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TOPIC 5: ENERGETICSGrade 11 IB CHEMISTRY IB Core Objective5.1.1Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)

Define: Give the precise meaning of a word, phrase or physical quantity.5.1.1Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)Thermochemistry = study of energy changes in chemical reactions

Most chemical reactions absorb or evolve energy (usually as heat, sometimes as light and mechanical energy)

Energy is measured in Joules (J)

5.1.1Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)Enthalpy (H, heat content) = the total energy of a systemSome is stored as chemical potential energy in the chemical bondsEnergy is absorbed to BREAK bondsEnergy is released when bonds are MADEThe potential energy of the bonds changes in chemical reactions this is ENTHALPY CHANGE

This quantity, H, is called the enthalpy of reaction, or the heat of reaction.5.1.1Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)5.1.1Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)1st Law of Thermodynamics:Energy can not be created nor destroyed, but it can change formEnergy lost = Energy gainedSystem + Surroundings = Universe (Which is constant)

ReactionsEndothermic: Energy is used by the system Exothermic: Energy is produced by the systemThe System represents the chemical reaction

The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).

The surroundings are everything else (here, the cylinder and piston).5.1.1Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts Hreactants

5.1.1Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)5.1.1Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (Ho)The change in enthalpy H = the net energy after bonds are broken and remade

H = (Negative) More energy was created than used. EXOTHERMIC

H = (Positive) More energy was used than created. ENDOTHERMIC

IB Core Objective5.1.2 State that combustion and neutralization are exothermic processes.

Combustion: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = - 882 kJ mol -1

Neutralization: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H= -57.3 kJ mol-1IB Core Objective5.1.3 Apply the relationship between temperature change, enthalpy change and the classification of a reaction a endothermic or exothermic.

What happens to the temperature when the reaction is exothermic?A: Temperature goes up.What happens to the enthalpy change?A: It is a negative value.If the same moles of substance were reacted in a larger container of water, would the temperature and enthalpy values stay the same?A: Enthalpy value would stay the same, the temperature however would not go up as much.IB Core Objective5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of the enthalpy change for the reaction.5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of the enthalpy change for the reaction.Exothermic

EndothermicProducts are more stable

Heat energy is RELEASED into the SURROUNDINGSReactants are more stable

Heat energy is ABSORBED into the SYSTEMH is NegativeH is PositiveEnthalpyEnthalpyStandard StateEnthalpy is affected by multiple factors (concentration, pressure, state of reactants, temperature))

H Under SATP (Standard Ambient Temperature and Pressure)P = 101.3kPaT = 25oC or 298 KC = 1mol/dm3 (Concentration for aqueous solutions aq)Standard state (physical state of element under these conditions)

(Theta) Means under SATPDifferent from STP (Temperature is not 0 oC)IB Core Objective5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed. 5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed. Specific Heat CapacityThe energy required to heat 1g of substance 1 degree CelsiusSpecific Heat Capacity of water1 g of H2O to heat 1 degree C requires 4.18 J

Therefore the specific heat capacity of water is 4180 Jkg-1K-14.18 Jg-1 K-1

Typically we use this for labs due to the small quantities used.5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed. CalorimetryTo measure the energy of a system is hardTo measure the energy of surroundings is easy

q = mcTq = Energy change in the surroundingsm = mass of surroundings (g)c = Specific Heat Capacity of surrounding substance (Sometimes s is used instead of c)T = Temperature change

Energy must have come from the systemFor us, this will be the mass of water used5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed. QuestionsHow much heat is needed to warm 250 g of water from 22C to near its boiling point 98C? A: 7.9 x 104 J or 79kJ.

Calculate the quantity of heat used when you mix 50cm3 of 1.0M HCl and 50cm3 of NaOH in a coffee-cup calorimeter, and the temperature increases from 21.0C to 27.5C. Assume the density is 1.00 g/cm3.A: 2.7 kJCalculate the enthalpy change for the above reaction in kJ/mol.A: -54 kJ/molQuestionsExcess solid Magnesium is added to a 100g of a 2M solution of Copper(II) Sulphate. The temperature increased from 20.0oC to 65.0oC. (Since the solution is largely water we will assume specific heat capacity as 4.18 Jg-1K-1.

What is the enthalpy change for the reaction?Dont forget it must be calculated as a ratio of a full mol.Push for answerH = -94.1 kJ/mol.

IB Core Objective5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions.5.2.4 Evaluate the results of experiments to determine enthalpy changes.5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions.5.2.4 Evaluate the results of experiments to determine enthalpy changes.

For liquids, a calorimeter should be well insulated and the heat capacity should be low.Calorimetry: Technique used to measure the enthalpy associated with a particular change.Calorimetry depends on the assumption that no heat is gained from or lost to the surroundings. Even with well insulated calorimeters, heat exchange with surroundings is a major source of error.

5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions.5.2.4 Evaluate the results of experiments to determine enthalpy changes.

In combustion experiments, where burning gas is used to heat liquid in a calorimeter, the error is quite large.Temperature rises are much less than expected and thus H values are less than literature values.

IB Core Objective5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions.Question: 20.0 cm3 of 2 mol dm-3 aqueous sodium hydroxide is added to 30.0 cm3 of hydrochloric acid of the same concentration, the temperature increases by 12.0 C. For dilute aqueous solutions, we can assume the density is 1.00 g cm-3.What is H?A: -62.7 kJ mol-1IB Core Objective5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.

Students should be able to use simple enthalpy cycles and enthalpy level diagrams and to manipulate equations. Students will not be required to state Hesss law.5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.

Hesss LawHeat of the whole = the sum of the parts

Reaction that take a direct route or multiple step route make no difference with Enthalpy.

ANALOGYI just bought a $1200 TV, I could either pay 12 equal instalments of $100, pay 2 instalments of $600 or pay it all $1200 at once. How I do it doesnt matter, in the end I still pay $1200. In enthalpy terms: H1 = H2 + H3

5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.Calculate Hf for C2H2(g)C2H2(g)

2O2(g)2CO2(g) + H2O(l)2C (s) + H2(g)

Hf =Hf =Hc =Hf = H Hc =2O2(g)CO2 -395 kJ mol-1H2O -287 kJ mol-1C2H2 -1301 kJ mol-1A: 224 kJ mol-1IB Core Objective5.4.1 Define the term average bond enthalpy.

Enthalpies are a measure of the strength of a covalent bond: the stronger the bond, the more tightly the atoms are joined together.Breaking of a chemical bond requires energy, and is thus and endothermic process.Average bond enthalpy: The bond enthalpy for a compound will be affected by surrounding bonds, therefore we use the average bond enthalpy.IB Core Objective5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic. Energy 5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic. It takes energy to break bonds

Energy is produced upon bond formationH = Energy IN (Bonds broken) Energy OUT(Bonds formed)H = 1(H-H 436) + (O=O 496) 2(O-H 463)H = 684 926H = -242 (Energy is left over, Exothermic)

OOHH

496 kJ/mol436 kJ/mol463 kJ/mol463 kJ/mol5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic. Sometimes Bond energy is effected by surrounding bonds so an average must be used.

Calculate the Enthalpy of change for the following:CH4(g) + O2(g) CO2(g) + H2O(g)Compare to enthalpy of combustion value (Data book)BondBond Enthalpy kJ/molH-H(g)436Cl-Cl(g)242F-F(g)158H-Cl(g)431H-F(g)562BondAvg. Bond Enthalpy kJ/molC-C(g)348C=C(g)612C=C(g) (in benzene)518C-H(g)412C=O(g)743O-H(g)463N-H(g)388O=O(g)496H = {Bonds broken Bonds formed}Why the difference?Difference calculating enthalpy using heat of formation/ combustion and calculating it using bond enthalpies is due to the average bond enthalpy used and the fact that heat of formation/combustion requires things in there standard state...ie H2O is liquid not gas...it takes more energy to vaporize this.315.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic.Given that the enthalpy change for the reaction N2(g) + 3Cl2(g) 2NCl3(g) is +688kJ/molCalculate the enthalpy of the N-Cl bond, given that the bond enthalpies in the nitrogen molecule and the chlorine molecule are 944kJ/mol and 242kJ/mol.

A: 164 kJ/molIB HL Objective15.1.1 Define and apply the terms standard state, standard enthalpy change of formation (Hf) and standard enthalpy change of combustion (Hc)

You have already learned about standard state. What are the conditions?A: Form normally found at 298K, 101.3 kPa, 1 mol dm-315.1.1 Define and apply the terms standard state, standard enthalpy change of formation (Hf) and standard enthalpy change of combustion (Hc)

Hc = Standard enthalpy change of combustionEnthalpy change when one mole of compound is burned in excess oxygen under standard conditions. (Always exothermic).

Hf = Standard enthalpy change of formationAmount of energy released or absorbed in the formation of one mol of compound from elements in their normal states.

By using the definition of standard state, the enthalpy of formation of an element in the standard state is zero.

IB HL Objective15.1.2 Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion.

H = Hf(products) - Hf(reactants)15.1.2 Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion.

For the following reaction, find the enthalpy of formation:2 C(graphite) + 3H2(g) + O2(g) C2H5OH(l)

A: Hf = -277 kJ mol-1 (All the reactants are in their elemental, standard state, so they would be zero!)Find the enthalpy of formation for the following reaction

15.1.2 Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion.

Find the enthalpy of formation for the following reaction:

NH4NO3 (s) N2O(g) +2H2O(l)The enthalpies for formation for the above compounds are: NH4NO3 (s) :-366 kJ mol-1, N2O(g) +82 kJ mol-1, and H2O(l) -285 kJ mol -1

A: -122 kJ mol-1

When we have learned more about Hesss Law, we will return to this later

15.1.2 Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion.

Find the formation of combustion of benzene using enthalpies of formation values.First, balance the equation:C6H6(l) + O2(g) CO2(g) + H2O(g)A: C6H6(l) + O2(g) 6CO2(g) + 3H2O(l)The enthalpies of formation are: CO2(g): -393.5 kJ mol-1, H2O(g): -285.8 kJ, benzene (find in your data booklet), O2???A: -3267 kJ mol-1 (Compare this answer with the data booklet).

IB HL Objective15.2.1 Define and apply the terms lattice enthalpy and electron affinity.

Lattice enthalpy: Enthalpy change to convert one mole of a solid ionic compound into gaseous ions or vice versa.

Electron affinity: Enthalpy change when one mole of gaseous atoms or anions gain electrons to form a mole of negatively charged gaseous ions.15.2.1 Define and apply the terms lattice enthalpy and electron affinity.

Standard enthalpy change of atomization (also known as standard enthalpy of vaporization). This is the enthalpy required to change one mole of atoms from the standard state to the gaseous state.Na(s) Na(g) H +103kJ mol-1

Exothermic or endothermic?15.2.1 Define and apply the terms lattice enthalpy and electron affinity.First ionization energy (remember this?)Na(g) Na+(g) + e- H = +494 kJ mol-1

Enthalpy atomization of ClCl2(g) Cl(g) H = +121 kJ mol-1First electron affinity of ClCl(g) + e- Cl-(g) H = -364 kJ mol-1

IB HL Objective15.2.3 Construct a Born-Haber cycle for Group 1 and 2 oxides and chlorides, and use it to calculate an enthalpy change.15.2.3 Construct a Born-Haber cycle for Group 1 and 2 oxides and chlorides, and use it to calculate an enthalpy change.Na(s) + Cl2(g) NaCl(S) Na+(g) + Cl-(g) LATTICE ENTHALPYHof =Na(g) + Cl (g) Cl-(g) Na+ (g)15.2.3 Construct a Born-Haber cycle for Group 1 and 2 oxides and chlorides, and use it to calculate an enthalpy change.AffinityFormation of NaCl(s) from its gaseous elements.

1) Na(s) Na(g) Hoat = 103 kJ/mol2) Cl2(g) Cl(g)Hoat = (242) kJ/mol3) Cl(g) + e- Cl(g)-Ho = -364 kJ/mol4) Na(g) Na+(g) Hoat = 500 kJ/mol

Find the enthalpy of formation. Literature value for lattice enthalpy can be found in the data booklet.A: -430 kJ/mol

15.2.3 Construct a Born-Haber cycle for Group 1 and 2 oxides and chlorides, and use it to calculate an enthalpy change.Draw a Born-Haber cycle for the formation of magnesium oxide, and calculate the enthalpy of formation. The enthalpy of atomization of magnesium is +150 kJ/mol, and for oxygen it is +249 kJ/mol.The second ionization energy of magnesium is +1450 kJ/mol. Use the data booklet to find other relevant information.A: -547 kJ/molIB HL Objective15.2.2 Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds.

The relative value of the theoretical lattice enthalpy increases with higher ionic charge and smaller ionic radius due to increased attractive forces. 15.2.2 Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds.

The greater the charge on the ions, the greater the electrostatic attraction and thus greater the lattice enthalpy, and vice versa.The larger the ions, then the greater the separation of the charges and the lower the lattice enthalpy, and vice versa. Lattice enthalpy of MgO > NaCl. Why?A: Increased ionic charge.Lattice enthalpy of KBrliquid>solid.Gas pressure increases, then entropy decreases (reduces volume for gas particles to move in). When a solid or liquid dissolves in a solvent, entropy increases.Increase the number of moles increases the entropy.Heating a substance increases the entropy, since this increases the movement of particles.

IB HL Objective15.3.2 Predict whether the entropy change (S) for a given reaction or process is positive or negative.

S = S(Products) S(Reactants)

So if the products have more disorder than the reactants, would the entropy change be positive or negative?A: Positive

15.3.2 Predict whether the entropy change (S) for a given reaction or process is positive or negative.

Consider the following equation:NH4Cl(s) NH3(g) + HCl(g)State S as positive or negative, and explain why.

A: Positive. (S=+285 J K-1 mol-1) More moles on product side, so more disorder. Also, solid has become a gas, so more disorder.IB HL Objective15.3.3 Calculate the standard entropy change for a reaction (S) using standard entropy values (S).

S = S(Products) S(Reactants)

15.3.3 Calculate the standard entropy change for a reaction (S) using standard entropy values (S).

Calculate the entropy change for the combustion of methane (CH4).First, write the equation.Make a prediction if it will be positive or negative, and explain why.Next, plug in the values and calculate. You can find the entropy value for methane in your data booklet. O2 is 205, CO2 is 214, and H2O is 70.

A: -242 J K-1 mol-1IB HL Objective15.4.1 Predict whether a reaction or process will be spontaneous by using the sign of G.

A reaction will be spontaneous when G has a negative value.Spontaneous: Once started, a reaction will continue without any extra energy having to be added. Remember redox? An electrochemical cell can create energy through a spontaneous reactions. So if Ecell is positive, what would G be?A: Negative15.4.1 Predict whether a reaction or process will be spontaneous by using the sign of G.

Is the combustion of coal spontaneous at room temperature?Yes. The activation energy is different than spontaneous reactions. You have to get the coal going, but once you do, it continues to react.What about the combustion of diamond at room temperature?Yes, although it has a very high activation energy. Once this is reached, it will be a spontaneous reaction. IB HL Objective15.4.2 Calculate G for a reaction using the equationG = H TS

and by using values of the standard free energy change of formation, Gf.15.4.2 Calculate G for a reaction using the equation G = H TS and by using values of the standard free energy change of formation, Gf.

Lets go back to the combustion of diamond at room temperature (298K):C (s) + O2(g) CO2(g)H = -395.4 kJ mol-1.S = +6.6 J K-1 mol-1.What is G?A: -397.4 kJ mol-1Would this reaction be spontaneous at all temperatures? Why or why not? Refer to the equation!A: If H is negative and S is positive, then G will always be negative. IB HL Objective15.4.3 Predict the effect of a change in temperature on the spontaneity of a reaction, using standard entropy and enthalpy changes and the equation:G = H TS

If both S and H are positive, then spontaneity depends on T. (Spontaneous at higher temperatures).If both S and H are negative, then spontaneity depends on T. (Spontaneous at lower temperatures).