Exothermic process is any process that gives off heat transfers thermal energy from the system to the surroundings. H 2 O (l) + energy

Transcription

1 Exothermic process is any process that gives off heat transfers thermal energy from the system to the surroundings. H 2 O (g) H 2 O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + H 2 O (s) H 2 O (l) 6.2

2 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 H products < H reactants DH < 0 Exothermic process H products > H reactants DH > 0 Endothermic process 6.3

3 Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > kj are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. H 2 O (s) H 2 O (l) DH = 6.01 kj 6.3

4 Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < kj are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) DH = kj 6.3

5 Reversing Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance H 2 O (s) H 2 O (l) DH = 6.01 kj If you reverse a reaction, the sign of DH changes H 2 O (l) H 2 O (s) DH = kj If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H 2 O (s) 2H 2 O (l) DH = 2 x 6.01 = 12.0 kj 6.3

6 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 kj H 2 O (l) H 2 O (g) DH = 44.0 kj 6.3

7 Solving Thermochemical Equations How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s) DH = kj 1 mol P g P 4 x g P kj x = 6470 kj 1 mol P 4 6.3

8 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 = ms Heat (q) absorbed or released: q = msdt q = CDt Dt = t final - t initial 6.4

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10 How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = J/g 0 C Dt = t final t initial = 5 0 C 94 0 C = C q = msdt = 869 g x J x 89 0 C g 0 C = -34,000 J

11 Constant-Volume Calorimetry q sys = q water + q bomb + q rxn q sys = 0 q rxn = - (q water + q bomb ) q water = msdt q bomb = C bomb Dt Reaction at Constant V DH = q rxn No heat enters or leaves! DH ~ q rxn 6.4

12 Constant-Pressure Calorimetry q sys = q water + q cal + q rxn q sys = 0 q rxn = - (q water + q cal ) q water = msdt q cal = C cal Dt Reaction at Constant P DH = q rxn No heat enters or leaves! 6.4

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14 Because there is no way to measure the absolute value of the enthalpy of a substance, do I have to 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 (O 2 ) = 0 f DH 0 (O 3 ) = 142 kj/mol f DH 0 (C, graphite) = 0 f DH 0 (C, diamond) = 1.90 kj/mol f 6.5

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16 The standard enthalpy of reaction (DH 0 a reaction carried out at 1 atm. rxn ) is the enthalpy of aa + bb cc + dd DH 0 rxn = [ cdh 0 f (C) + ddh 0 (D)] - [ adh 0 f (A) + bdh 0 f (B)] f DH 0 rxn = SnDH 0 f (products) - SmDHf 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.) 6.5

17 Calculate the standard enthalpy of formation of CS 2 (l) given that: C(graphite) + O 2 (g) CO 2 (g) DH 0 = kj rxn S(rhombic) + O 2 (g) SO 2 (g) DH 0 = kj rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) DH 0 = kj rxn 1. Write the enthalpy of formation reaction for CS 2 C(graphite) + 2S(rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. C(graphite) + O 2 (g) CO 2 (g) DH 0 rxn = kj 2S(rhombic) + 2O 2 (g) 2SO 2 (g) DH 0 rxn = x2 kj + CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g) DH 0 = kj rxn C(graphite) + 2S(rhombic) CS 2 (l) DH 0 rxn = (2x-296.1) = 86.3 kj

18 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 kj/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) DH 0 rxn = SnDH 0 f (products) - SnDH 0 f (reactants) DH 0 rxn = [ 12DH 0 (CO 6DH 0 2 ) + (H 2 O) ] - [ 2DH 0 (C 6 H 6 )] f f f DH 0 rxn = [ 12x x ] [ 2x49.04 ] = kj kj 2 mol = kj/mol C 6 H 6 6.5

19 The enthalpy of solution (DH soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DH soln = H soln - H components Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? 6.6

20 The Solution Process for NaCl DH soln = Step 1 + Step 2 = = 4 kj/mol 6.6

21 Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 6.7

22 Thermodynamics 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 6.7

23 Enthalpy and the First Law of Thermodynamics DE = q + w At constant pressure, q = DH and w = -PDV DE = DH - PDV DH = DE + PDV 6.7