General Chemistry I

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1 Thermochemistry General Chemistry I Dr Rick Attrill Oice MHMK 1405/5 The Nature o Energy and Types o Energy Energy Changes in Chemical Reactions Introduction to Thermodynamics Enthalpy o Chemical Reactions Calorimetry Standard Enthalpy o Formation and Reaction Heat o Solution and Dilution 1 2 Energy is the capacity to do work Thermal energy is the energy associated with the random motion o atoms and molecules Chemical energy is the energy stored within the bonds o chemical substances Nuclear energy is the energy stored within the collection o neutrons and protons in the atom Energy Changes in Chemical Reactions Heat is the transer o thermal energy between two bodies that are at dierent temperatures. Temperature is a measure o the thermal energy. Temperature = Thermal Energy Electrical energy is the energy associated with the low o electrons Potential energy is the energy available by virtue o an object s position Law o conservation o energy the total quantity o energy in the universe is assumed constant C 40 0 C greater thermal energy 4

2 Thermochemistry is the study o heat change in chemical reactions. The system is the speciic part o the universe that is o interest in the study. SYSTEM Exothermic process is any process that gives o heat transers thermal energy rom the system to the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy SURROUNDINGS Endothermic process is any process in which heat has to be supplied to the system rom the surroundings. energy + 2HgO (s) 2Hg (l) + O 2 (g) open closed isolated energy + H 2 O (s) H 2 O (l) Exchange: mass & energy energy nothing 5 6 Enthalpy (H) is used to quantiy the heat low into or out o a system in a process that occurs at constant pressure. H = H (products) H (reactants) H = heat given o or absorbed during a reaction at constant pressure Thermodynamics Thermodynamics is the scientiic study o the interconversion o heat and other kinds o energy. In thermodynamics, we study changes in the state o a system, which is deined by the values o all relevant macroscopic properties, e.g. composition, energy, temperature, pressure and volume. State unctions (energy, pressure, volume, temperature) are properties that are determined by the state o the system, regardless o how that condition was achieved. H products < H reactants H < 0 H products > H reactants H > 0 7 Potential energy o hiker 1 and hiker 2 is the same even though they took dierent paths. 8

3 The First Law o Thermodynamics The irst law o thermodynamics states that energy can be converted rom one orm to another, but cannot be created or destroyed. The change in the internal energy o a system E is given by E = E -E i where E i and E are the internal energies o the system in the initial and inal states, respectively. Thermodynamics E = q + w E is the change in internal energy o a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -P V when a gas expands against a constant external pressure For a chemical reaction the change in energy content E is given by E = E(products) E(reactants) As energy is always conserved: E sys + E surr = Enthalpy and the First Law o Thermodynamics E = q + w At constant pressure, q = H and w = -P V E = H - P V H = E + P V Thermochemical Equations Is H negative or positive? System absorbs heat Endothermic H > kj are absorbed or every 1 mole o ice that melts at 0 0 C and 1 atm. 11 H 2 O (s) H 2 O (l) H = 6.01 kj 12

4 Thermochemical Equations Thermochemical Equations Is H negative or positive? System gives o heat Exothermic H < kj are released or every 1 mole o methane that is combusted at 25 0 C and 1 atm. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) H = kj 13 The stoichiometric coeicients always reer to the number o moles o a substance H 2 O (s) H 2 O (l) H = 6.01 kj I you reverse a reaction, the sign o H changes H 2 O (l) H 2 O (s) H = kj I you multiply both sides o the equation by a actor n, then H must change by the same actor n. 2H 2 O (s) 2H 2 O (l) H = 2 x 6.01 = 12.0 kj 14 Thermochemical Equations The physical states o all reactants and products must be speciied in thermochemical equations. H 2 O (s) H 2 O (l) H = 6.01 kj H 2 O (l) H 2 O (g) H = 44.0 kj How much heat is evolved when 266 g o white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s) H = kj 1 mol P g P 4 x g P 4 x 3013 kj 1 mol P 4 = 6470 kj The speciic heat (s) o a substance is the amount o heat (q) required to raise the temperature o one gram o the substance by one degree Celsius. The heat capacity (C) o a substance is the amount o heat (q) required to raise the temperature o a given quantity (m) o the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = ms t q = C t t = t inal - t initial 15 16

5 How much heat is given o when an 869 g iron bar cools rom 94 0 C to 5 0 C? Constant-Volume Calorimetry s o Fe = J/g 0 C t = t inal t initial = 5 0 C 94 0 C = C q = ms t = 869 g x J/g 0 C x 89 0 C = -34,000 J q sys = q water + q bomb + q q sys = 0 q = - (q water + q bomb ) q water = ms t q bomb = C bomb t Reaction at Constant V H = q 17 No heat enters or leaves! H ~ q 18 Constant-Pressure Calorimetry q sys = q water + q cal + q q sys = 0 q = - (q water + q cal ) q water = ms t q cal = C cal t Reaction at Constant P H = q No heat enters or leaves! 19 20

6 Because there is no way to measure the absolute value o the enthalpy o a substance, must I measure the enthalpy change or every reaction o interest? Establish an arbitrary scale with the standard enthalpy o ormation ( H 0 ) as a reerence point or all enthalpy expressions. Standard enthalpy o ormation ( H 0 ) is the heat change that results when one mole o a compound is ormed rom its elements at a pressure o 1 atm. The standard enthalpy o ormation o any element in its most stable orm is zero. H 0 (O 2 ) = 0 H 0 (O 3 ) = 142 kj/mol H 0 (C, graphite) = 0 H 0 (C, diamond) = 1.90 kj/mol The standard enthalpy o reaction ( H 0 ) is the enthalpy o a reaction carried out at 1 atm. H 0 aa + bb cc + dd = [ c H 0 (C) + d H 0 (D)] - [ a H 0 (A) + b H 0 (B)] H 0 = n H 0 (products) - m H 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 o steps. (Enthalpy is a state unction. It doesn t matter how you get there, only where you start and end.) 23 Calculate the standard enthalpy o ormation o CS 2 (l) given that: C(graphite) + O 2 (g) CO 2 (g) H 0 = kj S(rhombic) + O 2 (g) SO 2 (g) H 0 = kj CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) H 0 = kj 1. Write the enthalpy o ormation reaction or CS 2 C(graphite) + 2S(rhombic) CS 2 (l) 2. Add the given s so that the result is the desired. + C(graphite) + O 2 (g) CO 2 (g) H 0 = kj 2S(rhombic) + 2O 2 (g) 2SO 2 (g) H 0 = x2 kj CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g) H 0 = kj C(graphite) + 2S(rhombic) CS 2 (l) 24 H 0 = (2x-296.1) = 86.3 kj

7 Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole o benzene combusted? The standard enthalpy o ormation o benzene is kj/mol. The enthalpy o solution ( H soln ) is the heat generated or absorbed when a certain amount o solute dissolves in a certain amount o solvent. H soln = H soln - H components 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H 0 = n H 0 (products) - m H 0 (reactants) H 0 = [ 12 H 0 (CO 6 H 0 2 ) + (H 2 O) ] - [ 2 H 0 (C 6 H 6 )] Which substance(s) could be used or melting ice? H 0 = [ 12x x ] [ 2x49.04 ] = kj kj 2 mol = kj/mol C 6 H 6 Which substance(s) could be used or a cold pack? The Solution Process or NaCl H soln = Step 1 + Step 2 = = 4 kj/mol 27