Thermochemistry Chapter 6 Dr. Ali Bumajdad

Study of heat change in a chemical reaction Chapter 6 Topics Thermochemistry Study of heat change in a chemical reaction Nature and Type of Energy Energy Changes in Chemical Reaction Intro. to Thermodynamics: 1st Law, Wark, Heat, Enthalpy Calorimetry, Specific Heat, Heat Capacity Sstandard enthalpy of formation and reaction Hess Law Enthalpy of solution Dr. Ali Bumajdad

Energy is the capacity to do work Work = Fd Radiant energy comes from the sun (also called solar energy) Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy object position relative to other objects (e.g. gravity result in pot. E to objects) Kinetic Energy is the energy of motion Unit = J = Kgm2s-2 Energy convert from one form to another (Law of Conservation of Energy)

Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Q) Water cup in the fridge, what is the direction of heat transfer? Temperature is a measure of the thermal energy. Temperature = Thermal Energy 900C 400C greater thermal energy

Thermochemistry is the study of heat change in chemical reactions. System is the specific part of the universe that is of interest in the study Surrounding is every things except the thing we are studying Open system mass & energy Exchange: Closed energy Isolated nothing Type:

2H2 (g) + O2 (g) 2H2O (l) + energy Exothermic process is any process that gives off heat or 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 + H2O (s) H2O (l) energy + 2HgO (s) 2Hg (l) + O2 (g)

‘The study of the interconversion of heat and other kinds of energy’ Thermodynamics ‘The study of the interconversion of heat and other kinds of energy’ State functions properties determined by the current state of the system, regardless of how that condition was achieved e.g. energy , pressure, volume, temperature DE = Efinal - Einitial Work and Heat Not State Function DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

First law of thermodynamics: 1) energy can be converted from one form to another, but cannot be created or destroyed DEsystem + DEsurroundings = 0 DEsystem = -DEsurroundings OR C3H8 + 5O2 3CO2 + 4H2O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings 2) Heat transfer to (+) or from (-) the system Work done on (+) or by (-) the system W=-PV Change in internal energy of a system DEsys = q + w (1) (2)

Combustion reaction release heat q=-ve Combustion reaction release heat Vaporization absorbed heat q=+ve Condensation release heat q=-ve Freezing release heat q=-ve Melting absorbed heat q=+ve

Work is not a state function! Work Done On the System w = F x d w = -P DV DV > 0 -PDV < 0 wsys < 0 P x V = x d3 = F x d = w F d2 pressure vol Work is not a state function! Dw = wfinal - winitial initial final

 Work is not state Function Q) A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? 1L atm =101.3 J w = -P DV (2) (a) DV = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 L•atm = 0 joules (b) DV = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = -14.1 L•atm w = -14.1 L•atm x 101.3 J 1L•atm = -1430 J  Work is not state Function

Enthalpy and the First Law of Thermodynamics Enthalpy (H) used to quantify heat flow into or out of a system in a process that occurs at constant pressure. DE = q + w At constant volume: w = 0 and DE = qv At constant pressure: DH = qp and w = -PDV DE = DH - PDV DH = DE + PDV (3) State Function

DH = H (products) – H (reactants) Enthalpy of Reactions DH = H (products) – H (reactants) (4) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants DH < 0 Hproducts > Hreactants DH > 0

Is DH negative or positive? H2O (s) H2O (l) System absorbs heat Endothermic DH > 0, H =+ve 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. H2O (s) H2O (l) DH = 6.01 kJ/mol

Is DH negative or positive? System gives off heat CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) Exothermic DH < 0, H =+ve 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/mol

Thermochemical Equations H2O (s) H2O (l) DH = 6.01 kJ/mol 1) If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = -6.01 kJ/mol  Hforward =-Hbackward 2) If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. H extensive property 2H2O (s) 2H2O (l) DH = 2 x 6.01 = 12.0 kJ/mol 3) The physical states of all reactants and products must be specified in thermochemical equations H2O (s) H2O (l) DH = 6.01 kJ/mol H2O (l) H2O (g) DH = 44.0 kJ/mol H H2O(g) > H H2O(l) > H H2O(l)

P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ/mol Q) How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ/mol = 6470 kJ

A Comparison of DH and DE 2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g) DH = -367.5 kJ/mol DE = DH - PDV At 25 0C, 1 mole H2 = 24.5 L at 1 atm and VVgas PDV = 1 atm x 24.5 L = 2.5 kJ DE = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol E > H Some of Internal energy released used to do gas expansion work, this is why For reactions involving gases E slightly different than H For reactions do not involve gases E  H (V is very small)

Calorimetry, Specific Heat, Heat Capacity Calorimetry The measurement of heat flow Specific heat (s) is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. q = m x s x DT s= q mT (5) J/g.°C Heat capacity (C) is the amount of heat (q) required to raise the temperature of a given quantity (m) by one degree Celsius. q = C x DT C= q T (6) J/°C C = m x s (7) From (5) and (6):

Q) How much heat is given off when an 869 g iron bar cools from 940C to 50C? s of Fe = 0.444 J/g • 0C q = m x s x Dt s= q mT (5) J/g.°C = -34,000 J

q = m x s x Dt

(1) Constant-Volume Calorimetry (Good for combustion reaction) Heat produced by combustion and absorbed by surrounding (calorimeter) qsys = qcal + qrxn qsys = 0 qrxn = - qcal qcal = Ccal x DT qrxn = -Ccal x DT (8) For Reaction at Constant V DH = qrxn DE = qrxn No heat enters or leaves (Adiabatic)

Molar heat of combustion: Heat of combustion of one mole qrxn = -Ccal x Dt (8b) n molar

(2) Constant-Pressure Calorimetry Heat produced (or absorbed) by reaction and absorbed (or produced) by the surrounding (solution) Endo. qsys = qsolution + qrxn Exoth. qsys = 0 qrxn = - qsolution qcal = Ccal x DT (8) qrxn = - Ccal x DT Reaction at Constant P DE  qrxn DH = qrxn No heat enters or leaves! (Adiabatic)

qrxn = -Ccal x Dt (8b) n molar

s= q mT J/g.°C =79kJ Cmolar= q nT n or Sa. Ex.5.6a: How much heat is needed to warm 250g of water from 22°C to 98°C. (water specific heat =4.184 J/g.°C) q = m x s x DT s= q mT (5) J/g.°C =79kJ Sa. Ex.5.6b: What is the molar heat capacity of water? Cmolar= q nT C = m x s (7) n or

qwater = m x s x DT =-qPb qPb = m x s x DT

Reaction is exothermic qrxn = - qsolution when density 1 volume = mass mol of NaOH = MNaOH VNaOH mol of HCl = MHCl VHCl =-56.2kJ/mol

Standard Enthalpy of Formation (H0f) (It is the heat change when 1 mole of a compound formed from its elements at 1 atm) It is a reference point for all enthalpy expressions (similar to the sea level for measuring heights) The standard enthalpy of formation of any element in its most stable form is zero DH0 (O2) = 0 f DH0 (C, graphite) = 0 f DH0 (O3) = 142 kJ/mol f DH0 (C, diamond) = 1.90 kJ/mol f

Hrxn= Hstep1 + Hstep2+ Hstep3+…… Standard Enthalpy of Reaction (H0 ) rxn (It is the enthalpy of a reaction carried out at 1 atm) aA + bB cC + dD DH0 rxn dDH0 (D) f cDH0 (C) = [ + ] - bDH0 (B) aDH0 (A) nDH0 (products) S mDH0 (reactants) (9) 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. Enthalpy is a state function It doesn’t matter how you get there, only where you start and end. If a reaction is carried out in steps, H will equal the sum of the enthalpy changes for the individual steps. Hrxn= Hstep1 + Hstep2+ Hstep3+…… (10)

C (graphite) + 1/2O2 (g) CO (g) CO (g) + 1/2O2 (g) CO2 (g) C (graphite) + O2 (g) CO2 (g)

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

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) Q) 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, for CO2 is -393.5 and for H2O is -285.8 kJ/mol 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - (9) DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6535 kJ -6535 kJ 2 mol = - 3268 kJ/mol C6H6

Hrxn= Hstep1 + Hstep2+ Hstep3+…… (10)

Hf°Fe2O3(s)=-822.2kJ/mol, Hf°Al2O3(s)=-1669.8 kJ/mol (9) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - (9) Divide by 2 to have the molar value then divide by the molar mass of Al to get it kJ/g

Enthalpy of solution (Hsoln) The heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents

The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol