as chemistry enthalpy changes. learning objectives candidates should be able to: explain that some...
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Learning ObjectivesCandidates should be able to:• Explain that some chemical reactions are
accompanied by energy changes, principally in the form of heat energy.
• Construct (and interpret) a reaction pathway diagram, in terms of the enthalpy change of the reaction.
• Calculate enthalpy changes from appropriate experimental results, including the use of the relationship: q=mcT.
• Explain and use the terms enthalpy change of reaction and standard conditions, with particular reference to formation and combustion.
Learning Objectives
Candidates should be able to:
• apply Hess’ Law to construct simple energy cycles
• carry out calculations involving such cycles
Standard enthalpy change of combustion, ΔHθ
c, 298 is the enthalpy change when 1 mole of a substance is burned completely in oxygen under standard conditions (100kPa and 298K), all reactants and products being in their standard states.
Standard enthalpy change of formation, ΔHθf,
298 is the enthalpy change when 1 mole of a compound is formed from its elements under standard conditions (100kPa and 298K), all reactants and products being in their standard states.
Hess’ Law The enthalpy change of a reaction depends only on the initial and final state of the reaction and is independent of the route by which the reaction may occur.
Learning Objectives
Candidates should be able to apply Hess’ Law to construct simple energy cycles, and carry out calculations involving such cycles and relevant energy terms, with particular reference to average bond energies.
The quantity of energy needed to break a particular bond in a molecule is called the bond dissociation enthalpy (H
diss), or bond enthalpy for short. It refers to the enthalpy change when one mole of bonds of the same type are broken in gaseous molecules under standard conditions.
Bond energy
H – H (g) H(g) + H(g)
Bond energy
Mean Bond Enthalpy
The mean bond enthalpy is the amount of energy needed to break a covalent bond.They are average values taken from many different molecules
Bond Numberbroken
Numberformed
Average MolarBond Enthalpy
kJ mol-1
C-C 1 0 +347
C-H 5 0 +413
C-O 1 0 +358
O-H 1 6 +464
O=O 3 0 +498
C=O 0 4 +805
Here bond enthalpies are defined endothermically.
Bond breaking:Total endothermic value = (+347 x 1) + (+413 x 5) + (+358 x
1) + (+464 x 1) + (+498 x 3) = +4728 kJ
Bond making:Total exothermic value = (-464 x 6) + (-805 x 4) = -6004 kJ
Sum total of bond breaking and bond making:DHc = +4728 + - 6004 = -1276 kJ mol-1
Learning Objectives
Candidates should be able to:
• Explain and use the terms rate of reaction and activation energy.
• Show understanding, including reference to the Boltzmann distribution, of what is meant by the term activation energy.
Seconds
Minutes
Hours
Days
Weeks
Months
Years
Decades
Centuries
Millennia
Different rates of reactiondynamite explodingmagnesium and acid
cake baking
fruit ripening
plants growingrusting of iron
erosion of rock
crude oil forming
The rate of a reaction is found by measuring the amount in moles of a reactant which is used up, or the amount of product produced, in a given time.
The units are often mol dm-3 s-1.
Rate of Reaction
There are five factors which can affect the rate of a reaction:
1. Surface Area2. Concentration3. Temperature4. Use of a catalyst5. Intensity of light
Changing the rate of a reaction
Reactions occur when the particles of reactants collide, provided they collide with a certain minimum amount of kinetic energy (and in the correct orientation).
Collision Theory
Enthalpy
Progress of Reaction
Enthalpy
Progress of Reaction
Enthalpy
Progress of Reaction
Enthalpy
Progress of Reaction
Reactants
Reactants
Products
Reactants
Reactants
Products
Products
Products
A
C D
B
Learning Objectives
Candidates should be able to:
• Explain qualitatively, in terms of both of the Boltzmann distribution and of collision frequency, the effect of temperature change on the rate of reaction.
Learning Objectives
Candidates should be able to:
• explain that, in the presence of a catalyst, a reaction has a different mechanism, i.e. one of lower activation energy, and interpret this catalytic effect in terms of the Boltzmann distribution.
• describe enzymes as biological catalysts (proteins) which may have specific activity.
"A catalyst provides an alternative route for the reaction with a lower activation energy."
Effect of a catalyst
Heterogeneous catalysis: where the reactants and catalyst are in different physical states. Common in industrial processes.
Homogeneous catalysis: where the reactants and catalyst are in the same physical state. E.g. enzyme-catalysed reactions in cells.
Types of catalysis
Enzymes are proteins that act as biological catalysts. Without them the reactions that make life possible would be too slow for life to exist.
Enzymes
Learning Objectives
Candidates should be able to:
explain qualitatively, in terms of collisions, the effect of concentration changes on the rate of a reaction.
1 4 0
1 2 0
1 0 0
8 0
6 0
4 0
2 0
0T im e
Vol
ume
of C
O /
cm3
2
Am
ount
of
prod
uct
2H2O2 (aq) 2H2O (l) + O2(g)
Candidates should be able to:
• describe and explain redox processes in terms of electron transfer and/or of changes in oxidation number (oxidation state)
Learning Objectives
Starter activity
In groups can you select the most appropriate term/s to describe the following chemical reactions?
Learning Objectives
Zn + CuO ZnO + Cu
Definitions• Oxidation is gain of oxygen.
• Reduction is loss of oxygen.
Blast Furnace
An oxidising agent is a substance which oxidises something else. It gains electrons and is reduced.
A reducing agent reduces something else. It loses electrons and is oxidised.
Consider the following reactions:
SO2 + H2O + HgO H2SO4 + Hg
SO2 + 2H2O + Cl2 H2SO4 + HCl
Clearly in both reactions there is an oxidation of SO2(g) to SO3(g) i.e. H2SO4.
Yet the second reaction does not involve oxygen!
We clearly need a more universal definition of oxidation and reduction.
Universal definition
Mg + ZnCl2 MgCl2 + Zn
Mg + CuCl2 MgCl2 + Cu
Definitions•Oxidation is loss of electrons.•Reduction is gain of electrons.
Electron transfer
•The oxidation state of an uncombined element is zero. •The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
•The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
•The more electronegative element in a substance is given a negative oxidation state. The less electronegative one is given a positive oxidation state.
•Some elements almost always have the same oxidation states in their compounds:
elementusual oxidation state
exceptions
Group 1 metals always +1
Group 2 metals always +2
Oxygen usually -2except in peroxides and F2O (see below)
Hydrogen usually +1
except in metal hydrides where it is -1 (see below)
Fluorine always -1
Chlorine usually -1except in compounds with O or F (see below
Candidates should be able to:
• describe and explain redox processes in terms of electron transfer and/or of changes in oxidation number (oxidation state).
Learning Objectives
Can you work out the oxidation states of the transition metal elements in the following compounds?
• KMnO4
• K2Cr2O7
Starter activity
Naming compounds
•SnO
•SnO2
•FeCl2
•FeCl3
•PbCl4
•Cu2O
•Mn(OH)2
•NO2-
•NO3-
•SO32-
•SO42-
•MnO4-
•CrO42-
•VO3-
1. 2ClO3- 2Cl- + 3O2
2. 2Br- + 2H+ + H2SO4 Br2 + SO2 + 2H2O
3. 8I- + 8H+ + H2SO4 4I2 + H2S + 4H2O
4. I2 + SO3- + H2O 2I- + SO4
2- + 2H+
What is oxidised, what’s reduced?
Writing ionic equations
In a reaction chlorine gas oxidises iron(II) ions to iron(III) ions. In the process, the chlorine is reduced to chloride ions.
Write a balanced equation for this redox process.
Manganate(VII) ions, MnO4-, oxidise hydrogen
peroxide, H2O2, to oxygen gas. The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulphuric acid.
The manganate(VII) is reduced to Mn2+.
Write a balanced equation for this redox process.
Writing ionic equations
Writing ionic equations
This technique can be used just as well in examples involving organic chemicals.
Potassium dichromate(VI) solution acidified with dilute sulphuric acid is used to oxidise ethanol, CH3CH2OH, to ethanoic acid, CH3COOH.
The Cr2O72-
is reduced to Cr3+.
Write a balanced equation for this redox process.
• Candidates should be able to explain, including the electrode reactions, the industrial process of:
◦ the electrolysis of brine, using a diaphragm cell;
◦ the extraction of aluminium from molten aluminium oxide/cryolite; and
◦ the electrolytic purification of copper.
Learning Objectives
Environmental concerns
Fort William
Anglesey
Candidates should be able explain, in terms of rates of the forward and reverse reactions, what is meant by a reversible reaction and dynamic equilibrium.
Learning Objectives
Starter activity
In pairs, consider the reaction given below. How many facts about this reaction can you write down? Try to use the correct scientific terminology.
N2(g) + 3H2(g) 2NH3(g)
H –ve
Candidates should be able to state Le Chatelier’s Principle and apply it to deduce qualitatively (from appropriate information) the effects of changes in temperature, concentration or pressure, on a system at equilibrium
Learning Objectives
Put simply, Le Chatelier’s Principle states that:
If a system is at equilibrium, and a change is made in any of the conditions, then the system responds to counteract the change as much as possible.
Le Chatelier’s Principle
Effect of concentration
Suppose you have an equilibrium established between four substances A, B, C and D.
What would happen if you changed the conditions by increasing the concentration of A?
Effect of temperature
What would happen if you changed the conditions by increasing the temperature?
Candidates should be able to
deduce expressions for equilibrium constants in terms of concentrations, Kc, and partial pressures, Kp.
deduce whether changes in concentration, pressure or temperature or the presence of a catalyst affect the value of the equilibrium constant for a reaction.
calculate the values of equilibrium constants in terms of concentrations or partial pressures from appropriate data.
calculate the quantities present at equilibrium, given appropriate data.
Learning Objectives
Starter activity
In pairs, consider the reaction given below. If you wanted to make as much ammonia as possible what conditions would you use?
N2(g) + 3H2(g) 2NH3(g)
H –ve
K – the equilibrium constant
Equilibrium constant
•to provide a quantitative measure of the extent of a reaction;
•to determine the position of equilibrium.
Kc
a A + b B c C + d D
ba
dc
c BA
DCK
][][
][][
Calculating Kc
CH3CH2OH + CH3COOH CH3COOCH2CH3 + H2O
n(start) 1.0 1.0 0 0
n(equil.) 0.34 0.34 0.66 0.66
[ ] 3.4 3.4 6.6 6.6
Kc = 6.6 x 6.6 / 3.4 x 3.4 = 3.7 (no units)
Kp
Partial pressure
The total pressure exerted by a mixture of gases is the sum of the partial pressure of the gases.
pA = xA x ptot
Kp
Partial pressure and mole fraction
Partial pressure terms are expressed in SI units as Pa or kPa.
The equilibrium constant Kp
a A + b B c C + d D
Kp = ba
dc
BpAp
DpCp
)()(
)()(
Calculating Kp
PCl5(g) PCl3(g) + Cl2
n(start) 2.0 0 0
n(equil.) 0.8 1.21.2
x 0.25 0.375 0.375
P 166 249 249
Kp = 249 x 249 / 166 = 374 kPa
Change in temperature
∆H for reaction
Change in Temp.
Shift of Equilibriu
m
Yield of Product
Equilibrium
constantExo Increase Decrease Decrease
Exo Decrease Increase Increase
Endo Increase Increase Increase
Endo Decrease Decrease Decrease
Candidates should be able to describe and explain the conditions used in the Haber process and the Contact process, as examples of the importance of an understanding of chemical equilibrium in the chemical industry.
Learning Objectives
Starter activityThe gases SO2, O2 and SO3 are allowed to reach equilibrium. The partial pressures of the gases are pSO2
= 0.050 atm, pO2 = 0.025 atm, pSO3
= 1.00
atm.
Find the values of Kp for the equilibria
a) SO2(g) + ½O2(g) SO3(g)
b) 2SO2(g) + O2(g) 2SO3(g)
Comment on your results!
Process Contact HaberMain product Sulphuric acid Ammonia
Main uses of product
Paints, detergents and soaps, fertilisers and dyes.
Fertilisers, explosives, nitric acid, polymers.
Balanced equation for main reaction/s
1.S + O2 SO2
2. 2SO2 + O2 2SO3
3. SO3 + H2SO4 H2S2O7
4. H2S2O7 + H2O 2H2SO4
3H2 + N2 2NH3
Catalyst Vanadium(V) oxide Porous iron (with metal oxide promoter)
Is the equilibrium reaction exothermic or endothermic?
Exothermic Exothermic
Optimum conditions for highest yield
Low temperatureHigh pressure
Low temperatureHigh pressure
Actual conditions used
400 – 600oC1-2 atm
400 – 500oC200 atm
Why are these conditions chosen?
Catalyst ineffective at low T.High pressure uneconomical.
Reaction slow at low T.High pressure too costly.
Points of interest Reaction mixture cooled after each exothermic stage.Vast excess of air.
Ammonia removed as produced.N2 and H2 recycled.Waste heat re-used.
Starter Activity
A white solid is formed at X. Can you explain what is happening in this reaction? What words would you use to describe it?
Learning objectives:
Candidates should be able to:
• show understanding of, and use the Bronsted-Lowry theory of acids and bases.
• explain qualitatively the differences in behaviour between strong and weak acids and bases and the pH values of their aqueous solutions in terms of the extent of dissociation.
Properties of Acids
React with metals to
produce H2
Taste sour
Turn litmus red
Have a pH < 7
React with carbonates to produce
CO2
Produce H+ in solution
Neutralise alkalis
In 1884 Arrhenius stated that:
•Acids are substances which produce hydrogen ions in solution.
•Bases are substances which produce hydroxide ions in solution.
Allowed an explanation for neutralisation:
Early theory of acid behaviour