VCE Chemistry UNIT 3 NOTES PDF

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VCE Chemistry UNIT 3 NOTES...

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Combustion Reactions KDP - combustion of fuels as exothermic reactions with reference to the use of the joule as the SI unit of energy, energy transformations and their efficiencies and measurement of enthalpy change including symbol (∆H) and common units (kJ mol-1, kJ g-1, MJ/tonne) KDP - the comparison of exothermic and endothermic reactions including their enthalpy changes and representations in energy profile diagrams KDP - the writing of balanced thermochemical equations, including states, for the complete and incomplete combustion of hydrocarbons, methanol and ethanol, using experimental data and data tables KDP - the use of specific heat capacity of water to determine the approximate amount of heat energy released in the combustion of a fuel.

Systems and surrounding

Endothermic and exothermic reactions The SI unit of energy The SI unit for energy is the joule, J Joule conversion flowchart;

Endothermic and Exothermic Reactions Exothermic Reaction O

Total chemical energy of the products is less than the total chemical energy of the reactants o excess energy is released to the surroundings in the form of heat, making surroundings warmer

Endothermic Reaction O

Total chemical energy of the products is greater than the total chemical energy of the reactants o Energy is absorbed from the surrounding environment in the form of heat, making the surroundings colder

Thermochemical equations and energy profile diagrams Enthalpy change O

A measure of the amount of energy absorbed or released during chemical reactions. o Enthalpy change = ∆H o The enthalpy of the reactants - Hr o The enthalpy of the products – Hp ΔH = Hp – Hr

Enthalpy Change in Exothermic Reactions (Fig. 2.2.1) O O

When Hr is greater than Hp, energy is released from the system into the surroundings, so the reaction is exothermic The system has lost energy - ΔH < 0

Enthalpy Change in Endothermic Reactions (Fig 2.2.2) O O

When Hp is greater than Hr, energy must be absorbed from the surroundings, so the reaction is endothermic. The system has gained energy - ΔH > 0

Thermochemical equations Definition: A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH.  The ΔH value in a thermochemical equation usually has the unit’s kJ/mol o o

The ΔH value (amount of energy in kJ) in a thermochemical equation corresponds to the mole amounts specified by the balanced equation – ΔH value changes according to mol If a chemical reaction is reversed the sign of the ΔH value will become the opposite (i.e. – will become + and vice versa) but value will remain the same

Effects of State on Enthalpy  The state of the species in a chemical reaction affects the enthalpy change of the reaction. o

For example, both of the following equations represent physical changes involving water. They have different ΔH values because the states are different. H2O(s) → H2O (l) ΔH = +6.00 kJ mol–1 H2O (l) → H2O (g) ΔH = +40.7 kJ mol–1

Hess’s Law  The overall enthalpy change from reactants to products is the same regardless of what or how many steps it takes are taken to get there  The overall enthalpy change is the sum of the individual steps required to get there Example: Calculate the enthalpy for this reaction: 2C(s) + H2(g)  C2H2(g) ΔH° = ??? kJ Given the following thermochemical equations: C2H2(g) + 5⁄2O2(g)  2CO2(g) + H2O(ℓ) ΔH° = -1299.5 kJ C(s) + O2(g)  CO2(g) ΔH° = -393.5 kJ H2(g) + 1⁄2O2(g)  H2O(l) ΔH° = -285.8 kJ 1) Determine what we must do to the three given equations to get our target equation: a) first eq: flip it so as to put C2H2 on the product side b) second eq: multiply it by two to get 2C c) third eq: do nothing. We need one H2 on the reactant side and that's what we have 2) Rewrite all three equations with changes applied:

2CO2(g) + H2O(l)  C2H2(g) + 5⁄2O2(g) ΔH° = +1299.5 kJ 2C(s) + 2O2(g)  2CO2(g)

ΔH° = -787 kJ

H2(g) + 1⁄2O2(g)  H2O(l) ΔH° = -285.8 kJ 3) Examine what cancels: 2CO2 ⇒ first & second equation H2O ⇒ first & third equation 5 ⁄2O2 ⇒ first & sum of second and third equation 4) Add up ΔH values for our answer: +1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ

Activation energy Definition: The minimum amount of energy required to break all bonds between reactants so that a reaction can begin

Energy Profile Diagrams  The energy changes that occur during the course of a chemical reaction can be shown on an energy profile diagram o The energy profile diagram for an exothermic combustion reaction indicates that the enthalpy of the products is always less than the enthalpy of the reactants – the ΔH is negative o The energy profile diagram for an endothermic reaction shows that the enthalpy of the products is greater than the enthalpy of the reactants – the ΔH is positive

Combustion reactions  Combustion reactions are exothermic reactions in which the reactant combines with oxygen to produce oxides (AKA oxidation reactions)  The combustion of a hydrocarbon produces carbon dioxide and water, provided there is enough oxygen present

o o

The enthalpy change is given in kJ mol–1 and is a negative value as the reaction is exothermic Water is released as a gas giving it the state (g)

Complete and Incomplete combustion  Complete combustion occurs when oxygen is in excess o The only products are carbon dioxide and water methanol: CH3OH (l) + 3/2O2 (g)  CO2 (g) + 2H2O (l) ethanol: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)  Incomplete combustion occurs when oxygen is limited o as less oxygen is available, not all of the carbon can be converted to carbon dioxide, hence, carbon monoxide and/or carbon are produced instead methanol: CH3OH (l) + O2 (g)  CO (g) + 2H2O (l) ethanol: C2H5OH(l) + 5/2O2(g)  2CO(g) + 3H2O(g)

Heat of combustion Definition: the total amount of energy released when a specific amount of fuel undergoes complete combustion  Water produced should be shown in the liquid state in the chemical equation  The heat of combustion can be given the symbol ΔHc (enthalpy of combustion) o Heat energy is released during combustion, so ΔHc always has a negative value.  Only fuels that exist as pure substances can have their heat of combustion measured in kJ mol-1

Calculating Energy Released The energy released when n mol of a fuel burns is calculated by the equation: Energy = n(fuel) × ΔHc

Calculating energy content per gram  The energy content/heat of combustion of a fuel is often expressed in units of kilojoules per gram.  For a pure substance, the heat of combustion per gram can be calculated by dividing the heat of combustion (kJ mol–1) by the molar mass of the substance For example, for ethanol o Heat of combustion per mole = –1367 kJ mol–1 o Molar mass = 46.0 g mol–1 o Heat of combustion per gram = –1367 / 46.0 = –29.7 kJ g–1

Common units for heat of combustion

  

1 kJ = 10–3 MJ 1 g = 10–6 tonne This can be simplified to 1 kJ g–1 = 103 MJ/tonne

Determining the heat of combustion of fuels Specific heat capacity of water Definition: a measure of the amount of energy, in joules, needed to increase the temperature of a one gram of water by 1°C.

   

Symbol = C It is usually expressed in joules per gram per degrees Celsius, i.e. J g–1 °C–1 The higher the specific heat capacity, the more effectively a material stores heat energy. Water has a specific heat capacity of 4.18 J g–1 °C–1. o This means that 4.18 joules of heat energy are needed to increase the temperature of 1 gram of water by 1°C o This relatively high value is due to the hydrogen bonds between the water molecules.

Specific Heat Capacity Equation  The heat energy required to increase the temperature of a given mass of water by a particular amount can be calculated through the equation:

q = C × m x ΔT o o o o

q = amount of heat energy in joules C = specific heat capacity of substance (in J g–1 °C–1) m = mass in grams ΔT = temperature change in °C

Gases KDP - the definition of gas pressure including units, the universal gas equation and standard laboratory conditions (SLC) at 25 °C and 100 kPa KDP - calculations related to the combustion of fuels including use of mass-mass, mass-volume and volume-volume stoichiometry in calculations of enthalpy change (excluding solution stoichiometry) to determine heat energy released, reactant and product amounts and net volume of greenhouse gases at a given temperature and pressure (or net mass) released per MJ of energy obtained

Volume and pressure Volume Definition: Volume is the quantity used to describe the space that a substance occupies  Because a gas occupies the whole container that it is in, the volume of a gas is equal to the volume of its container  Usually measured in either mL or L

Pressure Definition: the force exerted on a unit area of a surface by the particles of a gas as they collide with the surface  The total pressure of a mixture of gases is the sum of partial pressures of all gases in mixture o Partial pressure – the force exerted on a unit area of surface by an individual gas particle (either O2 or N2) as it collides with the surface

Units of pressure Pressure is determined by the rule:

Pressure = force/area P = F/A  Commonly measured in kilopascals (kPa)  Pascal (Pa) – 1 kPa = 1000 Pa  

Atmosphere (atm) – 100kPa = 0.987 atm Millimetres of mercury (mmHg) – 100kPa = 750 mmHg

Universal Gas equation Gas laws Boyle’s Law Volume and pressure of a gas are inversely proportional (at constant temperature and mole) o As volume increases, pressure decreases o As volume decreases, pressure increases V ∝ 1/P (for constant T and n)

Avogadro’s law Mole and volume are proportional to each other (at constant pressure and temperature) o As number of moles increase, volume increases by the same factor o As number of moles decreases, volume decreases by the same factor

V ∝ n (for constant P and T)

Charles’ Law Temperature and volume are proportional (at constant pressure and mole)  Temperature and volume are proportionally related o As number of moles increase, volume increases by the same factor o As number of moles decreases, volume decreases by the same facto V ∝ T (for constant P and n)

Molar volume of gas Definition: molar volume is the amount of space occupied by 1 mole of any gas at a particular pressure and temperature  Molar volume is represented through the symbol Vm  The amount of volume one mole of gas occupies is entirely dependent on its temperature and pressure  Molar volume of any gas under identical temperature and pressure is the same

Vm = V/n or n = V/Vm At SLC (298 K and 100kPa) an ideal gas has a molar volume of a gas is 22.7 L

The universal gas equation  The following three laws describe the behaviour of gases under different conditions: Use molar volume equation when V ∝ 1/P (for constant T and n) the temperature and pressure is at V ∝ T (for constant P and n) SLC – Use universal gas equation V ∝ n (for constant P and T) when temperature and pressure  We can combine all three laws by writing: are at any other random value

PV= nRT

o o o o o

R - Universal gas constant = 8.31 J K-1 mol-1 P – Pressure = kilopascals, kPa V – Volume = litres, L n – Mole = mol T – Temperature = kelvin scale, K

Chapter 4 – Redox Reactions KDP - redox reactions with reference to electron transfer, reduction and oxidation reactions, reducing and oxidising agents, and use of oxidation numbers to identify conjugate reducing and oxidising agents KDP - the writing of balanced half-equations for oxidation and reduction reactions and balanced ionic equations, including states, for overall redox reactions

Oxidation and Reduction Redox reactions Redox reactions involve the transfer of electrons from one chemical species to another

Oxidation = loss of electrons o o

Metals have low electronegativities so they loose electrons to become stabilised Reductants/reducing agents are substances that cause reduction but are themselves oxidised (i.e. lose electrons)

Reduction = gain of electrons o o

Non-metals have high electronegativities, so they gain electrons to become stabilised Oxidants/oxidising agents are substances that cause oxidation but are themselves reduced (i.e. gain electrons)

 Oxidation and reduction occur simultaneously as electrons are transferred between the reactants

Reduction and Oxidation Half-Equations  Half-equations show what is happening as electrons are transferred in a redox reaction o One half-equation describes the oxidation (loss of electrons) would be shown as: X-  X + eo The other half-equation describes the reduction (gain of electrons) would be shown as : X + e– → X–

Balancing half equations in acidic environments: Balancing half equations in acidic conditions occurs in the following steps; Cr2O72-(aq)  Cr3+ (aq) HNO2(aq)  NO3− (aq) 1. Balance all elements except hydrogen and oxygen in the half-equation Cr2O72-(aq)  2Cr3+ (aq) HNO2(aq)  NO3− (aq)

2. Balance the oxygen atoms by adding water molecules Cr2O72-(aq)  2Cr3+ (aq) + 7H2O (l) HNO2(aq) + H2O (l)  NO3− (aq) 3. Balance the hydrogen atoms by adding H+ ions (which are present in acidic solution). Cr2O72-(aq) + 14H+ (aq)  2Cr3+ (aq) + 7H2O (l) HNO2(aq) + H2O (l)  NO3− (aq) + 3H+ (aq)

4. Balance the charge in the equation by adding electrons (does not have to be zero, just has to be the same on either side) o The chromium reaction has (14+) + (2-) = 12+ on the left side and a (2 * 3+) = 6+ on the right side. To balance, add 6 electrons to the left side to make both sides have the charge of 6+ 6e- + Cr2O72-(aq) + 14H+ (aq)  2Cr3+ (aq) + 7H2O (l) o The Nitric acid reaction has no charge on the left and a (3+) + (-1) = 2+ charge on the right. To balance add 2 electrons to the right side to make both sides have no charge HNO2(aq) + H2O (l)  NO3− (aq) + 3H+ (aq) + 2e5. Scale the reactions so that the electrons are equal in both half equations - You may need to multiply one or both the half-equations (all coefficients) by a factor to ensure that the electrons balance and can be cancelled out in the overall equation o The chromium reaction has 6e- and the nitric acid reaction has 2e-, so it needs to be multiplied by 3 6e- + Cr2O72-(aq) + 14H+ (aq)  2Cr3+ (aq) + 7H2O (l) − HNO2(aq) + H2O (l)  NO3 (aq) + 3H+ (aq) + 2e- x3  3HNO2(aq) + 3H2O (l)  3NO3− (aq) + 9H+ (aq) + 6e6. Add the two reactions and cancel out common terms 6e- + Cr2O72-(aq) + 14H+ + 3HNO2(aq) + 3H2O (l)  2Cr3+ (aq) + 7H2O (l) + 3NO3− (aq) + 9H+ (aq) + 6eo All of the electrons cancel out as well as 3 water molecules and 9 protons. This leaves the balanced net reaction of: 3HNO2(aq) + 5H+(aq) + Cr2O72- (aq)  3NO3− (aq) + 2Cr3+(aq) + 4H2O(l)

Balancing half equations in basic environments: Balancing half equations in basic conditions occurs in the following steps; Ag (s)  Ag2O (aq) Zn2+ (aq)  Zn (s)

1. Balance all elements except hydrogen and oxygen in the half-equation 2Ag (s)  Ag2O (aq) Zn2+ (aq)  Zn (s)

2. Balance the oxygen atoms by adding water molecules H2O (l) + 2Ag (s)  Ag2O (aq) Zn2+ (aq)  Zn (s)

3. Balance the hydrogen atoms by adding H+ ions (which are present in acidic solution) H2O (l) + 2Ag (s)  Ag2O (aq) + 2H+ (aq) Zn2+ (aq)  Zn (s)

4. Balance the charge in the equation by adding electrons (does not have to be zero, just has to be the same on either side) H2O (l) + 2Ag (s)  Ag2O (aq) + 2H+ + 2eZn2+ (aq) + 2e-  Zn (s)

5. Scale the reactions so that the electrons are equal in both half equations – in this case they are already equal

6. Add the reactions and cancel the electrons H2O (l) + 2Ag (s) + Zn2+ (aq)  Ag2O (aq) + 2H+ + Zn (s)

7. Add OH- ions to balance the H+ ions o

There are 2 net protons in this equation, so add 2 OH- ions to each side. H2O (l) + 2Ag (s) + Zn2+ (aq) + 2OH- (aq)  Ag2O (aq) + 2H+ + Zn (s) + 2OH- (aq)

8. Combine OH- ions and H+ ions that are present on the same side to form water H2O (l) + 2Ag (s) + Zn2+ (aq) + 2OH- (aq)  Ag2O (aq) + Zn (s) + 2H2O (l)

9. Cancel common terms and complete full equation 2Ag (s) + Zn2+ (aq) + 2OH- (aq)  Ag2O (aq) + Zn (s) + H2O (l)

Oxidation numbers Definition: a number assigned to an element representing number of electrons lost (+) or gained (-) by the element’s atom

Rules when assigning oxidation numbers 1. Oxidation number of a free element is 0 2. Sum of oxidation numbers of an ion is equal the charge of the ion (applies to both polyatomic and simple ions) 3. Oxidation number of a metal in a compound is equal to its ionic charge 4. Oxidation number of hydrogen is +1, except when it forms metal hydrides (-1) 5. Oxidation number of oxygen is -2 except when it is bonded to fluorine (+2) or is in a peroxide (-1) 6. The most electronegative element is assigned the negative oxidation number (i.e. Fluorine is the most electronegative element, so it always is -ve)

Using oxidation numbers Using oxidation numbers to identify oxidation and reduction  Redox only occurs if a change in oxidation numbers is seen from reactants to products  If redox has occurred, you can tell what has happened to each element as; o oxidation involves an increase in oxidation number o reduction involves a decrease in oxidation number

Using oxidation numbers to identify conjugate redox pairs Includes:  Oxidant and its reduced form (gained electrons and therefore decreased in oxidation number)  Reductant and its oxidised form (lost electrons and therefore increased in oxidation number)

Electrochemical series Series containing a set of reduction half equations o

o

The strongest oxidising agent, F2, is at the top left of the table  strong oxidising agents accept electrons more readily than weak ones  strong oxidising agents have weak conjugate reducing agents The strongest reducing agent, Li, is at the bottom right of the table  Strong reducing agents donates electrons more readily than weak ones  Strong reducing agents have weak conjugate oxidising agents

Eo value  a measure of the relative spontaneity (in volts) of a particular reduction reaction relative to the standard hydrogen electrode (H +/H2 reduction reaction) at standard conditions Eocell  maximum potential of two half-cells indicating the extent of reaction cell voltage = E° of oxidising agent – E° of half-cell of reducing agent or cell voltage = higher half-cell E° – lower half-cell E°

o If Eocell is positive the reaction is spontaneous o If Eocell is negative the reaction is non-spontaneous o The further apart two reactions are on the electrochemical series, the more voltage it can produce Limitations to electrochemical series

Galvanic Cells KDP - galvanic cells as primary cells and as portable or fixed chemical energy storage devices that can produce electricity (details of specific cells not required) including common design features (anode, cathode, electrolytes, salt bridge and separation of half-cells) and chemical processes (electron and ion flows, half-equations and overall equations) KDP - the comparison of the energy transformations occurring in spontaneous exothermic redox reactions involving direct contact between reactants (transformation of chemical energy to heat energy) compared with those occurring when the reactants are separated in galvanic cells (transformation of chemical energy to electrical energy) KDP - the use of the electrochemical series in designing and constructing galvanic cells and as a tool for predicting the products of redox reactions, deducing overall equations fr...