Chapter 9 Chemical Reactions

I. Classification of Chemical Reactions

1. Combination reactions

heat

I2 + H2 2HI

2Na + Cl2 2NaCl

A + B C

2H2 + O2 2H2O

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2. Decomposition Reactions

A B + C

H2CO3

heat H2O + CO2 (gas)

2H2O2light 2H2O + O2 (gas)

3. Single Replacement Reactions

Zn + 2HCl H2 + ZnCl2

2NaI + Br2 2NaBr + I2

Zn + CuSO4 Cu + ZnSO4

2

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4. Double Replacement Reactions – Exchanging partners

AgNO3(aq) + NaCl(aq)

Solid (precipitation)

NaNO3(aq) + AgCl(s)

2KI(aq) + Pb(NO3)2(aq) 2KNO3(aq) + PbI2(s)

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5. Combustion Reactions - Reactions with O2

CH4 + 2O2 CO2 + 2H2O

C3H8 + 5O2 3CO2 + 4H2O

2Mg + O2 2MgO

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6. Oxidation-reduction (redox) reactions

Consider 2Cu + O2 2CuO

oxidized reducedreducing agent Oxidizing agent

Oxidation must be accompanied by reduction

2Cu 2Cu2+ + 4e- Oxidized (loss of electrons)(increase of oxidation number)

O2 + 4e- 2O2- Reduced (gaining of electrons)(decrease of oxidation number)

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Oxidation number (Page 226 - 227)

1. Ox.# of an element in its element state is zero.2. Ox.# of monatomic ion = charge of the ion.3. Ox.# of H is +1 in most hydrogen containing compounds.4. Ox.# of O is –2 in most O containing compounds5. Ox.# of F = -1 in F containing compounds.6. Ox.# of Cl, Br & I containing compounds are –1 except in

a compound with F or O.7. Sum of ox.#’s in a compound = 0

Sum of ox.#’s in a polyatomic ion = charge of the ion

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Oxidation number1. element =02. monatomic ion = charge of the ion.3. H = +1 in a compounds.4. O = –25. F = -1.6. Cl, Br & I = –1 except in a compound with F or

O.7. Sum of ox.#’s = charge of the species

Examples

CuCl2

Ox#

Cu

Cl

HNO3

H?

O?

N?8

More examples

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Examples of Redox Reactions

Zn + CuSO4 ZnSO4 + Cu

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Examples of Redox Reactions

2H2 + O2 2H2O

2Na + 2H2O 2NaOH + H2

3H2S + 2HNO3 3S + 2NO + 4H2O

3H2SO3 + 2HNO3 2NO + H2O + 3H2SO4

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Important of redox reactions

1. Combustion reaction (see examples 1 and 2 above)

2. Resperation Similar to combustion but at low temperature

4. Rusting – reaction with oxygen

4Fe + 3O2 2Fe2O3

3. Bleaching

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Page 151

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

e-

Batteries

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II. Reaction Rates

A. The kinetic theoryA + B C + D

A collides with B

A + B (AB) C + D

Transition state orActivated complex

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a) Collision orientation

+ +

H2O + HCl H3O+ Cl-+

Figure 8.3

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OH

H

Cl H

Incorrect orientation

OH

H

ClH

Correction orientation

O

H

H

+ HCl O

H

H

H + Cl-+

H+

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Energy diagram for an exothermic reaction

b). Activation energy (Ea) – energy needed for reaction to occur

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Energy diagram for an endothermic reaction.

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c) Exothermic reaction: A reaction in which energy is released

Endothermic reaction: A reaction in which an input of energy is needed (absorbed)

A + B C + D + heat

A + B + heat C + D

Origin of heat of reaction:Bond breaking – heat is absorbedBond formation – heat is released

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Reaction rate depends on 1) collision frequency2) velocity (kinetic energy) of reactant molecules.

Only those with kinetic energy > activation energy can undergo reaction

B. Factors that affect reaction rates

1) Nature of the reactants2) Concentration

Rate increases as concentration increases – more collisions/s

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3) Temperature

Reaction rate increases as T increasesAs T increases, kinetic energy of the reactants increases(more reactants have E > activation energy, Ea)

Optimum body temperature = 37oC 21

4) CatalystA catalyst decreases the activation energy

- reaction rate increases

Enzymes are catalysts

A + B + X C + D + X

Figure 8.9

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5. Surface area The larger the surface, the faster the reaction.

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III. Chemical equilibrium

A. Chemical EquilibriumMany reactions are reversible

After some time, forward rate = reverse rate

The system is at a state of chemical equilibrium (or just equilibrium)

H2(g) + I2(g) 2HI(g)forward

reverse

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B. Equilibrium constant

H2(g) + I2(g) 2HI(g)

At equilibrium

eqKIH

HI

]][[

][

22

2

At constant T

In general

aA + bB cC + dD

bBaA

dDcCKeq

][][

][][ Equilibrium (constant)

expression

[A] = concentration of A

Equilibrium constant

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Example

H2(g) + I2(g) 2HI(g)

At equilibrium: [H2] = 0.86 M, [I2] = 0.86 M, [HI] = 0.27 M

099.0)86.0)(86.0(

27.0

]][[

][ 2

22

2

IH

HIKeq

Large Keq more products than reactants at equilibriumSmall Keq more reactants than products at equilibrium

3H2 + N2 2NH3

][][

][

23

2

23

NH

NHKeq

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PbCl2(s) Pb2+(aq) + 2Cl-(aq)

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C. Factors that affect the equilibrium constant

Le Chatelier’s Principle If a stress is applied to a system in equilibrium, a new equilibrium will be established in which the equilibrium position has been shifted in such a way as to relieve the applied stress.

Equilibrium position – a quantitative indication of the relative amounts of reactants and products at equilibrium

a) Effect of concentration

N2 + 3H2 2NH3

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Concentration changes that result when H2 is added to an equilibrium mixture involving the system.

Equilibrium position shifts to the right

N2 + 3H2 2NH3

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In general

A + B C + D

Adding A or B will cause the equilibrium position to shift to the right.Adding C or D will cause the equilibrium position to shift to the left.

b) Effect of pressure (for reactions involving gases)

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

4 moles of gas 2 moles of gas

A increase of pressure causes the equilibrium to shift to the right

Adding A: [A] [B] [C] [D]

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c) Effect of Temperature

A + B C + D + heat exothermic

If T increases, shift left[A] and [B] increase, [C] and [D] decrease, Keq decreases

If T decreases, shift right[A] and [B] decrease, [C] and [D] increase, Keq increases

heat + A + B C + D

If T increases, shift

If T decreases, shift

endothermic

]][[

]][[

BA

DCKeq

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d) Catalyst

Adding a catalyst speeds up both the forward and reverse reaction rates but does not change the equilibrium position.

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Example: 2NO(g) + O2(g) 2NO2(g) + heat

Adding NO

Equilibrium position

Shift

Removing O2 Shift

Adding NO2 shift

Increasing T

Decreasing T

Increasing P by decreasing V

Decreasing P by increasing V

Adding a catalyst34

2NO(g) + O2(g) 2NO2(g) + heat

An increase of temperature will cause Keq to

An increase of pressure by decreasing volume will cause Keq to

Only temperature will can change Keq of a given equilibrium

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