10/23/10. Thermodynamics and Kinetics. Chemical Hand Warmers
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1 10/23/10 CHAPTER 6 Thermochemistry 6-1 Chemical Hand Warmers Most hand warmers work by using the heat released from the slow oxidation of iron 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) The amount your hand temperature rises depends on several factors the size of the hand warmer the size of your glove, etc. mainly, the amount of heat released by the reaction 6-2 Thermodynamics and Kinetics SiO2 --x--> Si + O2 Diamond ---> graphite Rusting is slow 6-3 1
2 Chemistry & Energy 6-4 Chemistry and Energy H 2 /O 2 Fuel Cell 6-5 Thermodynamics is the study of heat and its transformations. Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes. Fundamental premise When energy is transferred from one object to another, it appears as work and/or as heat. For our work we must define a system to study; everything else then becomes the surroundings. The system is composed of particles with their own internal energies (E or U). Therefore the system has an internal energy. When a change occurs, the internal energy changes
3 A chemical system and its surroundings. the surroundings the system 6-7 Figure 6.1 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. E tot = E pot + E kinetic ΔE = E final - E initial = E products - E reactants 6-8 Figure 6.2 Energy and Work 6-9 3
4 Forms of Energy Electrical kinetic energy associated with the flow of electrical charge Heat or thermal energy kinetic energy associated with molecular motion Light or radiant energy kinetic energy associated with energy transitions in an atom Nuclear potential energy in the nucleus of atoms Chemical potential energy due to the structure of the atoms, the attachment between atoms, the atoms positions relative to each other in the molecule, or the molecules, relative positions in the structure Manifestations of Energy Example of Work Heat transfers from surroundings to system in endothermic process
5 Molecular Picture of Process CO 2 (s, -78 o C) ---> CO 2 (g, -78 o C) A regular array of molecules in a solid -----> gas phase molecules. Gas molecules have higher kinetic energy Table 6.1 The Sign Conventions* for q, w and ΔE q + w = ΔE depends on sizes of q and w depends on sizes of q and w - * For q: + means system gains heat; - means system loses heat. * For w: + means word done on system; - means work done by system Units of Energy The amount of kinetic energy an object has is directly proportional to its mass and velocity KE = ½mv
6 Units of Energy 6-16 State Functions Two different paths for the energy change of a system Figure 6.6 State Function You can travel either of two trails to reach the top of the mountain. One is long and winding, the other is shorter but steep. Regardless of which trail you take, when you reach the top you will be 10,000 ft above the base. The distance from the base to the peak of the mountain is a state function. It depends only on the difference in elevation between the base and the peak, not on how you arrive there!
7 Pressure-volume work. Figure Specific Heat Capacity and Measurement of Heats of Reactions Specific heat capacity = heat lost or gained by substance (J) (mass, g)(t change, K) q = C spc x mass x ΔT 6-20 Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials (at 298 K) Substance Specific Heat Capacity (J/g*K) Substance Specific Heat Capacity (J/g*K) Elements aluminum, Al graphite,c iron, Fe copper, Cu gold, Au Solid materials wood cement glass granite steel Compounds water, H 2 O(l) ethyl alcohol, C 2 H 5 OH(l) ethylene glycol, (CH 2 OH) 2 (l) carbon tetrachloride, CCl 4 (l)
8 Finding the Quantity of Heat from Specific Heat Capacity PROBLEM: q = C spc x mass x ΔT 6-22 Coffee-cup calorimeter. -q sys = -q H2O -(C s x mass s x ΔT s = -(C H2O x mass H2O x ΔT H2O ) C s = C H2O x mass H2O x ΔT H2O mass s x ΔT H2O 6-23 Figure 6.9 Enthalpy w
9 The Meaning of Enthalpy w = - PΔV H = E + PV where H is enthalpy ΔH = ΔE + PΔV q p = ΔE + PΔV = ΔH ΔH ΔE in 1. Reactions that do not involve gases. 2. Reactions in which the number of moles of gas does not change. 3. Reactions in which the number of moles of gas does change but q is >>> PΔV Enthalpy diagrams for exothermic and endothermic processes. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H 2 O(s) H 2 O(l) CH 4 + 2O 2 H initial H 2 O(l) H final Enthalpy, H Enthalpy, H ΔH < 0 heat out ΔH > 0 heat in CO 2 + 2H 2 O H final H 2 O(s) H initial A Exothermic process B Endothermic process 6-26 Figure 6.8 Some Important Types of Enthalpy Change heat of combustion (ΔH comb ) C 4 H 10 (l) + 13/2O 2 (g) 4CO 2 (g) + 5H 2 O(g) heat of formation (ΔH f ) K(s) + 1/2Br 2 (l) KBr(s) heat of fusion (ΔH fus ) NaCl(s) NaCl(l) heat of vaporization (ΔH vap ) C 6 H 6 (l) C 6 H 6 (g)
10 Enthalpy of Reaction The enthalpy change in a chemical reaction is an extensive property the more reactants you use, the larger the enthalpy change By convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) ΔH = 2044 kj Relationships Involving ΔH rxn When reaction is multiplied by a factor, ΔH rxn is multiplied by that factor because ΔH rxn is extensive C(s) + O 2 (g) CO 2 (g) ΔH = kj 2 C(s) + 2 O 2 (g) 2 CO 2 (g) ΔH = 2( kj) = kj If a reaction is reversed, then the sign of ΔH is changed CO 2 (g) C(s) + O 2 (g) ΔH = kj Sample Problem 6.4 Determining the Heat of a Reaction PROBLEM: 50.0 ml of M NaOH is placed in a coffee-cup calorimeter at o C and carefully add 25.0 ml of M HCl, also at o C. After stirring, the final temperature is o C. Calculate q soln (in J) and ΔH rxn (in kj/mol). d = 1.00 g/ml and c = J/g*K PLAN: SOLUTION: Use volumes and density to find the mass of the water. Calculate the moles of NaOH and HCl. Write balanced equations and determine the limiting reactant. Calculate q soln using the mass, c, and ΔT and finally dividing by of water gives the heat per mole. Mass( ) ml x 1.00 g/ml= 75.0 g of water ΔT = ( ) o C = 2.21 o C = 2.21 K HCl(aq) + NaOH(aq) H + (aq) + OH - (aq) H 2 O(l) NaCl(aq) + H 2 O(l) NaOH M x L = mol OH - HCl M x L = mol H + HCl is the limiting reactant, so mol H 2 O q = mass x specific heat x ΔT = 75.0 g x J/g*K x 2.21 K = 693 J (693 J/ mol H 2 O)(kJ/10 3 J) = 55.4 kj/ mol H 2 O formed 10
11 Heating/Cooling Curve for Water" Isothermal Titration Calorimetry
12 Figure 6.11 Summary of the relationship between amount (mol) of substance and the heat (kj) transferred during a reaction Sample Problem 6.6 Using the Heat of Reaction (ΔH rxn ) to Find Amounts PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by Al 2 O 3 (s) 2Al(s) + 3/2O 2 (g) ΔH rxn = 1676 kj If aluminum is produced this way, how many grams of aluminum can form when 1.000x10 3 kj of heat is transferred? PLAN: heat (kj) 1676 kj = 2 mol Al mol of Al x M g of Al SOLUTION: 1.000x10 3 kj x 2 mol Al 1676 kj g Al 1 mol Al = g Al 6-35 Hess's Law The enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps. 1/2 O 2 (g) + C(s) CO(g) CO(g) + 1/2 O 2 (g) CO 2 (g) O 2 (g) + C(s) CO 2 (g) net reaction
13 Hess's Law S(s) + 3/2 O 2 (g) ΔH 3 = kj ΔH 1 = kj S(s) + O 2 (g) SO 2 (g) 2SO 2 (g) + O 2 (g) 1/2 (ΔH 2 = kj ) 1/2 ( 2SO 2 (g) + O 2 (g) 2SO 3 (g) ) SO 3 (g) 6-37 S(s) + 3/2 O 2 (g) SO 3 (g) Practice Hess s Law Given the following information: Cu(s) + Cl 2 (g) CuCl 2 (s) ΔH = 206 kj 2 Cu(s) + Cl 2 (g) 2 CuCl(s) ΔH = 36 kj Calculate the ΔH for the reaction below: Cu(s) + CuCl 2 (s) 2 CuCl(s) ΔH =? kj Sample Problem 6.7 Using Hess s Law to Calculate an Unknown ΔH PROBLEM: PLAN: SOLUTION: 6-39 Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation: CO(g) + NO(g) CO 2 (g) + 1/2N 2 (g) ΔH =? Given the following information, calculate the unknown ΔH: Equation A: CO(g) + 1/2O 2 (g) Equation B: N 2 (g) + O 2 (g) CO 2 (g) ΔH A = kj 2NO(g) ΔH B = kj Equations A and B have to be manipulated by reversal and/or multiplication by factors in order to sum to the first, or target, equation. Multiply Equation B by 1/2 and reverse it. CO(g) + 1/2O 2 (g) NO(g) CO(g) + NO(g) CO 2 (g) 1/2N 2 (g) + 1/2O 2 (g) CO 2 (g) + 1/2N 2 (g) ΔH A = kj ½ ΔH B = ½ ( kj) = kj ΔH rxn = kj 13
14 Standard Conditions The standard state is the state of a material at a defined set of conditions pure gas at exactly 1 atm pressure pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest usually 25 C substance in a solution with concentration 1 M The standard enthalpy change, ΔH, is the enthalpy change when all reactants and products are in their standard states The standard enthalpy of formation, ΔH f, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements the elements must be in their standard states the ΔH f for a pure element in its standard state = 0 kj/mol by definition Table 6.3 Selected Standard Heats of Formation at 25 o C (298K) 6-41 Formula Calcium Ca(s) CaO(s) CaCO 3 (s) Carbon C(graphite) C(diamond) CO(g) CO 2 (g) CH 4 (g) CH 3 OH(l) HCN(g) CS s (l) Chlorine Cl(g) ΔH 0 f (kj/mol) Formula Cl 2 (g) HCl(g) Hydrogen H(g) H 2 (g) Nitrogen N 2 (g) NH 3 (g) NO(g) Oxygen O 2 (g) O 3 (g) H 2 O(g) H 2 O(l) ΔH 0 f (kj/mol) Formula ΔH 0 f (kj/mol) Silver Ag(s) AgCl(s) Sodium Na(s) Na(g) NaCl(s) Sulfur S 8 (rhombic) 0 S 8 (monoclinic) 0.3 SO 2 (g) SO 3 (g) Formation Reactions Reactions of elements in their standard state to form 1 mole of a pure compound if you are not sure what the standard state of an element is, find the form in Appendix IIB that has a ΔH f = 0 because the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions
15 Writing Formation Reactions Write the Formation Reaction for CO(g) The formation reaction is the reaction between the elements in the compound, which are C and O C + O CO(g) The elements must be in their standard state there are several forms of solid C, but the one with ΔH f = 0 is graphite oxygen s standard state is the diatomic gas C(s, graphite) + O 2 (g) CO(g) The equation must be balanced, but the coefficient of the product compound must be 1 use whatever coefficient in front of the reactants is necessary to make the atoms on both sides equal without changing the product coefficient C(s, graphite) + 43 ½ O 2 (g) CO(g) 6-43 Sample Problem 6.8 Writing Formation Equations PROBLEM: PLAN: SOLUTION: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include ΔH 0 f. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. Use the table of heats of formation for values. (a) Ag(s) + 1/2Cl 2 (g) AgCl(s) (b) Ca(s) + C(graphite) + 3/2O 2 (g) (c) 1/2H 2 (g) + C(graphite) + 1/2N 2 (g) CaCO 3 (s) HCN(g) ΔH o f = kj ΔH o f = kj ΔH o f = 135 kj 6-44 Calculating Standard Enthalpy Change for a Reaction Any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products The ΔH for the reaction is then the sum of the ΔH f for the component reactions ΔH reaction = Σ n ΔH f (products) Σ n ΔH f (reactants) Σ means sum n is the coefficient of the reaction
16 Example: Calculate the enthalpy change in the reaction 2 C 2 H 2 (g) + 5 O 2 (g) 4 CO 2 (g) + 2 H 2 O(l) 1. Write formation reactions for each compound and determine the ΔH f for each 2 C(s, gr) + H 2 (g) C 2 H 2 (g) ΔH f = kj/mol C(s, gr) + O 2 (g) CO 2 (g) ΔH f = kj/mol H 2 (g) + ½ O 2 (g) H 2 O(l) ΔH f = kj/mol Example of ΔH o rxn from ΔH o f TiCl 4 (l) + 2H 2 O(g) TiO 2 (s) + 4HCl(g) TiCl4(l) Ti(s) + 2Cl 2 (g) -ΔH o f[ticl 4 (l)] 2H2O(g) 2H2(g) + O2(g) -2ΔH o f[h 2 O(g)] Ti(s) + O2(g) TiO 2 (s) ΔH o f[ti2o2(s)] 2H2(g) + 2Cl2(g) 4HCl(g) 4ΔH o f[hcl(g)] TiCl 4 (l) + 2H 2 O(g) TiO 2 (s) + 4HCl(g) net reaction 6-47 Figure 6.12 The general process for determining ΔH o rxn from ΔHo f values. Elements Enthalpy, H Reactants decomposition -ΔH o f formation ΔH o f H initial ΔH o rxn Products H final ΔH o rxn = Σ mδho f(products) - Σ nδho f(reactants)
17 Sample Problem 6.9 Formation PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: 4NH 3 (g) + 5O 2 (g) Calculating the Heat of Reaction from Heats of Calculate ΔH o rxn from ΔHo f values. 4NO(g) + 6H 2 O(g) PLAN: Look up the ΔH o f values and use Hess s Law to find ΔH rxn. SOLUTION: ΔH rxn = Σ mδh o f (products) - Σ nδho f (reactants) ΔH rxn = [4(ΔH o f NO(g) + 6(ΔHo f H 2 O(g)] - [4(ΔHo f NH 3 (g) + 5(ΔHo f O 2 (g)] = (4 mol)(90.3 kj/mol) + (6 mol)( kj/mol) - [(4 mol)(-45.9 kj/mol) + (5 mol)(0 kj/mol)] ΔH rxn = -906 kj
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