to what extent will the reaction proceed

• Solubility -the amount of solute that can be dissolved to form a solution.

• Solvent – the substance in a solution present in the greatest amount.

• Solute – the substance in a solution present in the least amount.

• Saturate – a solution that has come to equilibrium. The rate of dissolving is equal to the rate of recrystalizing.

• High solubility – more than .10 mol/L will dissolve (AgNO3)

• Low solubility – less than .10 mol/L will dissolve (AgCl)

Dissociation (ionization) ReactionsWhen ionic compounds dissolve in water, they separate into ions – one positive and one negative.

Examples:

Ca(NO3)2(s) → Ca2+(aq) + 2 NO3

–(aq)

Ca3(PO4)2 → 3 Ca2+(aq) + 2 PO4

3-(aq)

Write dissociation reactions for each of the following:

K2SO4(s) →

Al(OH)3(s) →

PbSO4(s) →

NH4NO3(s) →

2 K+(aq) + SO4

2-(aq)

Al3+(aq) + 3 OH – (aq) (slightly soluble)

Pb2+(aq) + SO4

2-(aq) (slightly soluble)

NH4+

(aq) + NO3–

(aq)

Combine solutions of NaCl(aq) and AgNO3(aq)

NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)

Na+(aq) + Cl-(aq) + Ag+

(aq) + NO3–

(aq) → AgCl(s) + Na+(aq) + NO3

–(aq)

Cl-(aq) + Ag+(aq)→ AgCl(s)

net ionic equation

A mathematical description of the number of collisions between molecules in a sample of matter per unit time, useful for predicting rates of reaction.

Reaction rate: the number of atoms, ions or molecules that react in a given period of time to form products

I2

H2

HI

Initial State Final State

THE STATES OF THE COLLISION THEORY

Collision Theory states that in order for reactions to occur between substances, their particles (molecules, atoms, or ions) must collide.

HI HI

These interactions, if effective will form two new molecules.

• This collision is not energetic enough to supply the required activation energy.

• Therefore the Collision is ineffective. HI HI

HI HI

HIHI

• The colliding molecules are not oriented in a way that enables them to react with each other.

• If the Collision doesn’t have the right orientation then the collision is not effective.

HIHI

HI HI

HI HI

• This collision has the right orientation.

• This collision is powerful enough to cause a good effect.

• Everything is satisfied, and the collision turns out to be effective.

HIHI

HI HI

I I

H2

• Eact - minimum energy a reactant must

possess in order to convert to products.

• The activation energy (Eact) can determine how fast a reaction occurs. In general, the higher the activation energy, the slower the reaction rate. The lower the activation energy, the faster the reaction.

Consider the process of someone trying to roll a boulder over a hill. The higher the hill, the slower the task. The lower the hill the faster the process. The height of the hill (a) correspond to the energy of activation (Eact).

E act

activated complex

• the atomic configuration at the top of the energy barrier• short life time (10-13 s)• breaks apart to form reactants or products both of which have lower potential energy

activated complex

• This explains why some reactions do not take place at room temperature.

• CH4(g) + O2(g) → no reaction at room temp.

• The molecules can not overcome the activation energy.

A reaction with a very low activation energy will occur spontaneously.

Types of Reactions

Spontaneous exothermic Spontaneous endothermic

Slow exothermic Slow endothermic

• Most reactions do not take place in a single step. A→B→C→D

• Each step is usually a simple one on one collision reaction.

• The set of steps is called the reaction mechanism.

• The slowest step in the reaction mechanism is the rate determining step.

En

erg

y

Reaction Progress

I

II

III

Intermediates

Activated Complex

Reactants

Products

rate determining step, highest activation energy and therefore the slowest rate.

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the reaction.

2H2(g) + O2(g) → 2H2O(l) Slow

2H2(g) + O2(g) → 2H2O(l) Fast

The Pt is not used up and does not appear as a reactant or product

Pt

Consider the task of moving coal over a barrier. A pathway with a lower barrier is analogous to a reaction affected by a catalyst. The task becomes easier for a pathway with a lower barrier .

• Enzymes are organic catalysts the allow chemical reactions in the human body to occur at a lower temperature than normal.

Inhibitors effect catalysts by rendering them useless.

(a) Cholinesterase catalyzes the hydrolysis of acetylcholine into acetic acid and choline. (b) An organic phosphate (as in the nerve gas Sarin) binds to cholinesterase, preventing it from breaking down acetylcholine.

http://upload.wikimedia.org/wikipedia/commons/f/fe/Competitive_inhibition.svg

A catalytic converter works by taking exhaust gases from the engine, including CO and NO, passing them through the catalytic converter, where they are converted to CO2 and N2 by catalyzed reactions.

Surface area- the more surface area, the greater the chance for reactants to encounter to form product.

CatalystCatalyst- lowers the activation energy for the reaction.

Temperature - the higher the temperature the faster the molecules will move:

• higher frequency of collisions• more energy in each collision

Nature of the reactants • solids and liquids cannot undergo a change in

concentration since they occupy a given space determined by intermolecular bonding.

• gases and solutions can alter their concentrations.

Concentration - The higher the concentration the more particles per unit area – higher probability of a collision

• The conversion of reactants into products and the conversion of products into reactants occur simultaneously.

• 2SO2(g) + O2(g) 2SO3(g)

• In a reversible reaction the rate of the reverse reaction is zero at the start.

•Eafwd → activation

energy for the forward reaction

•Earev → activation

energy for the reverse reaction

• Exothermic • Endothermic

EpEp

ΔH

Earev

Eafwd

ΔH

Earev

Eafwd

•The catalyst lowers the E act for the forward and the reverse reaction because the reaction takes place through a different set of steps.

Many reactions do not convert 100% of reactants to products. There is often a point in a reaction when the products will back react to form reactants.

The extent of the reaction, 20% or 80%, can be determined by measuring the concentration of each component in solution.

In general the extent of the reaction is a function of temperature and concentration which is monitored by some constant value called the equilibrium constant (Kequilibrium constant (Keqeq).).

•Chemical Equilibrium is a dynamic state in which the rates of the forward and the reverse reaction are equal.

• For any general chemical process at equilibrium.

aA + bB pP + qQ

eqA

••

a b

p

Keq B

P Qq

After the time, te, the reaction is at equilibrium, and the concentrations of reactant and products undergo no further change.

Concentration vs time graph for the reversible reaction 2Hl(g) ↔ H2(g) + I2(g)

[ NH3 ] 2

[ N2 ] ● [ H2 ] 3Keq =

1) Write the equilibrium expressions for the following reactions.

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

[ CO2 ] 12 ● [ H2O ] 6

[ O2 ] 15 ● [ C6H6 ] 2Keq =

2 C6H6 (g) + 15 O2 (g) 12 CO2 (g) + 6 H2O (g)

[ NH3 ] 2

[ N2 ] ● [ H2 ] 3Keq =

2) At equilibrium the concentration of nitrogen is 0.50 M, hydrogen 0.26 M and ammonia 0.15 M. Calculate the equilibrium constant for the reaction.

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

[ 0.15 M ] 2

[ 0.5 M ] ● [ 0.26 M ] 3Keq =

2.6 Keq =

[ NH3 ] 2

[ N2 ] ● [ H2 ] 3Keq =

3) In a 2.00 L reaction vessel there are 6.00 mol of ammonia, 2.00 mol of nitrogen and 8.00 mol of hydrogen at equilibrium. Calculate the equilibrium constant for the reaction.

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

[ 3.00 M ] 2

[1.00 M ] ● [ 4 M ] 3Keq =

0.141 Keq =

[ NH3 ] 2

[ N2 ] ● [ H2 ] 3Keq =

4) Ammonia is formed in a 1.00 L reaction vessel. If the equilibrium constant is 11.1 at a given temperature and the mixture contains one mole of nitrogen, calculate the number of moles of ammonia present at equilibrium.

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

[ x ] 2

[1 M] ● [ 3M ] 311.1 =

17.3 Mx =

n

Vc =

n

1.0 L17.3 =

n = 17.3 mol

For the system A(g) + 2 B(g) → 2 C(g) 0.500 moles of A and 1.00 mole of B were placed in 500mL reaction vessel. The equilibrium concentration of A was found to be 0.500 mol/L. Complete the following:

a. Use I.C.E to calculate the concentrations of B and C

A(g) + 2 B(g) → 2 C(g)

I 1.0 M 2.0 M 0.0 MC 0.5 M 1.0 M 1.0 ME 0.5 M 1.0 M 1.0 M

b. What is the equilibrium constant expression and the value of k?

Determine the [ ] of H+ a 0.500 mol/L CH3COOH (aq) solution.

CH3COOH(aq) → CH3COO -(aq) + H+

(aq)

I

C

E

0.500 mol/L 0 0

x x x

(0.500 – x) x x

Ka = [CH3COO-] [H+]

[CH3COOH]

1.8 x 10-5 = (x) (x)

(.500 – x)

X2 = (1.8 x 10-5)(0.500)

X = 3.00 x 10-3

X = [H+]

Keq > 1 Product is favoredProduct is favored

Keq = 1 Product and Reactant are equalProduct and Reactant are equal

Keq < 1 Reactant is favoredReactant is favored

P.419 #11 & 12

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

As NH3 is formed, some of it back reacts and forms N2 and H2. This takes place until the amount consumed is equal to the amount produced.

eq

23

32 2

Law of Mass Action:

NHK

N • H

2. Compressor

N2 H2

1. Gases are mixed and scrubbed

3. Converter

4. Cooler

N2 H2 NH3

NH3 to storage

unreacted N2 and H2 are recycled

iron catalyst

200 atmosphere

s 450°C

•Substances are in different phases at equilibrium •i.e., solid and aqueous.

Which solid is more concentrated?

100 g (1 cup)

200 g (2 cup)

Concentration of a solid (and pure liquid) is always a constantConcentration of a solid (and pure liquid) is always a constant

CaCO3(s) CaO (s) + CO2 (g)

2eq

3

3

1 2eq

2

eq 2

CaO • COK

CaCO

but, CaO = constant (solid)

and CaCO constant (solid)

Constant • COK

Constant

K CO

2eq

3

3

1 2eq

2

eq 2

CaO • COK

CaCO

but, CaO = constant (solid)

and CaCO constant (solid)

Constant • COK

Constant

K CO

1) Write the equilibrium expressions for the following reactions.

[ CO2 ] 12

[ O2 ] 15 Keq =

2 C6H6 (l) + 15 O2 (g) ↔ 12 CO2 (g) + 6 H2O (l)

If a system at equilibrium is disturbed by a change in temperature, pressure or the concentration of one of the components, the system will shift its equilibrium position so as to counter-act the effect of the disturbance.

If a chemical system is at equilibrium and then a substance is added (either a reactant or product), the reaction will shift so as to re-establish equilibrium by subtracting part of the added substance. Conversely, removal of a substance will result in the reaction moving in the direction that forms more of the substance.

Consider the Haber reaction at equilibrium:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

If some H2 is added to the reaction which was at equilibrium, the system self-adjust to remove the excess H2 by converting it to NH3 until equilibrium is re-established. In the process, some N2 is also consumed.

•Increasing the pressure of a gaseous mixture causes the system to shift in the direction that reduces the number of moles of gas.

N2O4(g) 2 NO2(g)

•The equilibrium shifts to the side that reduces the total number of moles of gas. In this example, 2 moles of product (NO2) will change to 1 mol of reactant (N2O4). The total moles in the reaction mixture is reduced to compensate for the increase in pressure.

• As the pressure increases, the amount of ammonia present at equilibrium increases.

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

• An increase in the temperature of a system at equilibrium will shift the reaction so that it will absorb the heat.

• The equilibrium constant changes with temperature. It will either increase or decrease depending on the exothermicity or endothermicity of the reaction.

The tubes in the photograph both contain a mixture of NO2 and N2O4 . As predicted by LeChatelier’s principle, the equilibrium favors colorless N2O4 at lower temperatures because the reaction N2O4 → 2 NO2 is endothermic. This is clearly seen in the tube at the right, where the gas in the ice bath at 0 °C is only slightly brown because there is only a small partial pressure of the brown gas NO2 . At 50 °C (the tube at the left), the equilibrium is shifted toward NO2 shown by the dark brown color.

The equilibrium constant changes with temperature.

Endothermic: reactant + E products

Temp increase→ Shift rxn to Right , (Keq ↑)

Exothermic: reactant products + E

Temp increase → Shift rxn to Left, (Keq ↓)

KKeqeq(old)(old) KKeqeq(new)(new)

KKeqeq(new)(new) KKeqeq(old(old))

Catalyst added to a system at equilibrium lowers the activation energy of a reaction and therefore accelerates the rate of the reaction in both directions. A catalyst does not change the value of the equilibrium constant.