Thermochemistry The branch of physical chemistry that deals with the heat changes accompanying various physical and chemical transformations. Heat is either evolved (exothermic) or absorbed (endothermic) in a chemical or physical change.

Thermochemistry System: A part of the universe which is under observation is called the system OR The part of the world in which we have interest. e.g. reaction vessel, calorimeter, and so on. Surroundings: The remaining portion of the universe which is not a part of the system is called the surroundings. e.g. rooms, buildings around etc. Open system: A type of system in which both mass and energy can move (enter or exit) freely. Example: a reaction vessel, or a beaker with reagents.

Conti….. Closed system: A type of system in which mass can not pass through the boundary but energy can move (enter or exit) freely. Example; Bomb calorimeter in which energy can move freely but no matter can enter or exit. Isolated system: Nothing can enter or leave. The system that has neither mass nor energy exchange with its surroundings. A substance contained in the ideal thermos flask is an example of the isolated system.

2H2 (g) + O2 (g) 2H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2 (g) Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO (s) 2Hg (l) + O2 (g) energy + H2O (s) H2O (l)

Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the degree of coldness or hotness of a body. 400C 900C

Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0C and 1 atm. H2O (s) H2O (l) DH = 6.01 kJ

Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = - 890.4 kJ

Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance. H2O (s) H2O (l) DH = 6.01 kJ If you reverse a reaction, the sign of DH changes. H2O (l) H2O (s) DH = -6.01 kJ HOWEVER; If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) 2H2O (l) DH = 2 x 6.01 = 12.0 kJ

ENTHALPY OF REACTION (HEAT OF REACTION) It is defined as the amount of heat evolved or absorbed when molar quantities of reactants shown by the balanced equation react completely at a given temperature. C (s) + O2 (g) CO2 (g) ∆H = - 94.03 k cal OR ∆H = - 393.5 KJ

STANDARD ENTHALPY CHANGES (∆Ho) Enthalpy of a reaction depends upon temperature and therefore for comparison of enthalpies of reactions of different substances a standard state is chosen. A convenient state for a substance is most stable state at 1 atmosphere and at specific temperature usually at 298 k. ∆Ho298 = Ho of Products ─ Ho of Reactants

For example Ho C (graphite) and Ho O2 (g) = 0 By convention enthalpy of every element in its standard State is assigned as zero value. For example Ho C (graphite) and Ho O2 (g) = 0 VARIOUS TYPES OF ENTHALPY CHANGES Enthalpy of Formation b) Enthalpy of Combustion or Heat of Combustion Enthalpy of Neutralization or Heat of Neutralization Enthalpy of Solution Heat of Solution

Enthalpy change (∆H) involved in the formation of a) Enthalpy of Formation Enthalpy change (∆H) involved in the formation of one mole of compound from its elements. If elements are at their standard state then enthalpy of formation is called standard enthalpy of formation (∆Hof) or standard heat of formation. Standard enthalpy of formation of CH4 from elements at 298K is ─ 74.81 kJ/mol. C (graphite) + 2H2 (g) CH4 (g) ∆Hof ,298 = ─ 74.81 kJ/mol

CH4(g)+ 2O2(g) CO2(g)+ 2H2O(l) ∆Hoc= ─ 890.40 b) Enthalpy of Combustion or Heat of Combustion Enthalpy change involved in the complete combustion of 1 mol of substance. If the substances are all in their respective standard states, then denoted by ∆Hoc . CH4(g)+ 2O2(g) CO2(g)+ 2H2O(l) ∆Hoc= ─ 890.40 kJ/mol

c) Enthalpy of Neutralization or Heat of Neutralization The change in enthalpy when 1 gram equivalent of an acid is neutralized by 1 gram equivalent of a base in dilute solution. HCl + NaOH NaCl + H2O ∆Ho = ─ 57.32 kJ HNO3 + NaOH NaNO3 + H2O ∆Ho = ─ 57.32 kJ H+ + OH- H2O ∆Ho = ─ 57.32 kJ

Enthalpy of Solution Heat of Solution Enthalpy change associated when 1 mole of the solute is dissolved in a specific amount of solvent. HCl (g) + 25 H2O(l) HCl. 25 H2O ∆Ho = ─ 69.5 kJ HCl (g) + 40 H2O(l) HCl. 40 H2O ∆Ho = ─ 73.18 kJ

The study of energy and its interconversions is called thermodynamics.

Thermodynamics First law of thermodynamics: “the internal energy of an isolated system is constant”. Mathematical statement of first law of thermodynamics: ∆u = q +w, w is the work done on a system and q is the energy transferred as heat to the system.

Conti… Thermodynamics if the system does work on the surrounding (energy flows out of the system), w is negative. if the surroundings do work on the system (energy flows into the system), w is positive. problem: calculate ∆u for a system undergoing endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ work done on the system. Using equation; ∆u = q + w= 15.6 + 1.4 =17.0 kJ

For a particular process q = 20 kJ and w = 15 kJ For a particular process q = 20 kJ and w = 15 kJ. Which of the following statements is true? Heat flows from the system to the surroundings. The system does work on the c) DE = 35 kJ. d) All of the above are true. e) None of the above are true

For a particular process q = 20 kJ and w = 15 kJ For a particular process q = 20 kJ and w = 15 kJ. Which of the following statements is true? Heat flows from the system to the surroundings. The system does work on the c) DE = 35 kJ. d) All of the above are true. e) None of the above are true

Thermodynamics A gas expanding against pressure p, w is a negative quantity as required, since work flows out of the system. when a gas is compressed, ∆v is a negative quantity, which makes w a positive quantity (work flows into the system). Problem: calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. solution: w = - p ∆v = -15x(64-46) =-270 l. atm note that since the gas expands, it does work on its surrounding.

Calculate the work for the expansion of CO2 from 1. 0 to 2 Calculate the work for the expansion of CO2 from 1.0 to 2.5 liters against a pressure of 1.0 atm at constant temperature. a) 1.5 liter.atm b) 2.5 liter.atm c) 0 d) –1.5 liter.atm e) –2.5 liter.atm

Calculate the work for the expansion of CO2 from 1. 0 to 2 Calculate the work for the expansion of CO2 from 1.0 to 2.5 liters against a pressure of 1.0 atm at constant temperature. a) 1.5 liter.atm b) 2.5 liter.atm c) 0 d) –1.5 liter.atm e) –2.5 liter.atm

Calculate the work associated with the expansion of a gas from 152 Calculate the work associated with the expansion of a gas from 152.0 L to 189.0 L at a constant pressure of 14.0 atm. a) 518 L . atm b) –518 L . atm c) –259 L . atm d) 171 L . atm e) 259 L . atm

Calculate the work associated with the expansion of a gas from 152 Calculate the work associated with the expansion of a gas from 152.0 L to 189.0 L at a constant pressure of 14.0 atm. a) 518 L . atm b) –518 L . atm c) –259 L . atm d) 171 L . atm e) 259 L . atm

Hess’s Law Hess’s law of heat summation: It states that the total enthalpy change for a reaction is the same whether the reaction takes place in a single step or in several steps (State function). OR Path Independent. OR It states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps. 3

Hess’s Law Consider the formation of carbon dioxide (CO2). Direct Method C (s) + O2 (g) CO2 (g) ∆H = - 393.5 KJ Indirect Method C (s) + ½ O2 CO (g) CO + ½ O2 CO2 (g) ___________________________________ ∆H = - 108.98 KJ ∆H = - 284.52 KJ ∆H = - 393.50 KJ Adding= C (s) + O2 (g) CO2 (g)

Hess’ Law: Example-2

Example (cont.) Our reaction of interest is: N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ • This reaction can also be carried out in two steps: N2 (g) + O2 (g) 2NO(g) DH = 180 kJ 2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ

Example (cont.) If we take the previous two reactions and add them, we get the original reaction of interest: N2 (g) + O2 (g) 2NO(g) DH = 180 kJ 2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ N2 (g) + 2O2 (g) 2NO2(g) DH = 68 kJ

H2O(g) + C(s) → CO(g) + H2(g) Example-3 H2O(g) + C(s) → CO(g) + H2(g) Use these equations to calculate the molar enthalpy change which produces hydrogen gas. C(s) + ½ O2(g) → CO(g) ∆H = -110.5kJ H2(g) + ½ O2(g) → H2O(g) ∆H = -241.8kJ

H2O(g) + C(s) → CO(g) + H2(g) Use these equations to calculate the molar enthalpy change which produces hydrogen gas. C(s) + ½ O2(g) → CO(g) ∆H = -110.5kJ H2O(g) → H2(g) + ½ O2(g) ∆H = +241.8kJ _____________________________________ C(s) + H2O(g) → H2(g) + CO(g) ∆H=+131.3kJ

Given: Cu2O(s) + (1/2)O2(g) 2CuO(s) DH = –144 kJ Cu2O(s) Cu(s) + CuO(s) DH = +11 kJ Calculate the standard enthalpy of formation of CuO(s). a) –166 kJ b) –299 kJ c) +299 kJ d) +155 kJ e) –155 kJ

Given: Cu2O(s) + (1/2)O2(g) 2CuO(s) DH = –144 kJ Cu2O(s) Cu(s) + CuO(s) DH = +11 kJ Calculate the standard enthalpy of formation of CuO(s). a) –166 kJ b) –299 kJ c) +299 kJ d) +155 kJ e) –155 kJ

Consider the following processes: 2A 1/2B + C DH1 = 5 kJ/mol (3/2)B + 4C 2A + C + 3D DH2 = –15 kJ/mol E + 4A C DH3 = 10 kJ/mol Calculate DH for: C E + 3D a) 0 kJ/mol b) 10 kJ/mol c) –10 kJ/mol d) –20 kJ/mol e) 20 kJ/mol

Consider the following processes: 2A 1/2B + C DH1 = 5 kJ/mol (3/2)B + 4C 2A + C + 3D DH2 = –15 kJ/mol E + 4A C DH3 = 10 kJ/mol Calculate DH for: C E + 3D a) 0 kJ/mol b) 10 kJ/mol c) –10 kJ/mol d) –20 kJ/mol e) 20 kJ/mol

At 25C, the following heats of reaction are known: DH (kJ/mol) 2ClF + O2 Cl2O + F2O 167.4 2CIF3 + 2O2 Cl2O + 3F2O 341.4 2F2 + O2 2F2O –43.4 At the same temperature, calculate DH for the reaction: ClF + F2 ClF3 a) –217.5 kJ/mol b) –130.2 kJ/mol c) +217.5 kJ/mol d) –108.7 kJ/mol e) none of these

At 25C, the following heats of reaction are known: DH (kJ/mol) 2ClF + O2 Cl2O + F2O 167.4 2CIF3 + 2O2 Cl2O + 3F2O 341.4 2F2 + O2 2F2O –43.4 At the same temperature, calculate DH for the reaction: ClF + F2 ClF3 a) –217.5 kJ/mol b) –130.2 kJ/mol c) +217.5 kJ/mol d) –108.7 kJ/mol e) none of these

Calorimetry (Measuring Heats of Reactions) Calorimetry, Latin words calor meaning heat, and the Greek metry meaning to measure, is the science of measuring the amount of heat. All calorimetric techniques are therefore based on the measurement of heat that may be generated (exothermic process), consumed (endothermic process) by a sample.

Calorimeter A calorimeter is a device used to measure heat of reaction. To measure heats of reactions, enclose reactants in a calorimeter, initiate the reaction, and measure the temperature difference before and after the reaction. The temperature difference enables us to evaluate the heat released in the reaction. The heat capacity is the amount of heat required to raise the temperature of the entire calorimeter by 1 K.

The Bomb Calorimeter

The Bomb Calorimeter For combustion reactions, all reactants are enclosed in an explosive-proof steel container, called the bomb whose volume does not change during a reaction. The bomb is then submerged in water or other liquid that absorbs the heat of reaction. The application of bomb calorimeter is called the bomb calorimetery.