with respect to A2 g ?

2. What is the average rate of reaction from 0 to 400 seconds?

3. What is the instantaneous rate of reaction at 200 seconds?

4. What is the rate constant for this reaction?

5. What is the instantaneous rate of disappearance of A2 g at 200 seconds?

0

0.5

1

1.5

0 150 300 450 600 750

time (s) time (s) time (s)

A2

3 A2 g ⇌ 2 A3 g , answer the questions below in terms of variables M, X or Y, adding units where needed:

0

5

10

15

20

0 150 300 450 600 750

1

A2 g

0

0.2

0.4

0.6

0.8

0 200 400 600 800

A3Slope = +M

Slope = XSlope = Y

The Arrhenius Equation k = Ae−EaRT the result of Collision Theory


Rate constant (k) is proportional to 𝒑 × 𝒛 × 𝒇p – the fraction of collisions with proper alignment for breaking and forming new bonds.z – the collision frequency, or the number of molecular collisions per unit time.f – the fraction of collisions with sufficient kinetic energy to overcome the activation energy of the reaction (EK must be > Ea)

Explanation of p:

lnk2k1

=Ea

R1T1

−1T2

19

k is the rate constant, A is the Arrhenius constant, Ea is the activation energy, R is the gas constant ( ) and T is the temperature in ____________.

Using the Arrhenius equation


20

For a reaction, if the rate constant (k) is 50.0 at 1000 K and 25.0 at 300 K, what is its activation energy?

lnk2k1

=Ea

R1T1

−1T2

Theoretical Model of Kinetics


The following general factors affect reaction rates:

• Reactant concentrations collision frequency increases as concentration increases.

• Temperature speed of the molecules, thus collision frequency increases as temperature increases.

• Physical state of the reactants – influences collision frequency. Reaction rate: (aq) > (g) > (l) > (s)

• Mechanism – reactions can occur at different rates for different pathways.

• Catalysts raises the reaction rate without being permanently consumed.

21

a catalyst lowers the activation energy!

Transition State Theory CH3Cl + I− → CH3I + Cl−


This reaction is a substitution of Cl with I.

The reaction has a high activation energy and proceeds through

an activated complex.

If you look at right, this complex is a carbon with what looks to

be five (5!!!) bonds – this is an extremely unstable species and

requires lots of energy to occur “activation energy” (Ea).

22

Reaction Coordinate (rxn. coord.)

Ep∆H=+ve(positive)∆H is unaffectedby a catalyst

+

+activation energy for uncatalyzedreaction

high energy activatedcomplex

Effects of Temperature and Catalysts on Rate


Catalysts lower the activation energy barrier.

Effects of Catalysis on Rxn. Rate

23

Kinetic Energy

Number of Molecu

les

Activation energy

low temperature

high temperature

As temperature increases, a greater number of moleculeshave sufficient kinetic energy to overcome the activation energy barrier; therefore, the rate of reaction increases.

Effects of Temperature on Rxn. Rate

Kinetic Energy

Number of Molecu

les

CatalyzedActivation energy

UnCatalyzedActivation energy

Step 1 NO2 g + NO2 g →𝑘1

NO3 g + NO g Slow, Rate = k

Step 2 NO3 g + CO g →𝑘2

NO2 g + CO2 g Fast, Rate = k

Net: Rate = k

Mechanisms


1. A mechanism describes how a reaction occurs in one or more elementary steps.

2. The sum of the individual steps in the mechanism gives the overall balanced chemical equation. The same elementary step

may occur more than once in the mechanism and a given step may be reversible.

3. The slowest elementary process is the ratedetermining step.

4. Any reaction mechanism must be consistent with the experimental rate law.

24

ELEMENTARY STEPS

ReviewQuestion


Consider a complex reaction and the proposed mechanism:

Overall Reaction: 2 NO2 g + F2 g → 2 NO2F g

NO2 g + F2 g → NO2F g + F g slow

NO2 g + F g → NO2F g fast

What is the rate law for the overall complex reaction?

a) Rate = k[NO2 g ][F2 g ]

b) Rate = k NO2 g2[F2 g ]

c) Rate = k NO2 g2[F g ]

d) Rate = k[NO2 g ][F g ]

*Elementary reactions in a mechanism must add up to give the overall chemical equation.*Intermediates must not appear in the overall chemical equation or rate law. (Products can if there is equilibrium)*The reaction mechanism must explain the experimentally determined rate law.*The reaction mechanism and overall rate law must be feasible: usually uni or bimolecular.

25

What is a catalyst, an intermediate and a reactant/product?


A + C → D + E ∆H = +100 kJ

D + A → F + C ∆H = −50kJ

2A → E + F ∆H = +50kJ

Energy

Reaction Coordinate Diagram

Reaction Coordinate

∆H = +100 kJA + C

D + E + A

E + F + C∆H = −50 kJ

∆H = +50 kJ

26

Catalysts are used and regenerated

Intermediates are created and destroyed

Which of the two elementary steps is the rate determining step (RDS)?What is the overall rate law?

Review


Net: 2 NO g + Br2 g → 2 NOBr g

NO g + Br2 g →k1

NOBr2 g slow

NOBr2 g + NO g →k2

2 NOBr g fast

What is the rate law for the overall complex reaction?

What role does NOBr2 g play?

27

Different Temperatures and Different Initial Concentrations


28

[A]

timeWhich experiments were performed at the same temperature?What’s the difference?

Performed at different temperatures?

The graphs below describe the same reaction under different conditions.

The graphs below show [A] vs. time.

[A]

timeWhat is the relationship between the 2 graphs?

Can you tell the order of reaction?

What extra information would be helpful?

BA

C

Pseudo first order rate constants


The rate of the reaction:16 H+ + 2 MnO4− + 10 I(aq)

− → 5 I2(aq) + 2 Mn2+ + 8 H2O depends on both the concentration of MnO4−

and I(aq)− . If the MnO4

−0 is set at 3.00 M for every reaction, can we find the order of reaction with respect to I(aq)

− ?

# I(aq)−

0(mol/L) Initial Rate (mol/L s)

1 0.020 3.28 × 10−2

2 0.010 8.20 × 10−3

3 0.030 7.38 × 10−2

29

Can we find the true rate constant of the reaction?

Equilibrium


time time


Rate of Rea

ction

Forward Rate

Reverse Rate

30

2.5

1.5

1.0

0.5

2.0

0

Forward reaction:N2 g + 3 H2 g ⇌ 2 NH3 g

Reverse reaction:2 NH3 g ⇌ N2 g + 3 H2 g

2 NH3 g

N2 g + 3 H2 g

Energy

Reaction Coordinate

∆H

N2 g + 3 H2 g ⇄ 2 NH3 g + 45.9 kJ

Initial 0 M 0 M 2 M

Change

Equilibrium 1 M

NH3 g

H2 g

N2 g

The Equilibrium Constant © Pavel SedachLearnfaster.ca

A s + 2 B l ⇄ 3 C aq + 4 D aq

K =productsreactants

=C aq

3D aq

4

A s B l2 = C aq

3D aq

4

Solids and Liquids are not included in equilibrium expressions as their activity is 1 (very low) UNLESS it’s a solution of all liquids

31

For all gas systems, K can be written as Kc (mol/L) or Kp (atm).

A system that is a mixture of both gases and aqueous is simply written as Keq or K and gases are given in atm and concentrations

in moles/L.

In all cases, K is unit-less (has no units).

A(aq)

B(aq)

reaction occurs in 3 dimensionsand rate is dependent on concentration

reaction occurs only at the interface/surface and reaction rate depends on the size/area

of the interface

C(l)

D(l)

Kconcentration andKpressure for an all Gas System


32

An‘allgas’systemissimplyonewherethereareno aqueousandonlygasreagents:

4C g + 3 E s → 2D g + 3F l

∆ngas = Kc= Kp=

The system above has a Kc of 1500 at 298 K. What is its Kp? (Use Kp = Kc RT ∆ngas)

PV = nRT

PRT

=nV

Kp

RT ∆ngas= Kc

OR

Kp = Kc RT ∆ngas

Will there ever be a situation where Kp and Kc have the same value? Give an example.

Le Chatelier’s Principle –Qualitative Analysis


Reactions maintain the equilibrium constant (K) at a specific Temp. by shifting right (to products) or shifting left (to reactants)

PCl3 g + Cl2 g ⇄ PCl5 g , K=1.00

If PCl3 g = Cl2 g = PCl5 g = 1.0 M and [Cl2 g ] is decreased to 0.5 M, which way does the system ‘shift’ to compensate

for the change in concentration?

1. Think of equilibrium as being on a wellbalanced seesaw:

The system is initially at equilibrium if we remove Cl2(g) then the seesaw is unbalanced

we shift left to correct for the lack of Cl2(g)

PCl3 g + Cl2 g ⇄ PCl5 g

33

Le Chatelier’s Principle


PCl3(g) + Cl2(g) ⇄ PCl5(g)Disturbance ↓

Shift/Response (change in concentration)

↑ ↑ ↓

2. Mathematically, at equilibrium: K =products

reactants=

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

=1mol/L

1mol/L 1mol/L= 1 =

P

R

After equilibrium is disturbed, Q = products

reactants=

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

= 1mol/L

1mol/L 0.5 mol/L= 2 = P

R

Q > K so we must restore equilibrium by making Products INTO Reactants

3. Alternatively, we can look at the change and reverse it :

34

Concentration and Pressure/Volume A g + B g ⇄ C g


DisturbanceAdd Reactants (A or B)OR Remove Products (C)

Shift Right

Add Products (C)OR Remove Reactants (A or B)

Shift Left

Raise PressureOR Lower Volume1

Shifts to the side with less moles of gas (think of it as relieving the pressure).

Shift Right

Lower PressureOR Increase Volume1

Shifts to the side with more moles of gas (think of it as making more moles to fill the new space available).

Shift Left

Concentration [ ]

Chan

ges

Pressure/Volume

Chan

ges

35

Explaining Pressure and Volume A g + B g ⇄ C g


Pressure and Volume are related. If the container is smaller (less volume) the space is tighter (there is more pressure).

Therefore the system makes less moles of gas. This works as follows:

𝐾 =𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠

=C g

A g B g=

1 mo ⁄l L1 mo ⁄l L 1 mo ⁄l L

= 1 =PR

After we halve the volume:

𝑄 =𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠

=C g

A g B g=

1 mo ⁄l 0.5 L1 mo ⁄l 0.5 L 1 mo ⁄l 0.5 L

=2molL

2molL × 2

molL

=24=

PR

Shift Right (to the products – the side with less moles of gas)

The reverse is also true; when the container size is increased, the system Shifts Left to make more moles of gas to fill the space (the math supports this).

What kinds of systems are unaffected by P/V Changes?

36

Le Chatelier’s Principle Temperature


Disturbance Exothermic (Exo.) SystemA + B⇄ C ∆H=negative

A + B⇄ C + heat

Endothermic (Endo.) SystemA + B⇄ C ∆H=positive

heat + A + B⇄ C Raise the Temperature

(Add Heat)Shift Left Shift Right

Lower the Temperature (Remove Heat)

Shift Right Shift Left

Add an inert gas

(any compound that does not react with the given system)

Add a catalyst

No Shift

Temperature

Chan

ges

37

K and temperature


38

If the K of Ca3(PO4)2 sH2O l

3 Ca aq2+ + 2 PO4 aq

3− is 15 at 100 °C and 47 at 300 °C, is the rxn endo or exothermic?

Is the forward or the reverse activation energy of this reaction greater?

If the K of H3O aq+ + OH aq

− ⇌ 2 H2O l is 1.1 × 10+11 at 90 °C and 1.0 × 10+14 at 25 °C, is the rxn endo or exothermic?

Is the forward or the reverse activation energy of this reaction greater?

Concentration/Time Graphs



Reaction Coordinate

39

5

3

2

1

4

0Addition of Hydrogen

Raising the Pressure

Raising the Temperature

all values in 𝐦𝐨𝐥𝐞𝐬

𝐋𝐢𝐭𝐞𝐫H2 g + I2 g ⇌ 2 HI g

Initial 2.0 3.0 0

Change

Equilibrium 2.0

H2 g

[I2 g ]

HI g

EquilibriumMathematics


Qisthe“Reaction Quotient”

Q =Productsnot at equilibriumReactantsnot at equilibrium

K =ProductsReactants

1. If K > Q, the system shifts to products

(shifts right)

2. If K = Q, the system is at equilibrium

(no shit)

3. If K < Q, the system shifts to reactants

(shifts left)

For the following reaction:PCl g + 2 Cl2 g ⇌ PCl5 g , K = 3

In which direction must the reaction proceed to reach equilibrium if:

(a) [PCl g ], [Cl2 g ], and [PCl5 g ] = 1.0 M

(b) [PCl g ] = 8.0 M, [Cl2 g ] = 0.25 M, and [PCl5 g ] = 5.0 M

40

[] vs. t Graphs and Bar Graphs


41all values in moles/L

CH3Br g + Cl g− → CH3Cl g +Br g

−

Initial 1.00 1.00 0.20 0.20

Change

Equilibrium

time


Initial [] Equilibrium []

1.00

0.60

0.40

0.20

0.80

0

If K=0.25, which way does the system shift and what is the concentration of reagents at equilibrium?

CH3Br g Cl g− CH3Cl g Br g

− CH3Br g Cl g− CH3Cl g Br g

−

CH3Cl g

Br g−

CH3Br g

[Cl g− ]

Bar Graphs © Pavel SedachLearnfaster.ca

42all values in moles/L

CH3Br g + Cl g− → CH3Cl g +Br g

−

Equilibrium 0.80 0.80 0.40 0.40

What change did we do to the system and, according to collision theory, why?

time

(mol/L)

CH3Cl g

Br g−

CH3Br g

[Cl g− ]



−

1.00

0.50

0.25

0.75

0CH3Br g Cl g

− CH3Cl g Br g−


time

(mol/L)

CH3Cl g

Br g−

CH3Br g

[Cl g− ]



−

1.00

0.50

0.25

0.75

0CH3Br g Cl g

− CH3Cl g Br g−


CH3Br g + Cl g− → CH3Cl g + Br g

− + heat © Pavel SedachLearnfaster.ca 43


time

(mol/L)

CH3Cl g

Br g−

CH3Br g

[Cl g− ]



−

1.00

0.50

0.25

0.75

0CH3Br g Cl g

− CH3Cl g Br g−

time

(mol/L)

CH3Cl g

Br g−

CH3Br g

[Cl g− ]



−

1.00

0.50

0.25

0.75

0CH3Br g Cl g

− CH3Cl g Br g−

CH3Br g + Cl g− → CH3Cl g + Br g

− + heat © Pavel SedachLearnfaster.ca 44


time

(mol/L)

CH3Cl g

Br g−

CH3Br g

[Cl g− ]



−

1.00

0.50

0.25

0.75

0CH3Br g Cl g

− CH3Cl g Br g−

CH3Cl g + Br g−

CH3Br g + Cl g−

Energy

Reaction Coordinate

∆H

Keq = kforward

kreverse, if kreverse increases by a greater degree, K is smaller, reaction shifts left!

Activation Energy

Manipulating Equilibrium Constants *(K can never be negative)


Reaction 1 A + B ⇌ C K1

Reaction 2 C ⇌ D K2

Net A + B ⇌ D K3 = K1 × K2

Reaction 1 A + B ⇌ C K1=

Reaction 2 C ⇌ D K2=

Net A + B + C ⇌ D + C K3=

× 1 = A + B ⇌ C K1

× 2 = 2A + 2B ⇌ 2C K2

× 3 = 3A + 3B ⇌ 3C K3

× −1 = C ⇌ A + B K−1 =1

K

× −1

2=

1

2C ⇌

1

2A +

1

2B K− 1

2 =1

K1/2 =1

K

Manipulation of Equation Effect on Equilibrium ConstantK =

45

Review


46

Given

12N2 g + O2 g → NO2 g , K1 = 1.2 × 10−4

2 NO2 g → N2O4 g , K2 = 52

What is the equilibrium constant for:

N2O4 g → N2 g + 2O2 g

Review


Phosgene, COCl2 g , is a toxic gas that dissociates according to the chemical equation:

COCl2 g ⇌ CO g + Cl2 g

For the reaction above, Kc = 8.05 × 10−4 at 673 K. Exactly 1.00 mol COCl2 g is placed in an empty 25.0L flask. The flask is

capped and then warmed to 673 K. What is the equilibrium concentration of CO at this temperature?

47

Acid and Bases


48

Water ionizes itself (autoionization)


49

2 H2O l ⇌ H3O aq+ + OH aq

−

Kwater = Kw = H3O aq+ OH aq

− = 1.00 × 10−14 (𝑎𝑡 298 𝐾)

In neutral water, H3O aq+ = OH aq

−

Where does Neutral pH come from?

1.00 × 10−14 = H3O aq+ OH aq

− = x2

x =

pH of neutral Water


50

At 90 ℃, the Kw of water is 9.00 × 10−12, what is the pH of neutral water at that temperature?

If the pH of neutral water at 2 ℃ is 8.255, what is Kw?

Is the dissociation of water endothermic or exothermic?

pH (potential of hydrogen) and other “p” Scales


51

pH is a measure of acidity. The lower the pH the more acidic the solution.

The letter “p” is a mathematical symbol like ×,÷,− or +.

p means log(x), for instance pH = −log[H3O+], pOH = −log[OH−], pKa = − log Ka

H3O aq+ pH = −log(H3O aq

+ )

10−14 M 14

10−10 M 10

10−7 = 0.0000001 M 7

10−4 = 0.0001 M 4

100 = 1 M 0

10+1 = 10 M 1

Commercial HClis 13.6 M, so its pH is even lower!

Question – what is p90?

H3O aq+ OH aq

−

14 = pH + pOHpH pOH

Kw = H3O aq+ OH aq

−

−log H3O aq+

− log OH aq−

10−pOH10−pH

Strong Acids *On Formula Sheet


52

Covalent compounds ionize 100% = No K value

HCl, HBr, HI, H2SO4, HClO4, HNO3hydrochloric acid, hydrobromic acid, hydroiodic acid, sulfuric acid, perchloric acid, nitric acid

What is the pOH of an aqueous solution of HI aq with an initial concentration of 1.0 × 10−3 M?

Acid = H3O+

if HClO4 = 1 mol/L, H3O+ = 1 mol/L

Strong Bases *On Formula SheetIonic compounds, Dissociate ~100% to give OH−. 1st column metal hydroxide: NaOH → Na+ + OH−

NaOH = OH−

If NaOH = 1 mol/L, then OH− = 1 mol/L

2nd column metal hydroxide: Mg OH 2 Mg2+ + 2 OH−

2 × Mg OH 2 = OH−

If Mg OH 2 = 0.0001 mol/L, then OH− = 0.0002 mol/L (times 2!)

With 2+ metals, you usually Perform a Ksp (Ksp is provided) calculation to determine pH.

Acidity and Basicity


53

The Conjugate of Strong Acid is a Neutral Salt (exception is H2SO4)

Strong AcidsIonize 100%

Strong Bases are Metal Hydroxides – their conjugate acid is water

Weak Acids Ionize Partially Because they Produce Weak Base

The Conjugate of a Weak Acid is a Weak Base

The conjugate of a strong acid is a

______________

⇌⇌⇌⇌⇌⇌

Relationship Between Acidity and Basicity (C) Pavel SedachLearnfaster.ca 54

CH3COOH aq + H2O l ⇄ H3O aq+ + CH3COO aq

− , Ka = 1.8 × 10−5

CH3COO aq− + H2O l ⇄ OH aq

− + CH3COOH aq , Kb = 5.5 × 10−10

Net Reaction 2 H2O l ⇌ H3O aq+ + OH aq

− , Kw = 1.0 × 10−14

∴ Ka × Kb = Kw

Conjugate Acid Strength

Ka

Conjugate Base Strength

Kb

Acid/Base Strength

Strong Acids ionize 100% their conjugate is neutral! Example: I−, Br−, ClO4−, NO3

−

HCl + H2O → H3O+ + Cl−

Weak acids/bases do not ionize 100% because on the other side there is a weak conjugate A/B

CH3COOH aq + H2O l ⇄ H3O aq+ + CH3COO aq

−

Calculating the pH of aWeak Acid Solution


55

What is the pH of a solution of 0.0017 moles of H2S aq (hydrosulfuric acid) added to 10 mL water? (Ka = 8.9 × 108)

H2S aq + H2O l⇄ H3O aq

+ + HS aq−

Initial

Change

Equilibrium

Ka =HS aq

− H3O aq+

H2S aq= 8.9 × 108

_________ ________________

= 8.9 × 108

_________ ________________

= 8.9 × 108 pH = −log H3O aq+

Calculating the pH of aWeak Base Solution


56

What is the pH of 0.1 mol/L CH3COO aq− ? (acetate ion) (Ka CH3COOH aq

= 1.8 × 10−5)

CH3COO aq−

+ H2O l⇄ OH aq

− + CH3COOH aq

Initial

Change

Equilibrium

Kb =Kw

Ka=

10−14

1.8 × 10−5= ______________ =

CH3COOH aq OH aq−

CH3COO aq−

_________ ________________

= ______________

_________ ________________

= ______________

pOH = − log OH aq−

Which of the following solutions has the lowest pH?


57

1. Which of the following solutions has the lowest pH?

a) 0.1 M HOOCCOOH, Ka = 5.6 × 10−2

b) 0.025 M Sr OH 2c) 0.1 M HF, Ka = 6.3 × 10−4

d) 0.0001 M HCl

2. Which of the following solutions has the lowest pH?

a) 0.1 M NaOHb) 0.1 M CH3COOH, Ka = 1.8 × 10−5

c) 0.1 M HOOCCOOH, Ka = 5.6 × 10−2

d) 0.01 M HCl

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Chemistry203 TermTest2 - Learnfaster.calearnfaster.ca/wp-content/uploads/chem203/TT2/C203-TT2...Chemistry203 –TermTest2 1 (C) Pavel Sedach Learnfaster.ca Questions? Suggestions?