CHM 1046 General Chemistry and Qualitative Analysis Unit

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) Thermochemistry • Module # 3 (sec. I-VIII)

Energy Potential Energy an object possesses by virtue of its position or chemical composition (bonds). 2 C 8 H 18 (l) + 25 O 2 (g) Chemical bond energy 16 CO 2(g) + 18 H 2 O(g) Energy (- E): work + heat Kinetic Energy an object possesses by virtue of its motion. 1 KE = mv 2 Thermochemistry 2

Chemical bond energy Energy 2 C 8 H 18 (l) + 25 O 2 (g) gas=25 34= +9 9 x 22. 4 L/ = 202 L 16 CO 2(g) + 18 H 2 O(g) H = - 5. 5 x 106 k. J/ Energy produced by chemical reactions = heat or work. Ø Heat (q): spontaneous transfer (flow) of energy (energy in transit) from one object to another; causes molecular motion & vibrations (kinetic energy); measured as temperature. Ø Work (w): Energy (force) used to cause an object that has mass to move (a distance) = kinetic energy. W = F x d Law of Conservation of Energy: E = q + w Thermochemistry

Units of Energy: Joule & calorie • The joule (J) = SI unit of energy (work) Energy (Work) = F x d Joule (J) = N·m kg m = m s 2 kg m 2 = s 2 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. • The calorie (cal) = heat required to raise 1 g of H 2 O 10 C 1 cal = 4. 184 J Thermochemistry {Nutritional Calorie = 1000 calories (1 kcal)=4, 184 J}.

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

First Law of Thermodynamics • Energy is neither created nor destroyed. • Also known as the Law of Conservation of Energy Work + Ew - Ew System Heat - Eq + Eq • 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. Esys + Esurr = 0 Esys = - Esurr or Surroundings Thermochemistry E sys = (q + w)surr

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 Work done on Syst >w -w Work done by Syst E = q + w + E = - E = ± E = +q -q ±q +w -w ±w Thermochemistry

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: Thermochemistry E = Efinal − Einitial

Energy as Work (w) of Gas Expansion F=PA Work = - (Force x distance) w = - (F x h) w = - (P x A x h) w = - P V (@ constant P) w = - n. RT (@ constant V) Thermochemistry {Work. Gas. Expansion}

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

Enthalpies of Reaction ( H) The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: 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 Thermochemistry

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. Thermochemistry

Calorimetry • The experimental methods of measuring of heats ( H) involved in chemical reactions 1 atm • Methods utilized: 1. Constant Pressure Calorimetry (open container in contact with atmosphere) 2. Constant Volume Calorimetry: closed container (bomb) Thermochemistry

Thermochemical Equations 2 C 8 H 18 (l) + 25 O 2 (g) 16 CO 2(g) + 18 H 2 O(g) H = - 5. 5 x 106 k. J/ C 8 H 18 1. H is per mole of reactant, at the indicated states (s, l, g). 2. Direct quantitative relationship between coefficients of balanced equation and H. 3. Reverse equation has H of opposite sign. Problem: Calculate H when 25 g of C 8 H 18 (l) (MM= 114) are burned in excess oxygen? Thermochemistry

(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 E = qp - P V 1 atm SOLVE FOR q = m c T E + P V = qp qp = Heat of Reaction = H or Enthalpy of Reaction Big Problem: calorimeter also absorbs heat E + P V = H H = E + P V Thermochemistry H = Epv and when the volume is constant

(a) Determining the Heat Capacity of a Calorimeter 50 m. L of H 2 O @ 650 C are added……. 50 m. L H 2 O …. . to 50 m. L of H 2 O @ 250 C inside the calorimeter The final temperature of the 100 m. L of H 2 O inside the calorimeter is 420 C Heat lost by hot water = heat gained by cold water qhot = qcold + + heat gained by calorimeter q calorimeter Thermochemistry

Determining the Heat Capacity of a Calorimeter (Ccal) qhot = qcold + q calorimeter (mc T)H O = (mc T)H O + (Ccal T) 2 Ccal = 2 (mc T)H O - (mc T)H O 2 2 Includes both the mass and heat capacity ( T)cal Thermochemistry

(b) Calculating the Heats of Reactions ( H) HCl(aq) + Na. OH(aq) Na. Cl(aq) + H 2 O(l) 50 m. L@25°C Heat of reaction + 50 m. L@25°C = heat gained by Na. Cl solution 100 m. L of Na. Cl(aq) @ 30°C + heat gained by calorimeter H(qrxn) = q. Na. Cl soln + qcal qrxn = (mc* T)Na. Cl + (Ccal T) qrxn = - Hrxn soln Thermochemistry = heat of neutralization rxn * How do you determine the c. Na. Cl soln?

How do you determine the c Na. Cl soln ? 50 m. L of H 2 O @ 650 C are added……. 50 m. L H 2 O 50 m. L Na. Cl soln The final temperature of the 100 m. L of solution inside the calorimeter is 410 C …. . to 50 m. L of Na. Cl solution @ 250 C inside the calorimeter Heat lost by = heat gained by + heat gained by hot water salt water soln calorimeter qh = q + qcal (mc T)H 2 O = (mc* T)Na. Cl + (Ccal T) Na. Clsoln Thermochemistry

(2) Calorimetry at Constant Volume (Bomb Cal. ) E = q + w w = P V V =0 w=0 E = qv q = m c T Thermochemistry 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. E = qv • For most reactions, the difference is very small. H = E + n. RT Thermochemistry

Review of Thermochemical Equations E = q + w E = qp - P V E = qp - n. RT since ∆H= qp 1 atm E = H - P V H = E + P V E = q + w E = qv Thermochemistry

Comparing H and E H = E + P V 2 C 8 H 18 (l) + 25 O 2 (g) 16 CO 2(g) + 18 H 2 O(g) H = - 5. 5 x 106 k. J/ E = H - P V E = H - n. RT (useful at const P) (useful at const V) Problem: calculate E @ 25°C for above equation @ const. V R= 8. 314 J/K. gas=34 -25 = +9 = 4. 5 C 8 H 18 Thermochemistry

Constant-Volume Calorimetry Problem: When one mole of CH 4 (g) was combusted at 250 C in a bomb calorimeter, 886 k. J of energy were released. What is the enthalpy change for this reaction? CH 4 (g) + 2 O 2 (g) H = E + P V = E + n. RT CO 2(g) + 2 H 2 O(l) E = qv = - 886 k. J R = 8. 31 J/ . K n = 1 – (1+2) = -2 Thermochemistry

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

Methods of determining H 1. Calorimetry (experimental) 2. Hess’s Law: using Standard ) of Enthalpy of Reaction ( Hrxn a series of reaction steps (indirect method). 3. Standard Enthalpy of Formation ( Hf ) used with Hess’s Law (direct method) 4. Bond Energies used with Hess’s Law Experimental data combined with theoretical concepts Thermochemistry

(2) Determination of H using Hess’s Law Ø Hrxn is well known for many reactions, but it is inconvenient to measure Hrxn for every reaction. Ø However, we can estimate Hrxn for a reaction of interest by using Hrxn values that are published for other more common reactions. ) of a Ø The Standard Enthalpy of Reaction ( Hrxn series of reaction steps are added to lead to reaction of interest (indirect method). Standard conditions (25°C and 1. 00 atm pressure). Thermochemistry (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 Thermochemistry

Calculation of H by Hess’s Law 3 C(graphite) + 4 H 2 (g) C 3 H 8 (g) H= -104 C 3 H 8 (g) 3 C(graphite) + 4 H 2 (g) H= +104 3 C(graphite) + 3 O 2 (g) 3 CO 2 (g) H=-1181 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) H=-1143 C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (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. Thermochemistry • Elimination of common terms that appear on both sides of the equation.

Calculation of H by Hess’s Law C 3 H 8 (g) 3 C(graphite) + 4 H 2 (g) H= +104 3 C(graphite) + 3 O 2 (g) 3 CO 2 (g) H=-1181 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) H=-1143 C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) Hrxn = + 104 k. J -1181 k. J - 1143 k. J - 2220 k. J Thermochemistry

Calculation of H by Hess’s Law Calculate heat of reaction W + C (graphite) WC (s) Given data: 2 W(s) + 3 O 2 (g) 2 WO 3 (s) ½(2 W(s) + 3 O 2 (g) 2 WO 3 (s) ) W(s) + 3/2 O 2 (g) WO 3 (s) ) ΔH = ? ΔH = -1680. 6 k. J ½(ΔH = -1680. 6 k. J) ΔH = -840. 3 k. J C (graphite) + O 2 (g) CO 2 (g) ΔH = -393. 5 k. J 2 WC (s) + 5 O 2 (g) 2 WO 3 (s) + CO 2 (g) ΔH = -2391. 6 k. J ½(2 WO 3 (s) + CO 2 (g) 2 WC (s) + 5 O 2 (g)) WO 3 (s) + CO 2 (g) WC (s) + 5/2 O 2 (g) W + C (graphite) WC (s) ½ (ΔH = +2391. 6 k. J) ΔH = + 1195. 8 k. J) ΔH = - 38. 0 Thermochemistry

Hess’s Law Problem: Chloroform, CHCl 3, is formed by the following reaction: Desired ΔHrxn equation: CH 4 (g) + 3 Cl 2 (g) → 3 HCl (g) + CHCl 3 (g) Determine the enthalpy change for this reaction (ΔH°rxn), using the following: 2 C (graphite) + H 2 (g) + 3 Cl 2 (g) → 2 CHCl 3 (g)ΔH°f = – 103. 1 k. J/mol CH 4 (g) + 2 O 2 (g) → 2 H 2 O (l) + CO 2 (g) ΔH°rxn = – 890. 4 k. J/mol 2 HCl (g) → H 2 (g) + Cl 2 (g) ΔH°rxn = + 184. 6 k. J/mol C (graphite) + O 2 (g) → CO 2(g) ΔH°rxn = – 393. 5 k. J/mol H 2 (g) + ½ O 2 (g) → H 2 O (l) ΔH°rxn = – 285. 8 k. J/mol answers: a) – 103. 1 k. J d) + 305. 2 k. J b) + 145. 4 k. J e) – 305. 2 k. J c) – 145. 4 k. J Thermochemistry f) +103. 1 k. J

Methods of determining H 1. Calorimetry (experimental) 2. Hess’s Law: using Standard ) of Enthalpy of Reaction ( Hrxn a series of reaction steps (indirect method). 3. Standard Enthalpy of Formation ( Hf ) used with Hess’s Law (direct method) 4. Bond Energies used with Hess’s Law Experimental data combined with theoretical concepts Thermochemistry

(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 + O 2 CO 2 ∆Hf = -393. 5 k. J/ Standard Enthalpy of formation Hf are measured under standard conditions (25°C and 1. 00 atm pressure). Thermochemistry

Calculation of H CH 4(g) + O 2(g) CO 2(g) + H 2 O(g) C + 2 H 2(g) CH 4(g) C(g) + O 2(g) CO 2(g) 2 H 2(g) + O 2(g) 2 H 2 O(g) ΔHf = -74. 8 k. J/ŋ ΔHf = -393. 5 k. J/ŋ ΔHf = -241. 8 k. J/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 CO 2(g) + n H 2 O(g) H - n CH 4(g) + n O 2(g) = [1(-393. 5 k. J) + 1(-241. 8 k. J)] - [1(-74. 8 k. J) + 1(-0 k. J)] = - 560. 5 k. J Thermochemistry

Calculation of H C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) H = n Hf(products) - m Hf(reactants) H = [3(-393. 5 k. J) + 4(-285. 8 k. J)] - [1(-103. 85 k. J) + 5(0 k. J)] = [(-1180. 5 k. J) + (-1143. 2 k. J)] - [(-103. 85 k. J) + (0 k. J)] = (-2323. 7 k. J) - (-103. 85 k. J) = -2219. 9 k. J Table of Standard Enthalpy of formation, Hf Thermochemistry

(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 k. J/mol. Thermochemistry

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, CH 4, will be a bit different than the C—H bond in chloroform, CHCl 3. Thermochemistry

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) Thermochemistry

Hess’s Law: Hrxn = (bonds broken) (bonds formed) CH 4(g) + Cl 2(g) CH 3 Cl(g) + HCl(g) Hrxn = [D(C—H) + D(Cl—Cl) [D(C—Cl) + D(H—Cl) = [(413 k. J) + (242 k. J)] [(328 k. J) + (431 k. J)] = (655 k. J) (759 k. J) = 104 k. J Thermochemistry

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 Thermochemistry increases, the bond length decreases.

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