Thermochemistry Lecture Notes PDF

Thermochemistry Thermochemistry is the study of heat change in chemical reactions. Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) +...

Thermochemistry Thermochemistry is the study of heat change in chemical reactions. Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO (s) 2Hg (l) + O2 (g) energy + H2O (s) H2O (l) 1 Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. Changes in Enthalpy (ΔH) DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 2  Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = -6.01 kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) 2H2O (l) DH = 2 x 6.01 = 12.0 kJ 3  Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ H2O (l) H2O (g) DH = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ 1 mol P4 3013 kJ 266 g P4 x x = 6470 kJ 123.9 g P4 1 mol P4 4 The specific heat (Cm) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius (1K). 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 (1K). C = mCm Heat (qp) absorbed or released: qp = CpDT qp = mCmDT DT= Tfinal - Tinitial 5  How much heat is given off when an 869 g iron bar cools from 940C to 50C? Cm of Fe = 0.444 J/g 0C DT= Tfinal – Tinitial = 50C – 940C = -890C qp = mCmDT = 869 g x 0.444 J/g 0C x –890C = -34,000 J The molar heat capacity of a substance is the amount of heat (q) required to raise the temperature of one mole of the substance by one degree Celsius (1K). 6 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 bar = 100Kpa ≈1 atm pressure and temperature of interest usually 25°C substance in a solution with concentration 1 M the standard enthalpy change, DH°, is the enthalpy change when all reactants and products are in their standard states the standard enthalpy of formation, DHf°, 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 DHf° for a pure element in its standard state = 0 kJ/mol by definition DH0 (C, diamond) = 1.90 kJ/mol What is ΔHfo of O2 (g), Hg(l), C(graphite)? 7  Methods of determining DH 1. Standard Enthalpy of Formation (DHf ) (direct method) Experimental data 2. Hess’s Law: using Standard Enthalpy of combined with  ) of a series of reaction steps Reaction (DHrxn theoretical concepts (indirect method) 3. Bond Energies 4. Calorimetry (experimental) 8  1- Direct method: by standard heat of formation DH°reaction = S n DHf°(products) - S n DHf°(reactants) S means sum n is the coefficient of the reaction Example: Calculate the enthalpy of the following reaction: 2Al (s) + Fe2O3 (s) Al2O3 (s) + 2Fe(l) ΔHfo of Fe2O3, Al2O3 and Fe(l) = - 822.2, - 1669.8 and 12.40 kJ/mol ΔHo = ∑ n ΔHfo (products) - ∑ n ΔHfo (reactants) ΔHo = [(ΔHfo (Al2O3))+ (2× ΔHfo (Fe))]-[(2×ΔHfo (Al))+(ΔHfo (Fe2O3))] ΔHo = [ (-1669.8)+ (2×12.40)] – [2×(0)+(-822.2)] = -822.8 kJ 9  2- Indirect method :(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. DH is a state function (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) 10 Example – Hess’s Law Given the following information: 2 NO(g) + O2(g) 2 NO2(g) DH° = -173 kJ 2 N2(g) + 5 O2(g) + 2 H2O(l) 4 HNO3(aq) DH° = -255 kJ N2(g) + O2(g) 2 NO(g) DH° = +181 kJ Calculate the DH° for the reaction below: 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) DH° = ? [2 NO2(g) 2 NO(g) + O2(g)] x 1.5 DH° = 1.5(+173 kJ) [2 N2(g) + 5 O2(g) + 2 H2O(l) 4 HNO3(aq)] x 0.5 DH° = 0.5(-255 kJ) [2 NO(g) N2(g) + O2(g)] DH° = -181 kJ 11  [3 NO2(g) 3 NO(g) + 1.5 O2(g)] DH° = (+259.5 kJ) [ N2(g) + 2.5 O2(g) + H2O(l) 2 HNO3(aq)] DH° = (-128 kJ) [2 NO(g) N2(g) + O2(g)] DH° = -181 kJ 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) DH° = - 49 kJ 12