Thermochemistry Chapter 5

5.1 Energy, Heat & Work Objectives Objectives Distinguish between system, surroundings, kinetic energy, potential energy, heat, work, and chemical energy Identify processes as exothermic or endothermic based on the heat of that process

5.1 Energy, Heat & Work Energy Nearly all chemical reactions involve the absorption or release of energy A complete chemical equation includes a quantitative measure of the energy produced or consumed Thermochemistry –the study of the relationship between heat and chemical reactions

5.1 Energy, Heat & Work cont’d Energy – –the ability to do work – –can take many forms including mechanical, electrical, chemical, heat, light, etc. – –can be classified as kinetic (KE) or potential (PE) Total energy = KE + PE Note the SI unit for energy is the joule (J)

5.1 Energy, Heat & Work cont’d Kinetic energy – –energy possessed by matter because of its motion – –KE =  mv 2 where m = mass and v = velocity

5.1 Energy, Heat & Work cont’d Kinetic energy – –Thermal energy is a type of kinetic energy and is due to random motion of particles in a sample of matter directly proportional to the Kelvin temperature of a sample Heat is the flow of energy from one object to another that causes a change in the temperature of the object

5.1 Energy, Heat & Work cont’d Work –the application of a force across some distance –Can take many forms Mechanical, gravitational, pressure-volume, electrical –Remember energy is required to perform work and the units of work are also expressed in joules

Converting chemical energy stored in gasoline, a battery, or food to kinetic energy

5.1 Energy, Heat & Work cont’d Key definitions Potential energy – –Energy possessed by matter because of its position or condition – –Chemical energy is a form of potential energy due to the forces that hold atoms together

5.1 Energy, Heat & Work cont’d Basic definitions Basic definitions System – –the sample of matter Surroundings – –all other matter besides the sample

5.1 Energy, Heat & Work cont’d Law of Conservation of Energy (aka the first law of thermodynamics…) – –Energy cannot be created or destroyed but can be converted or transferred or the textbook definition: – –The total energy of the universe is constant during a chemical or physical change. Energy can be transferred between the system and the surroundings, but the total energy before and after the change is constant

5.1 Energy, Heat & Work cont’d Key definitions Transfer of energy can take the form of light, sound, or work, but most often energy is transferred as heat during chemical reactions We generally express the change in energy as an equation: Energy change = work + heat

5.1 Energy, Heat & Work cont’d Exothermic reactions – –System releases heat to the surroundings – – CH 4 (g) + 2 O 2 → CO 2 (g) + 2 H 2 O (l) + energy Endothermic reactions – –System absorbs heat from the surroundings – –N 2 (g) + O 2 + energy → 2 N O (g)

5.1 Energy, Heat & Work cont’d Exothermic & Endothermic Reactions

5.1 Enthalpy Changes in Chemical Reactions cont’d Key definitions Joule (J) – –the SI unit for energy 1 J = 1 kgm 2 /s 2 Calorie (cal) – –also used as a measure of heat – –Defined as the amount of heat necessary to raise the temperature of 1 g of water by 1°C – –Conversion factor is 1 cal = J

5.2 Enthalpy and Thermochemical Equations Objectives Objectives Define enthalpy Express energy changes in chemical reactions Calculate enthalpy changes from stoichiometric relationships

5.2 Enthalpy and Thermochemical Equations Enthalpy (H) –A measure of the total energy of a system at a given temperature & pressure –Cannot really be measured but changes in enthalpy can Change in enthalpy (ΔH) –A measure of the quantity of heat absorbed or released by the system at constant temperature and pressure

5.2 Enthalpy and Thermochemical Equations Since ΔH = H final – H initial –A chemical reaction which gives off heat is exothermic and ΔH is negative –A chemical reaction which absorbs heat is endothermic and ΔH is positive –Please note that ΔH is determined experimentally

5.2 Enthalpy and Thermochemical Equations Thermochemical Equation –A chemical equation for which ΔH is given: Ex. of an exothermic reaction: CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2 H 2 O (l) ΔH = -890 kJ Ex. of an endothermic reaction: N 2 (g) + O 2 (g) → 2 NO (g) ΔH = kJ Note: These are all molar relationships // Note that the physical state must be given for every substance in the reaction

5.2 Enthalpy and Thermochemical Equations Why is the physical state so important? 2 H 2 (g) + O 2 (g) → 2 H 2 O (l) ΔH = kJ 2 H 2 (g) + O 2 (g) → 2 H 2 O (g) ΔH = kJ

Consider the thermochemical equation Consider the thermochemical equation The thermochemical equivalencies are –1 mol C 3 H 8 (g) react to give off 2.22×10 3 kJ –5 mol O 2 (g) react to give off 2.22×10 3 kJ –3 mol CO 2 (g) are formed and give off 2.22×10 3 kJ –4 mol H 2 O( ) are formed and give off 2.22×10 3 kJ C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O( )  H = -2.22×10 3 kJ Mass of CO 2 Molar mass of CO 2 Moles of CO 2 Thermochemical equivalent Enthalpy change

5.2 Enthalpy and Thermochemical Equations cont’d Since ΔH is part of a chemical equation, it can be treated as a stoichiometric quantity of the reaction Given the reaction: CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2 H 2 O (l) ΔH = -890 kJ CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2 H 2 O (l) ΔH = -890 kJ 1 mole CH 4 reacts, 890 kJ of energy released 1 mole CH 4 reacts, 890 kJ of energy released 2 mole CH 4 reacts, 1780 kJ of energy released 2 mole CH 4 reacts, 1780 kJ of energy released 8 mol H 2 O (l) produced, 3560 kJ of energy released 8 mol H 2 O (l) produced, 3560 kJ of energy released

5.2 Enthalpy and Thermochemical Equations cont’d Stoichiometry of Enthalpy Change in Chemical Reactions –Calculate the enthalpy change observed in the combustion of 1.00 g of ethane 2 C 2 H 6 (g) + 7O 2 (g) → 4 CO 2 (g) + 6 H 2 O (l) ΔH = kJ ΔH = kJ A: kJ

5.2 Enthalpy and Thermochemical Equations cont’d Stoichiometry of Enthalpy Change in Chemical Reactions –Calculate the enthalpy change observed when 5.00 g of O 2 are consumed by reaction with N 2, forming NO N 2 (g) + O 2 (g) → 2 NO (g) ΔH = kJ A: kJ

5.2 Enthalpy and Thermochemical Equations cont’d Stoichiometry of Enthalpy Change in Chemical Reactions –Calculate the mass of methane that must be burned to produce 4.00 x 10 3 kJ of heat CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O (l) ΔH = -890 kJ A: 72.1 g

5.3 Calorimetry Objectives Define heat capacity and use it to Relate heat flow to temperature change Determine changes in enthalpy from calorimetry experiments

5.3 Calorimetry Knowing the amount of heat absorbed or released in chemical reactions is a critical component to design of manufacturing facilities Calorimetry = the process by which the amount of heat released or absorbed in a chemical reaction is measured Calorimeter = the device used to measure the amount of heat absorbed or released in a chemical reaction –Ex: coffee cup calorimeter to detect changes in energy for reactions carried out in solution

Calorimeter Calorimeter A simple calorimeter can be made from styrofoam coffee cups.

5.3 Calorimetry The energy change of the surroundings is the heat, q surr, that caused a change in the temperature of the contents of the calorimeter According to the law of conservation of energy, the sum of the heat transferred to the surroundings and the ΔH of the system must be zero q surr + ΔH = 0 and ΔH = -q surr q surr + ΔH = 0 and ΔH = -q surr

5.3 Calorimetry How is heat related to the change in temperature? Heat Capacity = the quantity of heat required to raise the temperature of an object 1K or 1°C Specific Heat = the heat needed to raise the temperature of a one gram sample of a material by 1K and is in the units of J/gK or J/g°C The relationship between heat and temperature change is given by the following equation q = mC s ΔT where q = heat in J, m = mass in grams, C s = the specific heat and ΔT = T final -T initial

5.3 Calorimetry Sample Calculations What quantity of heat must be added to a 120 g sample of aluminum to change its temperature from 23.0°C to 34.0°C? What quantity of heat is needed to raise the temperature of 120 g of water by 6°C? When a 60.0 g sample of metal at 100.0°C is added to 45.0 g of water at 22.60°C, the final temperature of both the metal and the water is 32.81°C. The specific heat of water is J/g°C. What is the specific heat of the metal?

5.3 Calorimetry To simplify calorimetry calculations…. –The heat, q surr, is evaluated from the mass, temperature change, and specific heat of the solution. –The heat required to change the temperature of the vessel, stirrer, and thermometer is sufficiently small to be ignored –The specific heat of the solution, as long as it is dilute, is the same as that of water, 4.184J/g°C –Once we have determined q surr, we can set ΔH equal to -q surr

5.4 Hess’s Law Objectives Define a state function Draw and interpret enthalpy diagrams to illustrate energy changes in a reaction Use Hess’s Law to combine thermochemical equations to find an unknown ΔH

5.4 Hess’s Law We can calculate the ΔH of a particular reaction by summing the ΔH’s of several related reactions that yield the final reaction Why? ΔH is a state function rather than a path function Because ΔH is a state function rather than a path function

5.4 Hess’s Law State function = any property of a system that is fixed by the present conditions and is independent of a system’s history Path function = a property that depends on how a particular change is carried out Ex: (1) T (1) The altitude of a mountain is a state function because the value for the altitude is the same whether you walk on the most direct path or take the winding path to the top. The distance traveled to reach the top is a path function (2) The balance of a bank account on a particular day is a state function because it is the same whether you deposited money in one lump sum or deposited small amounts over several days

Fig. 5-5, p. 187

5.4 Hess’s Law Properties of Thermochemical Equations –Hess’s Law: The change in enthalpy for an equation obtained by adding two or more thermochemical equations is the sum of the enthalpy changes of the added equations. –When a thermochemical equation is written in reverse direction the enthalpy change is numerically the same but has the opposite sign –The enthalpy change is an extensive property that depends on the amounts of the substances that react. For example, if coefficients in a thermochemical equation are doubled, the enthalpy change doubles.

An energy-level diagram can represent the enthalpy change for a reaction. Energy-Level Diagrams H 2 (g) + ½O 2 (g) → H 2 O( )  H =  kJ  But …..  H 2 O (l) → H 2 (g) + ½O 2 (g)  ΔH = kJ

5.4 Hess’s Law Q: Find the enthalpy for the reaction C (s) + ½O 2 (g) → CO (g) ΔH = ? C (s) + ½O 2 (g) → CO (g) ΔH = ? Based on the following two reactions: C (s) + O 2 (g) → CO 2 (g) ΔH 1 = kJ C (s) + O 2 (g) → CO 2 (g) ΔH 1 = kJ CO (g) + ½O 2 (g) → CO 2 (g) ΔH 2 = kJ CO (g) + ½O 2 (g) → CO 2 (g) ΔH 2 = kJA: C (s) + O 2 (g) → CO 2 (g) ΔH 1 = kJ C (s) + O 2 (g) → CO 2 (g) ΔH 1 = kJ CO 2 (g) → CO (g) + ½O 2 (g) ΔH 2 = kJ CO 2 (g) → CO (g) + ½O 2 (g) ΔH 2 = kJ C (s) + ½O 2 (g) → CO (g) ΔH 1 = kJ

C(s) + ½O 2 (g) → CO(g)  H = kJ C(s) + ½O 2 (g) → CO(g)  H = kJ Energy Level Diagram: Hess’s Law

5.4 Hess’s Law Q: Find the enthalpy for the reaction C 2 H 4 (g) + H 2 O(l) → C 2 H 5 OH (l) ΔH = ? C 2 H 4 (g) + H 2 O(l) → C 2 H 5 OH (l) ΔH = ? Based on the following two reactions: C 2 H 5 OH (l) + 3O 2 (g) → 2CO 2 + 3H 2 O (l) ΔH 1 = kJ C 2 H 5 OH (l) + 3O 2 (g) → 2CO 2 + 3H 2 O (l) ΔH 1 = kJ C 2 H 4 (g) + 3O 2 (g) → 2CO 2 + 2H 2 O (l) ΔH 2 = kJ C 2 H 4 (g) + 3O 2 (g) → 2CO 2 + 2H 2 O (l) ΔH 2 = kJ A: -44 kJ

5.4 Hess’s Law Given the thermochemical equations: 2Cu (s) + Cl 2 (g) → 2CuCl (s) ΔH = kJ 2Cu (s) + Cl 2 (g) → 2CuCl (s) ΔH = kJ 2CuCl (s) + Cl 2 (g) → 2CuCl 2 (s) ΔH = kJ 2CuCl (s) + Cl 2 (g) → 2CuCl 2 (s) ΔH = kJ Find the enthalpy change for: Cu (s) + Cl 2 (g) → CuCl 2 (s) ΔH = ? Cu (s) + Cl 2 (g) → CuCl 2 (s) ΔH = ? A: kJ

5.5 Standard Enthalpy of Formation Objectives Identify formation reactions and their enthalpy changes Calculate the enthalpy change of a reaction from standard enthalpies of formation

5.5 Standard Enthalpy of Formation Standard Enthalpy of Formation (ΔH f ° ) = the standard enthalpy change for the formation of one mole of a single substance in its standard state from the most stable forms of its elements in their standard states Examples: C (graphite) + O 2 → CO 2 (g) ΔH f ° = -393 kJ/mol C (graphite) + O 2 → CO 2 (g) ΔH f ° = -393 kJ/mol H 2 + ½O 2 → H 2 O (l) ΔH f ° = -286 kJ/mol H 2 + ½O 2 → H 2 O (l) ΔH f ° = -286 kJ/mol H 2 + ½O 2 → H 2 O (g) ΔH f ° = -242 kJ/mol H 2 + ½O 2 → H 2 O (g) ΔH f ° = -242 kJ/mol O 2 (g) → O 2 (g) ΔH f ° = 0 O 2 (g) → O 2 (g) ΔH f ° = 0

5.5 Standard Enthalpy of Formation Standard state = 1 atm and K Please note the following…. –Fractional coefficients are common because the product is always one mole of a single substance –Only one mol of a single substance appears on the product side of each reaction –Although it may be impractical to carry out certain reactions in real life at K, that is the temperature used to determine the standard enthalpy of formation –Only the most stable form of an element has a standard enthalpy of formation of zero

5.5 Standard Enthalpy of Formation For any chemical reaction: reactants → products ΔH ° reaction = ΣnΔH f ° [products] – ΣmΔH f ° [reactants] ΔH ° reaction = ΣnΔH f ° [products] – ΣmΔH f ° [reactants] where m = the # of mol of each reactant and n = the # of mol of each product where m = the # of mol of each reactant and n = the # of mol of each product Note: The standard enthalpies of formation have units of kJ/mol whereas the ΔH ° reaction has the units of kJ

Example: Enthalpy of Reaction Use the standard enthalpies of formation to calculate the enthalpy change for the reaction P 4 O 10 (s) + 6H 2 O(g) → 4H 3 PO 4 (s) Substance∆H° f (kJ/mol) P 4 O 10 (s)-1640 H 2 O(g)-242 H 3 PO 4 (s)-1279

5.5 Standard Enthalpy of Formation What is the ΔH ° reaction for the following reactions… 4NH 3 (g) + 5O 2 (g) → 4NO (g) + 6H 2 O (g) 4NH 3 (g) + 5O 2 (g) → 4NO (g) + 6H 2 O (g)

Table 5-2, p. 195

Use of Enthalpies of Formation 4NH 3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(g)

5.5 Standard Enthalpy of Formation What happens if a standard enthalpy of formation is not known. Can it be calculated? The answer is yes. Let’s investigate how….. The standard enthalpy change when 1 mole rubbing alcohol (C 3 H 7 OH ( l )) burns to form carbon dioxide and liquid water at K is (-) kJ. Calculate the standard enthalpy of formation of rubbing alcohol A: (-) kJ A: (-) kJ