Unit 13 Thermochemistry

CHM 1046: General Chemistry and Qualitative Analysis Unit 13Thermochemistry Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL • Textbook Reference: • Chapter # 15 (sec. 1-11) • Module # 3 (sec. I-VIII)

1 KE =  mv2 2 Energy Potential Energyan object possesses by virtue of its position or chemical composition (bonds). 2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g) Chemical bond energy Energy (-E): work + heat Kinetic Energyan object possesses by virtue of its motion.

Chemical bond energy Energy  gas=2534= +9 9 x 22.4L/ = 202 L 2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g) • Work (w): Energy (force) used to causean objectthathas mass to move (a distance) =kinetic energy. H = - 5.5 x 106 kJ/ Energy produced by chemical reactions = heat or work. • Heat (q): spontaneous transfer (flow) of energy (energy in transit) from one object to another; causes molecular motion & vibrations (kinetic energy); measured as temperature. W = F x d q + w E = Law of Conservation of Energy:

kg m2 kg m =  =  s2 s2 Units of Energy: Joule & calorie Energy (Work) = F x d • The joule (J) = SI unit of energy (work) • The calorie (cal)= heat required to raise 1 g of H2O 10C 1 cal = 4.184 J {Nutritional Calorie = 1000 calories (1kcal)=4,184 J}. Joule (J) = N·m m The Newton (N) is the amount of force that is required to accelerate a kilogram of mass at a rate of one meter per second squared.

System and Surroundings Work • The system includes the molecules we want to study. • The surroundings are everything else (here, the cylinder and piston). +Ew -Ework Heat System -Eq(heat) +Eq E = q + w Surroundings

First Law of Thermodynamics • Energy is neither created nor destroyed. • Also known as the Law of Conservation of Energy Work • In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. +Ew -Ew UNIVERSE Heat System -Eq Esys + Esurr = 0 +Eq Esys = -Esurr or E sys = (q + w)surr Surroundings

Changes in Internal Energy (E) : energy released/absorbed as either work or heat, is looked at from the perspective of the system (the chemicals) E, q, w, and Their Signs E = q + w + q - q > q Syst looses heat Syst gains heat + w > w - w Work donebySyst Work done on Syst + E = + q + w - E = - q - w ± E = ± q ± w

Why consider Energy Change (ΔE) and not the total Internal Energy (E)? Esys + Esurr = 0 Esys + Esurr = ETotal(constant) • Total Internal Energy (E) = the sum of all kinetic and potential energies of all components of the system.How can it be determined? • But when a change occurs we can measure the change in internal energy: E = Efinal−Einitial • Usually we have no way of knowing the total internal energy of a system; finding that value is simply too complex a problem.

Energy as Work (w) of Gas Expansion F = P A Work = - (Force x distance) w = - (F x h) w = - (P x A x h) w = - P V (@ constant P) w = - nRT (@ constant V) {WorkGasExpansion}

Exchange of Heat (H)by chemical systems • When heat is released by the system to the surroundings, the process is exothermic. - H q • When heat is absorbed by the system from the surroundings, the process is endothermic. + H q

Enthalpies of Reaction (H) Reactants: The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts−Hreactants This quantity, H, is called the enthalpy of reaction, or the heat of reaction. Products Notice it is the same value, but of opposite sign for the reverse reaction

State Functions • Are E, q and w all state functions? A state function depends only on the present state of the system, not on the path by which the system arrived at that state. ∆E= q+w • However, q and w are not state functions. • Other state functions are P, V and T.

Calorimetry • The experimental methods of measuring of heats (H) involved in chemical reactions • Methods utilized: • Constant Pressure Calorimetry (open container in contact with atmosphere) • Constant Volume Calorimetry: closed container (bomb) 1 atm

Thermochemical Equations 2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g) H = - 5.5 x 106 kJ/ C8H18 • H is per mole of reactant, at the indicated states (s, l, g). • Direct quantitative relationship between coefficients of balanced equation and H. • Reverse equation has H of opposite sign. Problem: Calculate H when 25g of C8H18 (l) (MM= 114) are burned in excess oxygen?

(1) Calorimetry at Constant Pressure • Most chemical reactions are carried out in beakers or flasks that are open to the atmosphere (i.e., at constant pressure, isobaric). E = q + w 1 atm E = qp - PV SOLVE FOR E + PV=qp q = m  c  T qp =Heat of Reaction =DH or Enthalpy of Reaction Big Problem: calorimeter also absorbs heat E + PV=H H=E + PV H=Epv  and when the volume is constant

(a) Determining the Heat Capacity of a Calorimeter 50 mL of H2O @ 650C are added……. 50 mL H2O 50 mL H2O …..to 50 mL of H2O @ 250C inside the calorimeter The final temperature of the 100 mL of H2O inside the calorimeter is 420C Heat lost by = heat gained by+ heat gained by hot watercold water calorimeter qhot = qcold+ qcalorimeter

Determining the Heat Capacity of a Calorimeter (Ccal) qhot = qcold+ qcalorimeter (mcT)H2O = (mcT)H2O + (Ccal T) Includes both the mass and heat capacity (mcT)H2O - (mcT)H2O (T)cal Ccal =

(b) Calculating the Heats of Reactions (H) HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 50 mL@25°C + 50 mL@25°C  100 mL of NaCl(aq) @ 30°C Heat of = heat gained by+ heat gained by reactionNaClsolution calorimeter H(qrxn) = qNaCl soln+ qcal qrxn = (mc*T)NaClsoln + (CcalT) qrxn = - Hrxn = heat of neutralization rxn * How do you determine the cNaCl soln?

How do you determine the cNaCl soln? 50 mL of H2O @ 650C are added……. 50 mL H2O 50 mL NaCl soln …..to 50 mL of NaCl solution @ 250C inside the calorimeter The final temperature of the 100 mL of solution inside the calorimeter is 410C Heat lost by = heat gained by + heat gained by hot watersalt water soln calorimeter qh = qNaClsoln+ qcal (mcT)H2O = (mc*T)NaClsoln + (CcalT)

(2) Calorimetry at Constant Volume (Bomb Cal.) E = q + w w = P V V =0 w = 0 E = qv q = m  c  T Bomb calorimeter

Bomb Calorimetry • Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. • For most reactions, the difference is very small. E = qv H=E + nRT

Review of Thermochemical Equations E = q + w E = qp - PV E = qp - nRT since ∆H= qp E = H - PV 1 atm H=E + PV E = q + w E = qv

Comparing H and E H=E + PV 2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g) H = - 5.5 x 106 kJ/ E=H - PV (useful at const P) E=H - nRT(useful at const V) Problem: calculate E @ 25°C for above equation @ const. V R= 8.314J/K.  gas=34-25 = +9 = 4.5 C8H18

Constant-Volume Calorimetry Problem: When one mole of CH4 (g) was combusted at 250C in a bomb calorimeter, 886 kJ of energy were released. What is the enthalpy change for this reaction? CH4 (g) + 2 O2 (g) CO2(g) + 2 H2O(l) E = qv = - 886 kJ R = 8.31 J/ .K n = 1 – (1+2) = -2  H=E + PV = E + nRT

Energy in Foods Most of the fuel in the food we eat comes from carbohydrates & fats.

Methods of determining H • Calorimetry (experimental) • Hess’s Law: using Standard Enthalpy of Reaction(Hrxn) of a series of reaction steps (indirect method). • Standard Enthalpy of Formation(Hf ) used with Hess’s Law (direct method) • Bond Energies used with Hess’s Law  Experimental data combined with theoretical concepts 

(2)Determination of H using Hess’s Law • Hrxn is well known for many reactions, but it is inconvenient to measure Hrxn for every reaction. • However, we can estimate Hrxn for a reaction of interest by using Hrxn values that are published for other more common reactions.  • The Standard Enthalpy of Reaction(Hrxn) of a series of reaction steps are added to lead to reaction of interest (indirect method). Standardconditions (25°C and 1.00 atm pressure). (STP for gases T= 0°C) 

Hess’s Law “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” - 1840, Germain Henri Hess (1802–50), Swiss

Calculation of H by Hess’s Law 3 C(graphite) + 4 H2 (g)  C3H8 (g)H= -104 3 C(graphite) + 3 O2 (g) 3 CO2 (g)H=-1181 4 H2 (g) + 2 O2 (g) 4 H2O (l)H=-1143 C3H8 (g)  3 C(graphite) + 4 H2 (g) H= +104 C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) • Appropriate set of Equations with their H values are obtained (or given), which containing chemicals in common with equation whose H is desired. • These Equations are all added to give you the desired equation. • These Equations may be reversed to give you the desired results (changing the sign of H). • You may have to multiply the equations by a factor that makes them balanced in relation to each other. • Elimination of common terms that appear on both sides of the equation .

Calculation of H by Hess’s Law 3 C(graphite) + 3 O2 (g)  3 CO2 (g)H=-1181 4 H2 (g) + 2 O2 (g)  4 H2O (l)H=-1143 C3H8 (g)  3 C(graphite) + 4 H2 (g) H= +104 • Hrxn = • + 104 kJ • 1181 kJ • 1143 kJ • 2220 kJ C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

Calculation of H by Hess’s Law Calculate heat of reaction W + C (graphite)  WC (s) ΔH = ? Given data: 2 W(s)+ 3 O2 (g)  2 WO3 (s) ΔH = -1680.6 kJ C (graphite) + O2(g)  CO2(g)ΔH = -393.5 kJ 2 WC (s)+ 5 O2(g)  2 WO3 (s) + CO2(g)ΔH = -2391.6 kJ ½(2 W(s) + 3 O2 (g)  2 WO3 (s) ) ½(ΔH = -1680.6 kJ) W(s) + 3/2 O2 (g)  WO3 (s) ) ΔH = -840.3 kJ C (graphite) + O2(g)  CO2(g)ΔH = -393.5 kJ ½(2 WO3 (s) + CO2(g) 2 WC (s) + 5 O2(g)) ½ (ΔH = +2391.6 kJ) WO3 (s) + CO2(g) WC (s) + 5/2 O2(g)ΔH = + 1195.8 kJ) W + C (graphite)  WC (s) ΔH = - 38.0

Hess’s Law Problem: Chloroform, CHCl3, is formed by the following reaction: Desired ΔHrxn equation: CH4 (g) + 3 Cl2 (g) → 3 HCl (g) + CHCl3 (g) Determine the enthalpy change for this reaction (ΔH°rxn), using the following: 2 C (graphite) + H2 (g) + 3Cl2 (g) →2CHCl3 (g)ΔH°f = – 103.1 kJ/mol CH4 (g) + 2 O2 (g) → 2 H2O (l) + CO2 (g) ΔH°rxn = – 890.4 kJ/mol 2 HCl (g) → H2 (g) + Cl2 (g) ΔH°rxn = + 184.6 kJ/mol C (graphite) + O2 (g) → CO2(g) ΔH°rxn = – 393.5 kJ/mol H2 (g) + ½ O2 (g) → H2O (l)ΔH°rxn = – 285.8 kJ/mol answers: a) –103.1 kJ b) + 145.4 kJ c) – 145.4 kJ d) + 305.2 kJ e) – 305.2 kJ f) +103.1 kJ This is a hard question. To make is easer give: C (graphite) + ½ H2(g) + 3/2 Cl2(g) → CHCl3(g) ΔH°f = – 103.1 kJ/mol

Methods of determining H Calorimetry (experimental) Hess’s Law: using Standard Enthalpy of Reaction (Hrxn) of a series of reaction steps (indirect method). Standard Enthalpy of Formation (Hf ) used with Hess’s Law (direct method) Bond Energies used with Hess’s Law  Experimental data combined with theoretical concepts 

(3) Determination of H using Standard Enthalpies of Formation (Hf )  Enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. StandardEnthalpy of formationHf are measured under standard conditions (25°C and 1.00 atm pressure).  C + O2 CO2∆Hf = -393.5 kJ/  

Calculation of H CH4(g) + O2(g) CO2(g) + H2O(g) C + 2H2(g) CH4(g)ΔHf = -74.8 kJ/ŋ We can use Hess’s law in this way: H = nHf(products) - mHf(reactants) where n and m are the stoichiometric coefficients. C(g) + O2(g) CO2(g) ΔHf = -393.5 kJ/ŋ 2H2(g) + O2(g) 2H2O(g) ΔHf = -241.8 kJ/ŋ   - n CH4(g) + n O2(g) n CO2(g) + n H2O(g) H = [1(-393.5 kJ) + 1(-241.8 kJ)] - [1(-74.8 kJ) + 1(-0 kJ)] = - 560.5 kJ

Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) H = nHf(products) - mHf(reactants) H = [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ Table of Standard Enthalpy of formation, Hf

(4) Determination of H using Bond Energies • Most simply, the strength of a bond is measured by determining how much energy is required to break the bond. • This is the bond enthalpy. • The bond enthalpy for a Cl—Cl bond, D(Cl—Cl), is measured to be 242 kJ/mol.

Average Bond Enthalpies (H) • Average bond enthalpies are positive, because bond breaking is an endothermic process. NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than the C—H bond in chloroform, CHCl3.

Enthalpies of Reaction (H ) • Yet another way to estimate H for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed. • In other words, • Hrxn = (bond enthalpies of bonds broken)  • (bond enthalpies of bonds formed)

Hess’s Law: Hrxn = (bonds broken)  (bonds formed) CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g) Hrxn = [D(C—H) + D(Cl—Cl)  [D(C—Cl) + D(H—Cl) = [(413 kJ) + (242 kJ)]  [(328 kJ) + (431 kJ)] = (655 kJ)  (759 kJ) = 104 kJ

Bond Enthalpy and Bond Length • We can also measure an average bond length for different bond types. • As the number of bonds between two atoms increases, the bond length decreases.

2003 B Q3

2005 B

2002

2002 #8

Unit 13 Thermochemistry