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Thermochemistry

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THERMOCHEMISTRY THERMOCHEMISTRY, is the study of the heat released or absorbed by chemical and physical changes. 1N = 1Kg.m/s 2, 1J = 1Nm =1Kg.m 2 /s 2 1cal = 4.184 J,

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Temperature, T & Heat, q Temperature is defined as the degree of hotness Conversion of a temperature from Fahernheit scale(t F ) to the Celsius scale (t C ) Conversion of a temperature from Fahernheit scale(t F ) to the Celsius scale (t C ) t C = (5/9)(t F - 32) t C = (5/9)(t F - 32)

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

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system The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothingSYSTEM SURROUNDINGS 6.2

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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 = mass x specific heat = ms Heat (q) absorbed or released: q = ms t q = C t t = t final - t initial Specific heats of some common substances

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Heat capacity (C) for water is 4.184 J/g.C, heat capacity for 500 g of water is C = 500 x 4.184 = 2090 J / C = 2.09 KJ/C This sample absorbs 2.09 KJ of heat for each degree that temperature increase. By raising temperature from 20 to 25C, the heat absorbed by the sample q is, q = 2.09 (25 - 20) = 10.4 KJ Calorimeter is used to measure the heat changes accompanying the chemical reactions

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Measuring H in the lab by bomb calorimetry

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Heats of Some Typical Reactions Measured at Constant Pressure

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in an open system at constant pressure, i.e. H is a measure of heat Note that the standard enthalpy, H f o, of formation for a pure element in its standard state is zero.

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H = H final - H initial heat in products reactants Heat released or absorbed or absorbed Enthalpy Change, H

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Find H for the reaction C (diamond) + O 2(g) CO 2(g) if T i = 20.00 o C, T f = 21.26 o C, for a 0.250 g sample of diamond, with m = 1560 g H 2 O q = mc T = 1560 g x 4.18 J o C -1 g -1 x (21.26 - 20.00) o C = 8.22 x 10 3 J = 8.22 kJ H per mole (12.011 g) = (8.22 kJ/0.25 g) x (12.011 g/mol) = -395 kJ/mol

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example... CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O (g) H = -890.3 kJ/mol reactants products + heat i.e. less H in products than reactants i.e. heat is released!!!

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

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H 2 O (s) H 2 O (l) H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. H 2 O (l) H 2 O (g) H = 44.0 kJ 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) H = -3013 kJ 266 g P 4 1 mol P 4 123.9 g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ

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How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = 0.444 J/g 0 C t = t final t initial = 5 0 C 94 0 C = -89 0 C q = ms t = 869 g x 0.444 J/g 0 C x 89 0 C= -34,000 J

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Standard enthalpy of formation ( H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. H 0 (O 2 ) = 0 f H 0 (O 3 ) = 142 kJ/mol f H 0 (C, graphite) = 0 f H 0 (C, diamond) = 1.90 kJ/mol f

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Standard Enthalpies of Formation of Some Inorganic Substances at 25 C

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The standard enthalpy of reaction ( H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn a A + b B c C + d D H0H0 rxn d H 0 (D) f c H 0 (C) f = [+] - b H 0 (B) f a H 0 (A) f [+] H0H0 rxn n H 0 (products) f = m H 0 (reactants) f - Hesss 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.

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Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g) H 0 = -393.5 kJ rxn S (rhombic) + O 2 (g) SO 2 (g) H 0 = -296.1 kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) H 0 = -1072 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. rxn C (graphite) + O 2 (g) CO 2 (g) H 0 = -393.5 kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g) H 0 = -296.1x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g) H 0 = +1072 kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l) H 0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn

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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 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n H 0 (products) f = m H 0 (reactants) f - H0H0 rxn 6 H 0 (H 2 O) f 12 H 0 (CO 2 ) f = [+] - 2 H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x393.5 + 6x187.6 ] [ 2x49.04 ] = -5946 kJ -5946 kJ 2 mol = - 2973 kJ/mol C 6 H 6

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The enthalpy of solution ( H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. H soln = H soln - H components Heats of Solution of Some Ionic Compounds

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Hesss law Hesss Law states that the heat of a whole reaction is equivalent to the sum of its steps. For example: C + O 2 CO 2 This reaction can occur as 2 steps C + O 2 CO H = 110.5 kJ CO + O 2 CO 2 H = 283.0 kJ C + CO + O 2 CO + CO 2 H = 393.5 kJ I.e. C + O 2 CO 2 H = 393.5 kJ Hesss law allows us to add equations. We add all reactants, products, & H values.

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Hesss law: We may need to manipulate equations further: 2Fe + 1.5O 2 Fe 2 O 3 H =?, given Fe 2 O 3 + 3CO 2Fe + 3CO 2 H = 26.74 kJ CO + O 2 CO 2 H = 282.96 kJ 1: Align equations based on reactants/products. 2: Multiply based on final reaction. 3: Add equations. 2Fe + 1.5O 2 Fe 2 O 3 3CO + 1.5 O 2 3CO 2 H = 848.88 kJ 2Fe + 3CO 2 Fe 2 O 3 + 3CO H = + 26.74 kJ CO + O 2 CO 2 H = 282.96 kJ H = 822.14 kJ