Thermochemistry: Energy Flow and Chemical Reactions

Thermochemistry: Energy Flow and Chemical Reactions

Outline thermodynamics internal energy definition, first law enthalpy definition, energy diagrams, calorimetry, theoretical calculation (heats of formation and bond energies), stoichiometry hess s law energy from foods

Thermodynamics Thermodynamics is the study of heat and its transformations. Thermochemistry is a branch of thermodynamics that deals withthe heat involved with chemical and physical changes.

Internal Energy What is the difference of the two pictures above? Energy state If we go from cold water to boiling water as our reaction: H2O (solid) H2O (gas) energy state 1 energy state 2

Internal Energy we define E = a variable to denote the energy of our reacting substances = specifically, the internal energy of the substance undergoing reaction E = change in energy = E (boiling water) E (cold water) since Energy (gas) > energy (solid) E = positive, + For the reaction to proceed, energy has to be supplied. But where will it come from?

Internal Energy A chemical system and its surroundings. the surroundings the system System = the reaction itself, the object of our study Surroundings = the rest of the universe

Internal Energy

Internal Energy Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. DE = E final - E initial = E products - E reactants Energy changes occurs because of the exchange of energy between the system and the surroundings!

Internal Energy Notice that all the equations/energies refer to the system! But we can also define it from the point of view of the surroundings. Case 1: surroundings supplied energy to the system E = E2, surr - E1, surr = negative We have to specify which we are referring to: system or surroundings E 1, sys State 1: State 2: E 2, sys E 1, surr E 2, surr E 2, sys - E1, sys = E system = positive, + E 2, surr - E1, surr = E surroundings = negative, -

Internal Energy So we have a discrepancy of notations and definitions. As per convention, whenever we have E1, E2, etc., we are always referring to the system. Henceforth, we are always referring to the system in all our studies of energies. Notice, however that E system + E surr = 0 since: system + surroundings = universe E universe = 0 = E2, universe - E1, universe = 0 E2, universe = E1, universe The energy of the universe is constant!

Internal Energy First Law of Thermodynamics: The energy of the universe is constant. Thus, energy can neither be created nor destroyed. It can be only be converted from one form to another.

Internal Energy Let s define formally E = internal energy = total energy of the system = summation of all the kinetic and potential energies of the system Experimentally: E = q + w q = heat = energy transferred from a hotter object to a colder one w = work = energy used to cause an object to move against a force That means, we can change the energy of the system by applying heat and doing work. Or, When energy is transferred from one object to another, it appears as work and/or as heat.

Internal Energy A system transferring energy as heat only.

Internal Energy A system losing energy as work only. Zn(s) + 2H + (aq) + 2Cl - (aq) Energy, E DE<0 work done on surroundings H 2 (g) + Zn 2+ (aq) + 2Cl - (aq)

Internal Energy The Sign Conventions* for q, w and DE q + w = DE + + - - + - + - + depends on sizes of q and w depends on sizes of q and w - * For q: + means system gains heat; - means system loses heat. * For w: + means word done on system; - means work done by system.

Internal Energy DE = (sign), (number), (unit) Sign = gives direction as to the transfer of energy + = E final > E initial = system gained energy from the surroundings - = E final < E initial = system lost energy to the surroundings

Internal Energy Units of Energy Joule (J) 1 J = 1 kg*m 2 /s 2 calorie (cal) 1 cal = 4.184 J British Thermal Unit 1 Btu = 1055 J Calorie (Cal) 1 Calorie = 1 000 calorie

Sample Problem Hydrogen and oxygen reacts in a cylinder. As the reaction occurs, the system loses 1150 J of heat to the surroundings and the expanding gas product does 480 J of work on the surroundings as it pushes against the atmosphere. What is the change in the internal energy of the system? q = heat = - 1150 J w = work = - 480 J E = q + w = (-1150 J) + (- 480 J) = - 1630 J

Sample Problem A system receives 425 J of heat and delivers 400 J of work to its surroundings. What is the change in internal energy of the system (in joules)? Answer: E = q + w = (+ 425) + (- 400 J) = + 25 J

Worksheet # 9-1 1. Complete combustion of 1.0 metric ton of coal (assuming pure carbon) to gaseous carbon dioxide releases 3.3 x 10 10 J of heat. Convert this energy to (a) kilojoules; (b) kilocalories and (c) british thermal unit. 2. A system conducts 225 cal of heat to the surroundings while delivering 428 cal of work. What is the change in internal energy of the system (in cal and J)? 3. What is the change in internal energy (in J) of a system that releases 675 J of thermal energy to its surroundings and has 525 calories of work done on it?

Worksheet # 9-1: Answers 1. Complete combustion of 1.0 metric ton of coal (assuming pure carbon) to gaseous carbon dioxide releases 3.3 x 10 10 J of heat. Convert this energy to (a) kilojoules; (b) kilocalories and (c) british thermal unit. Answers: a) 3.3 x 10 7 kj b) 7.9 x 10 6 kcal c) 3.1 x 10 7 Btu 2. A system conducts 225 cal of heat to the surroundings while delivering 428 cal of work. What is the change in internal energy of the system (in cal and J)? Answer: - 653 cal = - 2732 J 3. What is the change in internal energy (in J) of a system that releases 675 J of thermal energy to its surroundings and has 525 calories of work done on it? Answer: 1522 J

Some interesting quantities of energy.

Enthalpy Two different paths for the energy change of a system.

Enthalpy Pressure-volume work.

Enthalpy There is another energy variable we call enthalpy: H = E + PV PV = work H = E + P V Formal definition: Enthalpy is the thermodynamic quantity that is the sum of internal energy and the product of the pressure-volume work. The change in Enthalpy equals the heat gained or lost at constant pressure. Most chemical reactions occur at constant pressure, so H is more relevant than E.

Enthalpy w = - PDV H = E + PV where H is enthalpy DH = DE + PDV q p = DE + PDV = DH DH DE in 1. Reactions that do not involve gases. 2. Reactions in which the number of moles of gas does not change. 3. Reactions in which the number of moles of gas does change but q is >>> PDV.

Enthalpy Let s examine enthalpy and internal energy more closely: We defined E = E2 E1 We can calculate E from the first law of thermodynamics. What about E2, E1 or the exact energy value of a system. E, internal energy = summation of all the energies of the system, including KE and PE For a number of particles (1, 2, 3 particles) energy can be calculated, but if we are dealing with macroscopic amounts (ex: a mole of water, 6.02 x 10 23 particles), the exact value is incalculable and impossible! But is it important?

Enthalpy We do not need to calculate the energy values exactly. But we assume that the energy of a system at a defined condition is exact, fixed and unchanging. Example: 1.0 mol of 2.5 L water at 90 C, 1.5 atm pressure has an internal energy E and an enthaly H whenever this condition is specified Mathematically, we call E and H = state functions state function = property of a system that is determined by specifying its conditions or its state (ex. T, P, etc.) = the value of a state function does not depend on the particular history of the sample, only on its present condition

Enthalpy State function analogy: scaling a mountain The height of the mountain is a state function. However, the way you can climb the mountain is not. H, E = state functions q, w = not state functions, path dependent (value depends on how the process took place)

Enthalpy Exothermic Reactions: H = H final H initial = negative, - energy is released by the system Endothermic reactions: H = H final H initial = positive, + energy is absorbed by the system The sign of enthalpy (like internal energy) indicates the direction of heat transfer during a process that occurs at constant pressure.

Enthalpy, H Enthalpy Enthalpy, H Enthalpy diagrams for exothermic and endothermic processes. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H 2 O(l) H 2 O(g) CH 4 + 2O 2 H initial H 2 O(g) H final DH < 0 heat out DH > 0 heat in CO 2 + 2H 2 O H 2 O(l) H final H initial A Exothermic process B Endothermic process

Enthalpy Video: endothermic reaction (endo final.wmv)

Enthalpy It turns out experimentally that: E = energy transferred at constant volume conditions H = energy transferred at constant pressure conditions In most chemical reactions, we are mostly concerned with enthalpy. As most chemical reactions occur at constant pressure. Thermochemical equations 2H 2 (gas)+ O 2 (gas) 2H 2 O (gas) H = - 483.6 kj The enthalpy value is given at the end of the chemical equation.

Enthalpy Some Important Types of Enthalpy Change heat of combustion (DH comb ) C 4 H 10 (l) + 13/2O 2 (g) 4CO 2 (g) + 5H 2 O(g) heat of formation (DH f ) K(s) + 1/2Br 2 (l) KBr(s) heat of fusion (DH fus ) NaCl(s) NaCl(l) heat of vaporization (DH vap ) C 6 H 6 (l) C 6 H 6 (g)

Enthalpy, H Sample Problem Enthalpy, H Write a balanced equation and draw an approximate enthalpy diagram for each of the following: a) Combustion of one mol of methane in oxygen b) The evaporation of liquid water Answer: a) CH 4 + 2O 2 CO 2 + 2H 2 O, exothermic b) H 2 O (l) H 2 O (g), endothermic CH 4 + 2O 2 H 2 O (g) heat out heat in CO 2 + 2H 2 O H 2 O (l)

Worksheet # 9-2 In each of the following cases, determine the sign of enthalpy, state whether the reaction is exothermic or endothermic and draw an enthalpy diagram. 1. H 2 (g) + ½ O 2 (g) H 2 O (l) + 285.8 kj 2. CO 2 (s) + 260 kj CO 2 (g) 3. The heat released for the combustion of liquid ethanol 4. The vaporization of liquid ammonia

Enthalpy, H Enthalpy, H Worksheet # 9-2: Answers H 2 (g) + ½ O 2 (g) C 2 H 5 OH + 3O 2 heat out - 285.8 kj H 2 O (l) 2CO 2 + 3H 2 O exothermic CO 2 (g) NH 3 (g) heat in + 260 kj endothermic CO 2 (s) NH 3 (l)

Enthalpy Properties of enthalpy: 1. Enthalpy is an extensive property. Thus its magnitude is directly proportional to the amount of substance reacted or produced. Implication: in a thermochemical equation, if the equation is multiplied by 2, the H is also multiplied by 2. If the equation is divided by 2, the H is also divided by two. Example: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) 2CH 4 (g) + 4O 2 (g) 2CO 2 (g) + 4H 2 O(l) H = - 890 kj H = 2(- 890) kj

Enthalpy 2. H reaction = - H reverse reaction CO 2 (g) + 2H 2 O (l) CH 2 (g) + 2O 2 (g) H = 890 kj 3. H is depends on the state of the chemical species used in the equation. The state of the substance has to be shown. Example: CO 2 (g) + 2H 2 O (l) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) CH 4 (g) + 2O 2 (g) H = 890 kj H = 978 kj

Enthalpy Enthalpy and chemical stoichometry -The calculation of the enthalpy for a particular reaction and its relation to the chemical equation is governed property 1 of enthalpy it is an extensive property. Example: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) H = - 890 kj relationships: 1 mol CH 4 is to H = - 890 kj 2 mol O 2 (g) is to H = - 890 kj 1 mol CO 2 is to H = - 890 kj 2 mol H 2 O is to H = - 890 k

Sample Problem Consider the equation: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) H = - 890 kj 1. How many moles of methane, CH4 (g) has to be burned to generate 1500 kj of energy? 2. How many grams of carbon dioxide is produced if the enthalpy generated in the combustion is 1830 kj?

Sample Problem Consider the following balanced thermochemical equation for a reaction sometimes used for H 2 S production: 1/8 S 8 + H 2 H 2 S ΔH = - 20.2 kj a) Is this an exothermic or endothermic reaction? b) What is the ΔH rxn for the reverse reaction? c) What is ΔH when 3.2 mol of S 8 reacts? d) What is ΔH when 20.0 g of S 8 reacts? Answers: a) Exothermic b) 20.2 kj c) -517 kj d) -12.6 kj

Enthalpy AMOUNT (mol) of compound A Summary of the relationship between amount (mol) of substance and the heat (kj) transferred during a reaction. molar ratio from balanced equation AMOUNT (mol) of compound B HEAT (kj) DH rxn (kj/mol) gained or lost

Worksheet # 9-3 1. Consider the following balanced thermochemical equation for the decomposition of the mineral magnesite: MgCO 3 MgO + CO 2 H = 117.3 kj (a) is heat absorbed or released in the reaction? (b) what is the H for the reverse reaction? (c) what is the H when 5.35 mol of CO2 reacts with excess MgO? (d) What is the H when 2.36 mol of magnesite reacts? (e) what is the H when 0.15 mol of MgO is produced from the reaction? 2. When 1.0 mol of NO forms from its elements, 90.29 kj of heat is absorbed. (a) write a balanced thermochemical equation for this reaction. (b) how much heat is involved when 1.50 mol of NO decomposes to its elements? (c) how much heat is involved when 1.75 grams of NO decomposes to its elements? (d) how much heat is involved when 3.50 grams of N2 is produced?

Worksheet # 9 3: Answers 1. Consider the following balanced thermochemical equation for the decomposition of the mineral magnesite: MgCO 3 MgO + CO 2 H = 117.3 kj (a) is heat absorbed or released in the reaction? Absorbed (b) what is the H for the reverse reaction? -117.3 kj (c) what is the H when 5.35 mol of CO2 reacts with excess MgO?-627.6 kj (d) What is the H when 2.36 mol of magnesite reacts? + 276.8 kj (e) what is the H when 0.15 mol of MgO is produced from the reaction? + 17.6 kj 2. When 1.0 mol of NO forms from its elements, 90.29 kj of heat is absorbed? (a) write a balanced thermochemical equation for this reaction? ½ N2 + ½ O2 NO H = 90.29 kj (b) how much heat is involved when 1.50 mol of NO decomposes to its elements? (c) how much heat is involved when 1.75 grams of NO decomposes to its elements? (d) how much heat is involved when 3.50 grams of N2 is produced? b) -135.4 kj c) -5.3 kj d) -22.6 kj

Enthalpy Calorimetry = measurement of heat flow Calorimeter = apparatus that measures heat flow H reaction can be determined experimentally using a calorimeter.

Coffee-cup calorimeter. Enthalpy

Enthalpy What did the thermometer say to the graduated cylinder? "You may have graduated but I've got many degrees."

Enthalpy In chemical equations, energy changes can be calculated by monitoring the temperature changes in the reaction. q is proportional to T In equation form: q α T q = C T where C = constant of proportionality = heat capacity = amount of heat required to raise the temperature of a substance by 1.0 Kelvin The larger the heat capacity the greater the heat required to produce a given rise in temperature. For 1.0 mol substance, C = C (mol)

Enthalpy Lets look at another experiment: 1.0 L water and 10.0 ml water are heated separately such that each would have its temperature increase by 10.0 C. What would be the result. The 10.0 ml water sample will have a greater increase in energy as some of its molecules can turn into gas. Implication: mass is also a factor in measuring enthalpies. q = C T As there is no mass in the equation, it can be in the heat capacity factor. C = m c where m = mass of the substance c = specific heat of the substance Thus, q = m c T

Enthalpy Heat Capacity unit: J/K C = q / ΔT Specific heat capacity (c) = q / (mass x ΔT) unit: J/gK Molar heat capacity (C mol) = q / (moles x ΔT) unit: J/mol K

Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials Substance Specific Heat Capacity (J/g*K) Substance Specific Heat Capacity (J/g*K) Elements Materials aluminum, Al 0.900 wood 1.76 graphite,c 0.711 cement 0.88 iron, Fe 0.450 glass 0.84 copper, Cu 0.387 granite 0.79 gold, Au 0.129 steel 0.45 Compounds water, H 2 O(l) ethyl alcohol, C 2 H 5 OH(l) ethylene glycol, (CH 2 OH) 2 (l) carbon tetrachloride, CCl 4 (l) 4.184 2.46 2.42 0.864

Enthalpy Relation of enthalpy and q (from calorimetry experiments) q = m c T = q solution q solution = - q reaction = H reaction H reaction = - m c T

Sample Problem A 295 g aluminum engine part at an initial temperature of 3.0 C absorbs 85.0 kj of heat. What is the final temperature of the part (specific heat of aluminum = 0.900 J/g K)? q = m c T 85 000 J = 295 g Al (0.9 J/gK) (Tf 3.0 C) Answer: Tf = 323 C

Sample Problem When 165 ml of water at 22 C is mixed with 85 ml of water at 82 C, what is the final temperature? Assume that no heat is lost to the surroundings (density of water is 1.0 g/ml). (m c T) water 1 = (m c T) water 2 165 ml ( Tf 22 C ) = 85 ml (Tf 82 C) Answer: Tf = 42 C

Worksheet # 9-4 1. Calculate q when 12.0 grams of water is heated from 20 degrees to 100 degrees? ( c water = 4.184 J/g C) 2. The specific heat capacity of silver is 0.24 J/g C. a) Calculate the energy required to raise the temperature of 150.0 grams Ag from 273 K to 298 K. b) Calculate the energy required to raise the temperature of 1.0 mol Ag by 1.0 C c) It takes 1.25 kj of energy to heat a sample of pure silver from 12.0 C to 15.2 C. Calculate the mass of the sample of silver. 3. A 30.0 g of water at 280 K is mixed with a 50.0 g water at 330 K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

Worksheet # 9 4: Answers 1. Calculate q when 12.0 grams of water is heated from 20 degrees to 100 degrees? Ans = 4016.6 J 2. The specific heat capacity of silver is 0.24 J/g C. a) Calculate the energy required to raise the temperature of 150.0 grams Ag from 273 K to 298 K. Ans = 900 J b) Calculate the energy required to raise the temperature of 1.0 mol Ag by 1.0 C. Ans = 25.9 J c) It takes 1.25 kj of energy to heat a sample of pure silver from 12.0 C to 15.2 C. Calculate the mass of the sample of silver. Ans = 1627.6 g 3. A 30.0 g of water at 280 K is mixed with a 50.0 g water at 330 K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. Ans = 311 K