Thermochemistry Chapter 6 Copyright The Mc GrawHill Companies

Thermochemistry Chapter 6 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

Energy is the capacity to do work • Radiant energy comes from the sun and is earth’s primary energy source • 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 available by virtue of an object’s position 6. 1

Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of thermal energy. Temperature = Thermal Energy 900 C 400 C greater thermal energy 6. 2

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. open Exchange: mass & energy closed isolated energy nothing 6. 2

Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2 H 2 (g) + O 2 (g) H 2 O (g) 2 H 2 O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + 2 Hg. O (s) energy + H 2 O (s) 2 Hg (l) + O 2 (g) H 2 O (l) 6. 2

Exothermic Endothermic 6. 2

Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, pressure, volume, temperature DE = Efinal - Einitial 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. 6. 3

First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. DEsystem + DEsurroundings = 0 or DEsystem = -DEsurroundings C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings 6. 3

Another form of the first law for DEsystem 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 = -PDV when a gas expands against a constant external pressure 6. 3

Work Done On the System w=Fxd w = -P DV Px. V= F 3 = F x d = w x d d 2 DV > 0 -PDV < 0 wsys < 0 Work is not a state function! Dw = wfinal - winitial final 6. 3

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? w = -P DV (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 101. 3 J = -1430 J w = -14. 1 L • atm x 1 L • atm 6. 3

Chemistry in Action: Making Snow DE = q + w q=0 w < 0, DE < 0 D E = CD T DT < 0, SNOW!

Enthalpy and the First Law of Thermodynamics DE = q + w At constant pressure: q = DH and w = -PDV DE = DH - PDV DH = DE + PDV 6. 4

Enthalpy (H) 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) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants DH < 0 Hproducts > Hreactants DH > 0 6. 4

Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 6. 01 k. J are absorbed for every 1 mole of ice that melts at 00 C and 1 atm. H 2 O (s) H 2 O (l) DH = 6. 01 k. J 6. 4

Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 890. 4 k. J are released for every 1 mole of methane that is combusted at 250 C and 1 atm. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) DH = -890. 4 k. J 6. 4

Thermochemical Equations • The stoichiometric coefficients always refer to the number of moles of a substance H 2 O (s) • DH = 6. 01 k. J If you reverse a reaction, the sign of DH changes H 2 O (l) • H 2 O (l) H 2 O (s) DH = -6. 01 k. J If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2 H 2 O (s) 2 H 2 O (l) DH = 2 x 6. 01 = 12. 0 k. J 6. 4

Thermochemical Equations • The physical states of all reactants and products must be specified in thermochemical equations. H 2 O (s) H 2 O (l) DH = 6. 01 k. J H 2 O (l) H 2 O (g) DH = 44. 0 k. J How much heat is evolved when 266 g of white phosphorus (P 4) burn in air? P 4 (s) + 5 O 2 (g) 266 g P 4 x P 4 O 10 (s) 1 mol P 4 123. 9 g P 4 x DH = -3013 k. J = 6470 k. J 1 mol P 4 6. 4

A Comparison of DH and DE 2 Na (s) + 2 H 2 O (l) DE = DH - PDV 2 Na. OH (aq) + H 2 (g) DH = -367. 5 k. J/mol At 25 0 C, 1 mole H 2 = 24. 5 L at 1 atm PDV = 1 atm x 24. 5 L = 2. 5 k. J DE = -367. 5 k. J/mol – 2. 5 k. J/mol = -370. 0 k. J/mol 6. 4

The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C=mxs Heat (q) absorbed or released: q = m x s x Dt q = C x Dt Dt = tfinal - tinitial 6. 5

How much heat is given off when an 869 g iron bar cools from 940 C to 50 C? s of Fe = 0. 444 J/g • 0 C Dt = tfinal – tinitial = 50 C – 940 C = -890 C q = ms. Dt = 869 g x 0. 444 J/g • 0 C x – 890 C = -34, 000 J 6. 5

Constant-Volume Calorimetry qsys = qwater + qbomb + qrxn qsys = 0 qrxn = - (qwater + qbomb) qwater = m x s x Dt qbomb = Cbomb x Dt Reaction at Constant V DH = qrxn No heat enters or leaves! DH ~ qrxn 6. 5

Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = m x s x Dt qcal = Ccal x Dt Reaction at Constant P DH = qrxn No heat enters or leaves! 6. 5

6. 5

Chemistry in Action: Fuel Values of Foods and Other Substances 6 CO 2 (g) + 6 H 2 O (l) DH = -2801 k. J/mol C 6 H 12 O 6 (s) + 6 O 2 (g) 1 cal = 4. 184 J 1 Cal = 1000 cal = 4184 J Substance DHcombustion (k. J/g) Apple -2 Beef -8 Beer -1. 5 Gasoline -34

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 (DH 0 f ) as a reference point for all enthalpy expressions. Standard enthalpy of formation (DH 0 f) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. The standard enthalpy of formation of any element in its most stable form is zero. DH 0 f (O 2) = 0 DH 0 f (C, graphite) = 0 DH 0 f (O 3) = 142 k. J/mol DH 0 f (C, diamond) = 1. 90 k. J/mol 6. 6

6. 6

0 ) is the enthalpy of The standard enthalpy of reaction (DHrxn a reaction carried out at 1 atm. a. A + b. B c. C + d. D DH 0 rxn = [ c. DH 0 f (C) + d. DH 0 f (D) ] - [ a. DH 0 f (A) + b. DH 0 f (B) ] DH 0 rxn = S n. DH 0 f (products) - S m. DHf 0 (reactants) 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. ) 6. 6

C (graphite) + 1/2 O 2 (g) CO (g) + 1/2 O 2 (g) C (graphite) + O 2 (g) CO 2 (g) 6. 6

Calculate the standard enthalpy of formation of CS 2 (l) given that: 0 = -393. 5 k. J C(graphite) + O 2 (g) CO 2 (g) DHrxn S(rhombic) + O 2 (g) CS 2(l) + 3 O 2 (g) SO 2 (g) 0 = -296. 1 k. J DHrxn CO 2 (g) + 2 SO 2 (g) 0 = -1072 k. J DHrxn 1. Write the enthalpy of formation reaction for CS 2 C(graphite) + 2 S(rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. C(graphite) + O 2 (g) 2 S(rhombic) + 2 O 2 (g) + CO 2(g) + 2 SO 2 (g) 0 = -393. 5 k. J CO 2 (g) DHrxn 0 = -296. 1 x 2 k. J 2 SO 2 (g) DHrxn CS 2 (l) + 3 O 2 (g) 0 = +1072 k. J DHrxn C(graphite) + 2 S(rhombic) CS 2 (l) 0 = -393. 5 + (2 x-296. 1) + 1072 = 86. 3 k. J DH rxn 6. 6

Benzene (C 6 H 6) 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 k. J/mol. 2 C 6 H 6 (l) + 15 O 2 (g) 12 CO 2 (g) + 6 H 2 O (l) DH 0 rxn = S n. DH 0 f (products) - S m. DHf 0 (reactants) DH 0 rxn = [ 12 DH 0 f (CO 2) + 6 DH 0 f (H 2 O)] - [ 2 DH 0 f (C 6 H 6)] DH 0 rxn = [ 12 x– 393. 5 + 6 x– 187. 6 ] – [ 2 x 49. 04 ] = -5946 k. J = - 2973 k. J/mol C 6 H 6 2 mol 6. 6

Chemistry in Action: Bombardier Beetle Defense C 6 H 4(OH)2 (aq) + H 2 O 2 (aq) C 6 H 4 O 2 (aq) + H 2 (g) DH 0 = 177 k. J/mol C 6 H 4(OH)2 (aq) H 2 O 2 (aq) C 6 H 4 O 2 (aq) + 2 H 2 O (l) DH 0 = ? H 2 O (l) + ½O 2 (g) DH 0 = -94. 6 k. J/mol H 2 (g) + ½ O 2 (g) H 2 O (l) DH 0 = -286 k. J/mol DH 0 = 177 - 94. 6 – 286 = -204 k. J/mol Exothermic!

The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? 6. 7

The Solution Process for Na. Cl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 k. J/mol 6. 7