Thermochemistry

1. Thermochemistry Chapter 8

2. THERMOCHEMISTRY • Thermochemistry is branch of chemistry concerned with the heat changes that occur during chemical reactions

3. THERMODYNAMICS • (from the Greek word therme, meaning “heat” and  dynamics,  meaning “moving”) • Thermodynamics is the study of energy conversion between heat and mechanical work, and subsequently the macroscopic variables suchas temperature, volume and pressure. 

4. Energyis the capacity to do work or supply heat • Kinetic energy is the energy because of movement of a body • EK = 1/2mv2 • Potential energy is the energy • available by virtue of an object’s position • Thermal energy is the kinetic energy associated with the random motion of atoms and molecules • Chemical energy is the energy stored within the bonds of chemical substances

5. Temperature = Thermal Energy 900C 400C Energy Changes in Chemical Reactions • Heat is the amount of thermal energy transferred between two bodies because of temperature difference.(depend on amount of substance) • Heat alone cannot be measured • Only changes caused by heat can be detected • Heat always flows from warmer to cooler Temperature is a measure of the thermal energy. greater thermal energy Low temperature

6. Energy & Chemistry All of thermodynamics depends on the law of CONSERVATION OF ENERGY. • The total energy is unchanged in a chemical reaction. • If E of products is less than reactants, the difference must be released as E.

7. James Joule 1818-1889 UNITS OF ENERGY EK = 1/2mv2 =SI units (kg m2)/s2 = J (JOULE common scientific units) Mostly expressed in terms of kJ Some scientists use calorie where 1 cal= heat required to raise temp. of 1 g of H2O by 1oC. 1 cal = 4.184 joules Nutritionists used Calorie (for food energy) 1Cal=1000 cal = 1 kcal

8. SURROUNDINGS SYSTEM Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. closed isolated open energy nothing Exchange: mass & energy

9. Internal Energy (E) • PE + KE = Internal energy (E) • Int. E of a chemical system depends on • number of particles • type of particles • temperature • The higher the T the higher the internal energy • So, use changes in T (∆T) to monitor changes in E (∆E).

10. FIRST LAW OF THERMODYNAMICS • The total internal energy of an isolated system is constant. OR • The change in the internal energy of a closed thermodynamic system is equal to the sum of the amount of heat energy supplied to or removed from the system and the work done on or by the system. ΔE = Efinal − Einitial

11. T(system) goes down T(surr) goes up Directionality of Heat Transfer • Heat always transfer from hotter object to cooler one. • EXOthermic: heat transfers from SYSTEM to SURROUNDINGS.

12. T(system) goes up T (surr) goes down Directionality of Heat Transfer • Heat always transfer from hotter object to cooler one. • ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM.

13. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings ΔE = -ve. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. ΔE = +ve.

14. State functions (written in capital letters) are properties that are determined by the present state of the system, regardless of path that condition was achieved. changes in state functions are Reversible. Examples energy E, pressure P, volume V, temperature T Parameters which are not state function are written in small letters Examples work w, heat q, time t Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

15. EXPANSION WORK w = F × d F= - P × A - as it is against external pressure w= - P × A × d w = -PDV 6.7

16. Enthalpy and the First Law of Thermodynamics DE = q + w = q – PDV q= DE + PDV At constant volume, qV = DE as DV=0 At constant pressure, qP = DE + PDV =DH Called Enthalpy So DE = DH - PDV Therefore DH = DE + PDV And DH = DE when DV=0

17. Thermodynamics DE = q + w DE is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -PDVwhen a gas expands against a constant external pressure

18. Enthalpy (H) (heat content) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) Standard enthalpy (DHo)= Enthalpy change measured under standard conditions 1 atm pressure, 25oC temperature 1molar concentration

19. H2O (s) H2O (l) DH = 6.01 kJ 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 00C and 1 atm.

20. DH = -890.4 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O (l) 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.

21. 2H2O (s) 2H2O (l) H2O (s) H2O (l) H2O (l) H2O (s) DH = -6.01 kJ DH = 6.01 kJ DH = 2 x 6.01= 12.0 kJ Thermochemical Equations • The stoichiometric coefficients always refer to the number of moles of a substance • If you reverse a reaction, the sign of DH changes • If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.

22. How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4(s) + 5O2(g) P4O10(s)DH = -3013 kJ x H2O (l) H2O (g) H2O (s) H2O (l) 3013 kJ 1 mol P4 x DH = 6.01 kJ DH = 44.0 kJ 1 mol P4 123.9 g P4 Thermochemical Equations • The physical states of all reactants and products must be specified in thermochemical equations. = 6470 kJ 266 g P4

24. Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero.

26. aA + bB cC + dD - [ + ] [ + ] = - S S = DH0 DH0 rxn rxn dDH0 (D) bDH0 (B) aDH0 (A) cDH0 (C) nDH0 (products) mDH0 (reactants) f f f f f f Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Remember: Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

27. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn 2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ C(graphite) + 2S(rhombic) CS2 (l) C(graphite) + 2S(rhombic) CS2 (l) rxn rxn C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ + CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ rxn DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn Calculate the standard enthalpy of formation of CS2 (l) given that: 1. Write the enthalpy of formation reaction for CS2 2. Add the given rxns so that the result is the desired rxn.

28. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - S S = DH0 DH0 DH0 - [ ] [ + ] = rxn rxn rxn [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = - 2973 kJ/mol C6H6 6DH0 (H2O) -5946 kJ f 2 mol mDH0 (reactants) nDH0 (products) f f Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

29. Bond Dissociation energies