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

Energy Potential Energy an 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 Energy an object possesses by virtue of its motion. 1 2 KE =  mv2

Energy q + w E = 2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g) Chemical bond energy  gas=2534= +9 9 x 22.4L/ = 202 L 2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g) 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. Work (w): Energy (force) used to cause an object that has mass to move (a distance) = kinetic energy. W = F x d q + w E = Law of Conservation of Energy:

Units of Energy: Joule & calorie 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}. Energy (Work) = F x d =  kg m2 s2 =  kg m s2 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 The system includes the molecules we want to study. The surroundings are everything else (here, the cylinder and piston). Work +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) E, q, w, and Their Signs : energy released/absorbed as either work or heat, is looked at from the perspective of the system (the chemicals) + q - q > q Syst looses heat Syst gains heat + w > w - w Work done by Syst Work done on Syst E = q + w + 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? Usually we have no way of knowing the total internal energy of a system; finding that value is simply too complex a problem. But when a change occurs we can measure the change in internal energy: E = Efinal − Einitial

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? ∆E= q+w 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 water cold 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 reaction NaCl solution 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 water salt 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 E = qv H = E + nRT 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 E = H - PV (useful at const P) 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). Standard conditions (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.  C + O2  CO2 ∆Hf = -393.5 kJ/   Standard Enthalpy of formation Hf are measured under standard conditions (25°C and 1.00 atm pressure).

Calculation of H CH4(g) + O2(g)  CO2(g) + H2O(g) - C + 2H2(g)  CH4(g) ΔHf = -74.8 kJ/ŋ C(g) + O2(g)  CO2(g) ΔHf = -393.5 kJ/ŋ 2H2(g) + O2(g)  2H2O(g) ΔHf = -241.8 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.   - 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

2003 A