Energetics and Enthalpy Changes All physical and chemical processes involve the energy changes. The variations in the strengths of forces of attraction (like intermolecular forces) are responsible for the energy changes in the physical processes.Q-1 Why the energy is absorbed in when water is converted into vapors? Is it a physical or chemical change? The breaking of older bonds and making of newer bonds are responsible for the energy changes in chemical processes. The study of exchange of energy between the reacting chemicals and surroundings is called energetics or thermochemistry. Exothermic and Endothermic ReactionsPropertyExothermic ReactionsEndothermic Reactions

Heat changeHeat is evolved to the surroundingsHeat is absorbed from the surroundings

Temperature changeTemperature rises above room temperature eventually and falls back to room temperature after a long timeTemperature falls below room temperature eventually and rises back to room temperature after a long time

Energies of reactants and productsDue to loss of heat energy, products now have less energyDue to gain of heat energy, products now have high energy

Stabilities of reactants and productsDue to loss of energy, products are more stableDue to gain of energy, products are less stable

Strengths of bondsBonds in products are strongerBonds in products are weaker

Sign of HNegative signPositive sign

Q-2 Draw a graph between time (start of reaction to a very long time) for an exothermic reaction and its temperature change. No need of units on the axis. Just label them. Temperature is the measurement of the kinetic energies not the potential energies of the particles. Q-3 An exothermic reaction releases energy, even then its temperature does not fall below room temperature. Explain in terms of potential energies and kinetic energies of the particles. Breaking forces of attraction either bonds or intermolecular forces and separating oppositely charged ions absorb energy (endo). Decomposition reactions and photosynthesis are common examples.

H2O(l) H2O(g)HCl(g) H(g) + Cl(g) Making forces of attractions, bonds and joining oppositely charged ions releases energy (exo). Combustion reactions are examples.

Na+(g) + Cl-(g) NaCl(s) A reaction is at the same time exo and endo because both the bond breaking and bond making takes place. It is the net change which makes it endo or exo.Explaining the reactions as endo or exo in terms of strengths of bonds of reactants and products.In an exothermic reaction the energy released in bond making is greater as compared to the energy absorbed in bond breaking. So bonds in products are stronger than the bonds in reactants. Stronger bonds are more stable and have less energy in them. But to break them more energy is required. e.g bonds of methane are store houses of energy (at high energy level)Homework: Write answers in the notebooks of questions on page 33 of the book (replace the word temperature oC with energy J in 3).Some important terminology in thermochemistrySystem: The assembly of reactants, products, solvent and catalyst of a reaction is called system.Surroundings: Outside the system everything is called surroundings. Container in which the reaction is taking place, air all are surroundings.Boundary: A real or imaginary surface which separates the system from surroundings is called boundary. It is the part of system like the outer surface of a system.Open System: A system which can exchange both the energy and matter with the surroundings.Closed System: A system which can exchange energy but not the matter with the surroundings.Isolated System: A system which neither exchanges energy nor matter with the surroundings.Principle of conservation of energy: Energy cannot be created or destroyed.Or energy of the universe is constant. Or change of energy of the system and surroundings equal but opposite. Energy change of a reaction depends upon the conditions of the reaction to some extent.Enthalpy (H): The total heat content of a system at constant pressure is called enthalpy of the system. The absolute enthalpy of system can never be measured.Enthalpy Changes (H): The energy change of a system at constant pressure is called enthalpy change of the reaction. It can be measured. H = Hproducts - Hreactants Stop! Enthalpy of reactants and products can never be measured. So what is the significance of the above formula?Answer: This formula is not used to calculate the enthalpy change of a reaction. This just gives us the idea of relative energies of reactants and products. IfHproducts > Hreactants reaction is endothermic and vice versa. We cannot measure the enthalpy of a cup of hot tea but when it gets cool, how much enthalpy is lost can be calculated by the change in its temperature.Enthalpy level diagrams for exothermic and endothermic reactions

fig. 1.2.5 enthalpy level diagrams show the enthalpy changes in reactions. The length of an arrow in the energy diagrams shows magnitude of energy transferred. Its direction downwards tells that the reaction is exothermic and vice versa. Energy absorbed or released brings ultimate change in potential energy which is linked with the strengths of bonds or forces of attraction.Homework: Write answers in the notebooks of questions (1, 2 and 3) on page 35 of the bookEnergy Demands The amount of energy needed by a substance depends upon the amount of substance, nature of substance and the amount of temperature change. Increasing anyone of these will increase the energy demand.Heat Capacity(C): Amount of heat required to raise the temperature of a substance by 1K or 1oC. Its SI unit is JK-1. For a temperature change T, Kelvin and Celsius have the same meanings. But for T the values are not same.Specific Heat Capacity(c): Amount of heat (in J) required raising the temperature of unit mass (Kg or g) of a substance through 1K or 1oC. Its unit is JKg-1K-1 or Jg-1K-1How much energy is transferred?H = mcTm is the mass of liquid in grams whose temperature is changing not of the masses of the solid reactants because their specific heat capacities are very less. Include the appropriate sign with the answer. Normally in dilute aqueous solutions it is equal to the volumes of solutions because their densities are 1.0gcm-3. If solids are heated it is difficult to measure T. OR reaction cannot be controlled. Like decomposition reactions or reactions in which evaporation of water of crystallization is involved.Measuring the energy transferred: T should be measured first of all.Using polystyrene calorimeter: This method is used when two solutions are to be mixed for reaction or for dissolving solids into liquids.Procedure:

1. A simplest calorimeter is an expanded H. M. Sulman Munir (salus82@hotmail.com)Bright Future Pakistani Intr. SchoolO / A Level Chemistry TeacherDoha, Qatar(two cups one inside the other)Notes Chapter 1.2 (Energetics and enthalpy changes) AS Level1

polystyrene coffee-cup calorimeter having an insulated lid on it.Its advantages are: itself it does not absorb much heat (has very low heat capacity) and does not allow heat to go (an excellent insulator) in the surroundings. So in it temperature changes are measured almost accurately. 2.Measure volume of water or solution with the help of measuring cylinder or pipette accordingly.3. Stir the reactants thoroughly after mixing to spread heat uniformly.4. Initial and final temperature is measured to calculate T (either positive or negative). Initial temperature is measured over a period of time to get the steady temperature. The final temperature is measured for a long time even after reaching the maximum change, it is done to compensate the heat losses by graphical method.Experimental results show variation from the standard data: Reasons:(1) Apparatus is not properly insulated (either lid is not put or walls conduct heat).(2) Heat capacity of the apparatus is not exactly zero.(3) Incomplete or slow reaction.(4) Laboratory conditions are not standard.(5) Density of all solutions is not exactly 1.0gcm-3.(6) If mixing is not thorough.(7) Solutions should be of moderate concentrations.

(8) Solids involved should be in their finely divided forms.Q-4 A student placed 25.0cm3 of 1.0M HCl solution in a calorimeter and measured the temperature which was 22.5oC. Then he added 25.0cm3 of 1.0M NaOH solution with temperature 22.5oC. The maximum temperature reached was 29.2oC. Densities of both the solutions were 1.00gcm-3. If the specific heat capacity of the solutions is 4.2Jg-1K-1, calculate the enthalpy change for this reaction per mole of HCl. (Assume no heat lost or gained)Calculating percentage error from maximum error in the apparatus.

Difference of the measured value and standard value (data book) is used to measure the percentage error..

Homework:Q-5 The standard enthalpy change, H for the decomposition of potassium hydrogencarbonate, KHCO3, is impossible to determine directly.2KHCO3(s) K2CO3(s) + CO2(g) + H2O(l)

The value of Hcan be calculated from the standard enthalpy changes which accompany the reactions below:

KHCO3(s) + HCl(aq) KCl(aq) + CO2(g) + H2O(l) H

K2CO3(s) + 2HCl(aq) 2KCl(aq) + CO2(g) + H2O(l) HProcedure: The solids were added to separate 30 cm3 portions of dilute hydrochloric acid. The acid was in excess for both solids. The maximum temperature change for each experiment was noted.Results: The following results were obtained with KHCO3(s).Mass of KHCO3 used = 2.00 gTemperature change = 4.9 C

The experiment with K2CO3(s) gave a H value of 34 kJ mol1.Assumption: The dilute hydrochloric acid solution has a density of 1 g cm3.(a) (i) Calculate the heat energy absorbed, in joules, by the reaction of the KHCO3(s) with the solution of dilute hydrochloric acid. Use the expressionenergy absorbed (J) = mass of solution 4.18 temperature change (1)(ii) Calculate the number of moles of KHCO3(s) used. Assume that the molar mass of KHCO3(s) is 100 g mol1. (1)

(iii) Use your answers to (a)(i) and (ii) to calculate, in kJ mol1, the enthalpy change when one mole of KHCO3(s) reacts completely with the acid (i.e. H ). Include a sign in your answer. (2)(c) The maximum errors for the apparatus used in the experiment with the KHCO3(s) were as follows:Balance + 0.01gMeasuring cylinder + 0.5 cm3(i) Calculate the maximum percentage error in using each of the following piecesof apparatus in the KHCO3(s) experiment:(2)

BalanceMeasuring cylinder(ii) Suggest a piece of apparatus that couldhave been used to measure the volume ofdilute hydrochloric acidmore accurately in this experiment.(1)Measuring the energy changeswhen fuels burn: H = mcT1. We need mass of the fuel burned.1. Mass of water being heated.1. Temperature change in water.

fig. 1.2.9 Another direct method of calorimetry which is useful for measuring the enthalpy changes when fuel burn.Precautions(i) Fuel should be burned completely.(ii) Calorimeter (beaker) should be of negligible Specific heat capacity(iii) Fuel should not be lost in evaporation.(iv) Heat losses should be minimized (insulation)A shield around the apparatus must be used.(v) The distance between burner and beaker should be minimum.(vi) evaporation of volatile fuel may change the results, so it should be minimized.(vii) evaporation of hot water takes some energy in the surroundings so it should be covered.If above four conditions are not met, your results are not accurate. In the exam questions you are asked to give the reason for the variation in calculated value from standard, all above are the reasons. While suggesting read out the text very carefully.Homework Q-6 (c) The enthalpy change of combustion of hexane, C6H14, was measured using a spirit burner to heat a known mass of water in a calorimeter. The temperature rise of the water was measured. The results of the experiment are shown below.Mass of hexane burned = 0.32gmass of water in the calorimeter = 50gInitial temperature of water = 22 oCFinal temperature of water = 68 oCThe specific heat capacity of water is 4.18 J g1C1.(i) Calculate the energy in joules produced by burning the hexane. Use the expressionenergy transferred = mass specific heat capacity temperature change. (1)(ii) Calculate the enthalpy change of combustion of hexane which is the enthalpy change for combustion of one mole of hexane (molar mass of hexane is 86gmol-1)Give your answer to TWO significant figures. Include a sign and units in your answer.(3)(iii) The value for the enthalpy change of combustion in this experiment is different from the value given in data books. Suggest TWO reasons for this difference.(2)(iv) A student suggested that the results would be more accurate if a thermometerwhich read to 0.1C was used. Explain why this would not improve the accuracy of the result. A calculation is not required. (1)Bomb CalorimeterThis is more accurate than the previous two methods. It is used to measure enthalpy change of combustion of fuels and calorific values (energy content) of food.

How can we improve accuracy in enthalpy change measurements? Good insulation minimizes heat losses Calculate the heat losses and make corrections by compensations Best way is to measure the temperature change time to time (after regular intervals) due to heat given by food or fuel is graphed. The maximum temperature change is measured by extrapolating the graph. Then an electrical heater is used to raise the same temperature in the calorimeter. The electrical energy can be measured easily which is equal to the heat energy from the food or fuel. So no need of insulation this way.Q-7

Homework: Write answers in the notebooks of questions on page 39 of the book.Representing Standard Enthalpy Changes of Reactions

The symbol for representing the enthalpy change of a reaction is HH = enthalpy changer = the type of reaction occurring

= standard conditions 1atm pressure, 298K (25oC), if concentration of a solution is to be considered, it is 1.0 moldm-3. We measure these enthalpy changes under standard conditions because the enthalpy change depends upon temperature, pressure, concentration, amount, state and form of a substance. In a reaction a substance should be written in its normal physical state and its most stable (like the most stable form of carbon is graphite not diamond) form until you are asked to write in some other state or form.Enthalpy change of a reaction depends upon the physical state of substance. It is clear from the following enthalpy diagram.

Water produced from H2 and O2 depends upon the physical state of water.It is an exothermic reaction. Water vapors have more enthalpy than the liquid water.

Q-8 Ethanol, C2H5OH, is used as a fuel which on combustion gives CO2(g) and H2O(l). Show on an enthalpy diagram the enthalpy changes for the combustion of liquid ethanol and gaseous ethanol.

Standard Enthalpy Change of Combustion HThe enthalpy change of a substance (elements and compounds both) when one mole of it is burnt completely in oxygen under standard conditions. Incomplete combustion is not considered in this process like formation of CO and SO2 from carbon and sulfur respectively but formation of CO2 and SO3. It is always exothermic.

C(s) + O2(g) CO2H[Cgraphite(s)] = -393kJmol-1

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) H[C2H5OH(l)] Energy released in a reaction depends upon the number of bonds formed and broken. If more bonds are formed, more energy is released, the reaction is exothermic and vice versa. The energy released by the combustion of methanol, CH3OH, is less as compared to methane, CH4. If in the reactants the bonds are stronger like O-H, less energy is released.Q-9 Graphite and diamond are the two allotropic forms of carbon. Graphite is more stable. On combustion both give CO2(g). Show the enthalpy changes of combustion of these two forms of carbon on an enthalpy level diagram.

Standard Enthalpy Change of formation H The enthalpy change when one mole of a compound is formed from its elements under standard conditions. It can be either exo or endo. The product always should be one mole.

Cgraphite(s) + 2H2(g) CH4(g)H[CH4(g)] = -74.8kJmol-1

2Cgraphite(s) + 3H2(g) + O2 C2H5OH(l)H[C2H5OH(l)]For elements under standard conditions this value is zero.

Standard Enthalpy Change of Atomization H The enthalpy change when one mole of atoms in the gaseous state are formed from the element under standard conditions. It is always positive and for noble gases it is zero.

Na(s) Na(g)H [Na(s)]

Br2(l) Br(g)H [Br2(l)]

Standard Enthalpy Change of Neutralization H Enthalpy change when one mole H+(aq) ions from an acid reacts with one mole of OH-(aq) ions to form one mole of liquid water under standard conditions.Its values for strong acids and strong alkalis is about -58kJmol-1 because reaction is same in all these cases.

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)The ionic equation is

H+(aq) + OH-(aq) H2O(l) H = -57.9kJmol-1If one of the acid or base is weak, the value will be bit less like -54 kJmol-1. If both are weak this value is near -50 kJmol-1.Q-10 For the following reaction we can use both the terms standard enthalpy change of combustion and standard enthalpy change of formation.

Srhombic(s) + O2(g) SO3(g)Explain why? Both terms need separate explanations. Mention the substances for which these terms will be used.Q-11

Q-12

Homework: Write answers in the notebooks of questions on page 42 of the book.

Q-13 Water can be formed according to the following equation

2H2(g) + O2(g) 2H2O(l)H = -572kJ(a) Suggest the enthalpy change for the reverse reaction. 2H2O(l) 2H2(g) + O2(g).Explain the reason for a value you have suggested.(2)(b) Calculate the enthalpy change of formation of water.(2)Q-14 We prepared CO2 by two different routes. One of them is direct (single step) reaction while other is indirect (two-step) reaction.

(a) Calculate H1 + H2. Draw a conclusion about the enthalpy change of a reaction when the different routes are adopted.(b) Measuring H and H2 is practically possible. So calculate the H1, enthalpy of formation of carbon monoxide.(H)route 1 = (H1 + H2)route 2(c) If we go round loop in the above cycle, the reactions involved will be

Add enthalpy change of these three reactions.Why do we need Hesss Law? Some of the reactions written cannot be carried out in actual like CO cannot be prepared from C and O2. So its enthalpy of formation cannot be measured but calculated. Some reactions do not undergo to completion like a hard coating of Al2O3 on the surface of Al stops the reaction. So we cannot measure the enthalpy of formation Al2O3. Most of the reactions do not take place under standard conditions but we need their enthalpy changes under standard conditions like fuels burn at higher than temperatures than the room temperature.Definition of Hesss Law: The total enthalpy change for a reaction is independent of the route taken.

Trick: In such questions (from Q-15-Q-24, except Q-18) try to draw a triangle of reactions from the given data and equation asked.Enthalpy changes of formation from enthalpy changes of combustions:In such exam questions we are given the standard enthalpy change of combustions of the reactants and products are given, we are to calculate the enthalpy change formation of a compound normally a fuel.Q-15 (a) Write equation for the standard enthalpy change of formation of sucrose, C12H22O11(s). Include the state symbols in your equation.(b) On combustion, the reactants and products of above equation give the same products i.e CO2(g) and H2O(l). Draw a triangle of reactions from the given data. Mark the two routes.The total enthalpy change in both the routes is same. Using this law (Hesss Law), calculate

H[C12H22O11(s)]. H[Cgraphite(s)] = -393.5kJmol-1 H[H2(g)] = -285.8kJmol-1

H[ C12H22O11(s)] = -5639.7kJmol-1Q-16 (c) Standard enthalpy changes of combustion can be used to calculate the standard enthalpy change of formation of a compound.(i) Define the term standard enthalpy change of formation, making clear the meaning of standard in this context.(3)(ii) Use the standard enthalpy changes of combustion, Hc, given in the table below to find the standard enthalpy change of formation for ethanoic acid, CH3COOH, in kJ mol1.

2Cgraphite(s) + 2H2(g) + O2(g) CH3COOH(l)(3)Q-17 The standard enthalpy changes of combustion of carbon, hydrogen and methane are shown in the table below.

Which one of the following expressions gives the correct value for the standard enthalpy change of formation of methane in kJ mol1?C(s) + 2H2(g) CH4(g)A 394 + (2 286) 891B 394 (2 286) + 891C 394 + 286 891D 394 286 + 891Q-18 The standard enthalpy changes of formation of some sulfur species are:

The enthalpy of atomization of sulfur is (in kJ mol1)A 103 8B 279 8C 279D (103 8) + 279Q-19 Enthalpy changes of combustion can be used to determine enthalpy changes of formation. The following equation represents the enthalpy change of formation of butane.4C(s) + 5H2(g) C4H10(g)By using the following standard enthalpy of combustion data, what is the value of the standard enthalpy change of formation for this reaction?

A 5883 kJmol1B 129 kJmol1C +129 kJmol1D +2197 kJmol1Q-20Sulphur is S(s)(3)

Trick: For above type of reactions the formula is

H[Product] = H[Reactants] - H[Products]Enthalpy changes of reactions from enthalpy changes of formations:In such exam questions an equation is given, enthalpy changes of formation of reactants and products are given, we are to calculate the enthalpy change of the reaction taking place in the given equation.Q-21 (a) Write the general form (mathematical) of Hesss Law for the following set of reactions.

(b) Calculate Hfrom the following data.

H[H2O2(l)] = -187.8kJmol-1H[H2O(l)] = -285.8kJmol-1

Trick: In such type of reactions H = H[Products] - H[Reactants]Q-22

Q-23 Calculate standard enthalpy change of the following reaction from the data given.

Q-24

Homework: Write answers In the notebooks of questions on page 45 of the book.Note: For Questions regarding Hesss Law the data may be given in the form of chemical reactions of combustion and formation.Q#25

Bond Enthalpy or bond energy: The enthalpy change when one mole of a particular bond in gaseous state is formed or broken under standard conditions.Note: In the equations showing the bond enthalpies, left side should have one mole of gaseous bonds an d right side should contain one mole of gaseous atoms of each type.This value bears negative sign for bond making and positive for bond breaking.The strength of the same bond slightly varies in different compounds. For example the bond enthalpy for C-C bond in ethane, C2H6, is slightly lower than C-C bond in ethanol, CH3CH2OH.Mean bond enthalpy: So due to above reason, for a particular bond, we take mean of the bond enthalpies of a bond in different compounds.

CH4(g) C(g) + 4H(g)

Enthalpy change in the above reaction is 4xE(C-H)orH[CH4(g)]The experimentally measured values may be different from the calculated values due to two reasons(1) The standard data is for the substances in gaseous state while in experiment a substance may be in some other state.(2) The standard data is the mean from different substances but in an experiment we have one substance.Calculating standard enthalpy of a reaction using standard bond enthalpies:

Applying Hesss LawHr = H1 + (-H2)As the bond making is an exothermic process, so H2 is given a negative sign. So simply we can writeHr = E(bond breaking) - E(bond making)In this formula just put the values of bond enthalpies given.Q-26 Calculate the standard enthalpy change of the above reaction from the data given below.E(C-H) = +412kJmol-1E(C=O) = +805kJmol-1E(O-H) = +463kJmol-1E(O=O) = +496kJmol-1Q-27 Which equation represents the reaction for which the enthalpy change, H, is the meanbond enthalpy of the CH bond?A CH4(g) C(g) + H(g)B CH4(g) C(s) + 2H2(g)C CH4(g) C(g) + 4H(g)D CH4(g) C(g) + 2H2(g)Q-28 The enthalpy change for the reactionCH4(g) C(g) + 4H(g)is +1648 kJ mol1. Hence the mean bond enthalpy for the CH bond isA +329.6 kJ mol1B +412.0 kJ mol1C +1648 kJ mol1D +6592 kJ mol1Q-29

E(C-H)=+412kJmol-1 E(CC)=840kJmol-1 E(C=O) =+805kJmol-1 E(O-H)=+463kJmol-1 E(O=O) = +496kJmol-1Q-30 The following diagram is for the formation of CO2. C labeled in the diagram is the standard bond enthalpy of the oxygen.

(a) Name or write the symbols for the enthalpy changes labeled as A and B.(b) Write the formulae of the species present at levels x and y.Homework: Write answers in the notebooks of questions on page 47 of the book.Q-31 A fuel the highest energy per gram is called the most efficient fuel. Some fuels are given with their enthalpy changes of combustions. Calculate the enthalpy change per gram for each fuel and the most efficient and the least efficient fuels among these.FuelFormulaEnthalpy change of combustion kJ mol-1Molar mass (g mol-1)

CarbonC-393.512

MethaneCH4-890.316

MethanolCH3OH-726.032

PropaneC3H8-2219.244

ButaneC4H10-2876.558

Energetic (thermal) stability and kinetic stability:(1) It is thought that the compounds having positive values of standard enthalpies of formation are energetically unstable. But in many cases they can be stored without decomposing.(2) The reactants of the exothermic reactions should be very energetically unstable and the reactions should be spontaneous.The decomposition of the substances of type 1 above and reactions of type 2 require very high temperature in many cases. They are energetically unstable but kinetically stable. Such substances once started by giving energies move forward rapidly and spontaneously. For example, combustion of fuels, fuels are energetically unstable but kinetically stable.Homework: Write answers in the notebooks of questions on page 51 of the book. In bond enthalpy equations reactants should be gaseous and products only atoms in gas states.

Happy End of the Chapter