Chapter 5. Thermochemistry

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1 Chapter 5. Thermochemistry THERMODYNAMICS - study of energy and its transformations Thermochemistry - study of energy changes associated with chemical reactions Energy - capacity to do work or to transfer heat Work- energy expended to move an object against a force (w = f x d) Heat - energy transferred from hotter to colder object; heat is associated with the motion of particles in a substance. 2 forms of energy 1) Kinetic Energy is the energy of motion: KE = 1/2 mv 2 2) Potential energy is the stored energy that object possesses due to its position relative to other objects or its composition. E.g. e- moving around nucleus has KE due to motion & PE due to attraction to protons in nucleus. ENERGY UNITS: SI unit is Joules: 1 J = 1 kg m 2 1 cal = J s 2 Calorie - energy required to raise the temperature of 1 g of water by 1 C. FIRST LAW OF THERMODYNAMICS is Law of Conservation of energy: Total Energy of a system and its surroundings is constant, but energy can be transferred between a system and its surroundings. E can also be converted from one form to another - heat, light, chemical, electrical, mechanical & nuclear. System - chemicals in a reaction Surroundings - container and environment around the system Internal Energy - total energy stored in a system; symbol is E. Sum of KE & PE for all particles (atoms, p, n, e-, etc.) in a system. We cannot measure absolute internal energy of a system because there are several interactions involved, but we can measure changes in energy. E = E final - E initial Mathematical form of 1st Law of Thermodynamics: E = q + w q = heat absorbed from surroundings or heat released to surroundings w = work done on or by system Most of the energy gained or lost is in the form of heat. EXOTHERMIC: Heat is released to surroundings; q < 0 ENDOTHERMIC: Heat is absorbed from surroundings; q > 0 Endothermic Processes melting, boiling decomposition of water Exothermic Processes freezing, condensation acid-base neutralization reactions combustion reactions

2 State function - property that defines a state & that depends only on initial and final conditions of system (temp, pressure, etc.); does not depend on path taken. Analogy: stock market profit doesn t depend on fluctuations in stock market; profit just depends on initial and final prices. H = Enthalpy: Heat transferred between a system and surroundings during a chemical reaction carried out at constant pressure ( P = 0) *Note: rxns carried out in open containers are carried out at constant P - atm P. H = q p this is heat gained or lost by system ENDOTHERMIC: q p > 0, H > 0, + value; EXOTHERMIC: q p < 0, H < 0, - value Similarities between H & E: H and E are state functions We can only measure changes in enthalpy ( H) Heat of Reaction: H = H(products) - H(reactants) enthalpy of reaction = heat content of products - heat content of reactants Each chemical reaction has an associated H E.g. 2H 2(g) + O 2(g) 2H 2 O (g) H = kj Thermochemical Rxn Can write heat as product in Exothermic reaction (reactant for Endothermic): 2H 2(g) + O 2(g) 2H 2 O (g) kj Enthalpy Diagram: 2H 2(g) + O 2(g) H = kj Exothermic 2H 2 O (g) In this reaction, reactants have a higher enthalpy than the products. Characteristics of Enthalpy 1) enthalpy is an extensive property - H depends on amount of reactant consumed or product formed. 2H 2(g) + O 2(g) 2H 2 O (g) H = kj kj are released to surroundings when 2 moles of water form. If 1 mole H 2 O forms, kj of heat evolve. E.g. Calculate the amount of heat evolved when 5.00 g of H 2 O 2 decompose. 2H 2 O 2(l) 2H 2 O 2(l) + O 2(g) H = -196 kj 1mol H 2O2 196 kj 5.00 g = KJ 34.0 g 2 mol H 2O2 2) If we write the reverse reaction, the sign of H changes. Exothermic: 2H 2(g) + O 2(g) 2H 2 O (g) H = kj Endothermic: 2H 2 O (g) 2H 2(g) + O 2(g) H = kj 3) The enthalpy depends on the physical state of the reactants and products. H 2 O (g) H 2 O (l) H = -44 kj

3 5.5 Calorimetry: Measurement of heat flow. Calorimeter: apparatus that measures heat flow ( temp). Constant pressure Calorimeter: Pressure of system doesn t change. E.g. coffee cup calorimeter Heat Capacity: amount of heat required to raise the temperature by 1 C (or 1 K). Units: J/ C (or J/K) Molar Heat Capacity: heat capacity of 1 mole of a substance. Units: J/mol C Specific Heat: amount of heat required to raise the temperature of 1 gram of a substance by 1 C(or 1 K). Units: J/g C (or J/g K) H 2 O has a high specific heat it takes a lot of heat to raise the T of water metals have low specific heats only takes a little heat to increase T Equation: q = m c T How do we find specific heat of a material? Design heat transfer process with substance that has known sp. heat. (e.g. H 2 O) Heat lost + Heat gained = 0 Heat lost = -Heat gained E.g. A g metal is heated to 78.4 C. The metal is placed in g H 2 O at 25.0 C. The final temperature of the mixture is C. What is the specific heat of the metal? For H 2 O, c = 4.18 J/g C. q metal = -q water m metal c metal T metal = - m H2O c H2O T H2O J c metal = mh OcH O T 2 2 H 2 O m metal T metal = g 4.18 g C g ( C) For solutions: q soln = m soln c soln T dilute solutions: use c soln = c H2O = 4.18 J/g K ( 8.43 C) = Since calorimeter prevents heat gain or loss to surroundings, heat of the reaction is transferred to the solution: H = q rxn = -q soln If temp increases ( T is +), Rxn is exothermic and q rxn is negative E.g. When a 4.25 g sample of ammonium nitrate dissolves in 60.0 grams of water, the temp. drops from 22.0 C to 16.9 C. Calculate H in kj/mol NH 4 NO 3 for the rxn: NH 4 NO 3 NH NO 3 - J g C Assume c = 4.18 J/g K H = - m soln c soln T m soln = 4.25 g g = g J 1kJ H = -(64.25 g)(4.18 )(16.9 C 22.0 C) = 1370 J = 1.37 kj g C 1000 J 1.37 kj 80.0 g NH 4 NO3 H = = 4.25 g NH 4NO3 mol NH 4 NO3 kj 25.8 mol

4 Hess' Law - If a reaction is carried out in a series of steps, then H for the overall process must be the sum of the H's for the individual steps. We can manipulate equations to arrive at desired eq. (multiply, reverse, etc.) If a reaction is multiplied or divided by a number, H must be multiplied or divided by the same number. Note that H does depend on the physical states of the reactants & products 1. Calculate H for 2S(s) + 2OF 2 (g) SO 2 (g) + SF 4 (g) Use the following reactions and enthalpy changes: OF 2 (g) + H 2 O(l) O 2 (g) + 2HF(g) H 1 = kj SF 4 (g) + 2H 2 O(l) SO 2 (g) + 4HF(g) H 2 = kj S(s) + O 2 (g) SO 2 (g) H 3 = kj 2OF 2 + 2H 2 O 2O 2 + 4HF 2 H 1 = kj SO 2 + 4HF SF 4 + 2H 2 O - H 2 = kj 2S + 2O 2 2SO 2 2 H 3 = kj 2S + 2OF 2 SO 2 + SF 4 H = kj 2. Calculate H for 2C + 3H 2 (g) C 2 H 6 (g) 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(l) H 1 = kj C(s) + O 2 (g) CO 2 (g) H 2 = -394 kj 2H 2 (g) + O 2 (g) 2H 2 O(l) H 3 = -572 kj 2CO 2 + 3H 2 O C 2 H 6 + 7/2 O 2-1/2 H 1 = kj 2C + 2O 2 2CO 2 2 H 2 = -788 kj 3H 2 + 3/2 O 2 3H 2 O 3/2 H 3 = -858 kj 2C + 3H 2 C 2 H 6 H = -86 KJ 3. Calculate the H f for solid Mg(OH) 2 given the following reactions. Hint: write the combination reaction for the formation of Mg(OH) 2 from its elements. 2Mg(s) + O 2 (g) 2MgO(s) H 1 = kj Mg(OH) 2 (s) MgO(s) + H 2 O(l) H 2 = 37.1 kj 2H 2 (g) + O 2 (g) 2H 2 O(l) H 3 = kj Mg(s) + O 2 (g) + H 2 (g) Mg(OH) 2 (s) H f =? Mg + 1/2O 2 MgO 1/2 H 1 = kj MgO + H 2 O (l) Mg(OH) 2 - H 2 = kj H 2 + 1/2 O 2 H 2 O (l) 1/2 H 3 = kj Mg + O 2 + H 2 Mg(OH) 2 H f = kj

5 Enthalpies of Formation H f = Enthalpy of formation: heat gained or lost when a compound is formed from its pure elements. H = enthalpy change at standard conditions: T = 298 K (25 C) and P = 1 atm. H f = Standard Molar Enthalpy of formation: enthalpy change for the formation of 1 mole of a compound from its elements in their most stable state at 298 K and 1 atm pressure. Appendix C provides values of H f (kj/mol) for several compounds. Units for Heats of formation are kj/mol. H f for a pure element in its most stable state is 0. E.g. 2C(graphite) + 3H 2(g) + 1/2 O 2(g) C 2 H 5 OH (l) H f = kj C(graphite), H 2(g) & O 2(g) are elements in most stable state at standard conditions Can use Heats of formation to calculate heat of reaction: H = = Σ[n H ) rxn ( products ] - Σ[n H ( reac tan ts) ] 1. Calculate H rxn for the following: f a. C 2 H 4(g) + 2H 2 O (g) 2CH 3 OH (l) H rxn = [2 H f (2CH 3OH (l) )] - [ H f ( C 2H 4(g) ) + 2 H f (H 2O (g) ) ] H rxn = 2( kj) - [ kj + 2( kj)] = kj f b. 4NH 3(g) + 5O 2(g) 4NO (g) + 6H 2 O (g) H rxn = [4 H f (NO (g)) - 6 H f (H 2O (g) )] - [ 4 H f ( NH 3(g)) + 5 H f (O 2(g)) ] H rxn = [4(90.37 kj) + 6( kj)] - [4( kj) + 0] = kj 2. Calculate the standard enthalpy change for the combustion of 15.0 g of glucose, C 6 H 12 O 6(s), to CO 2(g) and H 2 O (l). Use standard Heats of Formation found in Appendix C. C 6 H 12 O 6(s) + 6O 2(g) 6CO 2(g) + 6H 2 O (l). H rxn = [6 H f (CO 2(g) ) + 6 H f (H 2 O (l) )] - [ H f (C 6 H 12 O 6(s) ) + 6 H f (O 2(g) ] = [6 ( kj) + 6( kj)] - [(-1273 kj) + 0] = kj kj heat given off for combustion of 1 mole of glucose 1mol C6H12O6 2803kJ 15.0 g C 6 H 12 O g C6H12O 6 1mol C6H12O 6 = -234 kj

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