Energy & Chemistry. Internal Energy (E) Energy and Chemistry. Potential Energy. Kinetic Energy. Energy and Chemical Reactions: Thermochemistry or

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1 Page III-5-1 / Chapter Five Lecture Notes Energy & Chemistry Energy and Chemical Reactions: Thermochemistry or Thermodynamics Chapter Five Burning peanuts supplies sufficient energy to boil a cup of water Chemistry 221 Professor Michael Russell These reactions are PRODUCT FAVORED They proceed almost completely from reactants to products, perhaps with some outside assistance. Potential Energy Energy and Chemistry ENERGY is the capacity to do Potential energy - energy a motionless body has by virtue of its position In chemistry, positive and negative particles (ions) attract one another. As the particles attract to make a bond, they have a lower potential energy work or transfer heat. HEAT is the form of energy that flows between 2 samples because of their difference in temperature. Other forms of energy light next chapter electrical - CH 223 kinetic and potential NaCl - composed of Na+ and Cl- ions. Kinetic Energy Internal Energy (E) Three forms of kinetic energy: Translational - from physics, KE = 1/2 mv2 = 0.5(mass)(velocity)2 Vibrational Rotational rotate vibrate translate Burning sugar (sugar reacts with KClO3, a strong oxidizing agent) PE + KE = Internal Energy (E or U) The higher the T the higher the internal energy So, use changes in T ( T) to monitor changes in E ( E). Page III-5-1 / Chapter Five Lecture Notes

2 Page III-5-2 / Chapter Five Lecture Notes Thermodynamics Thermodynamics is the science of heat (energy) transfer. Directionality of Heat Transfer Heat always transfers from the hot object to the cooler object. EXOthermic: heat transfers from SYSTEM to SURROUNDINGS. Heat energy is associated with molecular motions (Kinetic Molecular Theory) Heat transfers until thermal equilibrium is established. T(system) goes down T(surr) goes up - this is what we usually measure Directionality of Heat Transfer Heat always transfers from the hot object to the cooler object. ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM. T(system) goes up T(surr) goes down - this is what we usually measure Energy & Chemistry All of thermodynamics depends on the law of CONSERVATION OF ENERGY. The total energy is unchanged in a chemical reaction. If the potential energy of products is less than reactants, the difference must be released as kinetic energy, etc. UNITS OF ENERGY 1 calorie = heat required to raise temp. of 1.00 g of H 2 O by 1.0 o C cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food "calorie") But we use the unit called the JOULE 1 cal = joules James Joule Thermochemistry is the science of heat (energy) flow. A difference in temperature leads to energy transfer. The heat, q, "lost" or "gained" is related to a) sample mass b) change in T and c) specific heat capacity Memorize 4.184! Page III-5-2 / Chapter Five Lecture Notes

3 Page III-5-3 / Chapter Five Lecture Notes : The heat required to raise an object's temperature by 1 C. Aluminum Substance Spec. Heat (J/g K) H 2 O(l) Ethylene glycol(l) 2.39 Al(s) glass(s) 0.84 Water q = mc T Important! q = heat "lost" or "gained" m = sample mass (usually grams) C = specific heat capacity T = change in temperature T = final T - initial T q is positive when heat flows in (cold surroundings) - endothermic q is negative when heat flows out (hot surroundings) - exothermic If 25.0 g of Al cool from 310. o C to 37 o C, how many joules of heat energy are lost by the Al? q = mc T CAl = J g -1 K -1 If 25.0 g of Al cool from 310. o C to 37 o C, how many joules of heat energy are lost by the Al? CAl = J g -1 K -1 heat gain/lost = q = (mass)(sp. ht.)( T) where T = T final - T initial q = (25.0 g)(0.902 J/g C)( ) C q = J Notice also that that the T negative can be sign either on K q or signals C - the heat difference "lost by" in or temperatures transferred is OUT the same! of Al. A piece of iron (88.5 g) at 77.8 o C is placed in 244 g of water at 18.8 o C. What is the final temperature of the mixture? By the law of conservation of energy: q hot + q cold = 0, or m Fe C Fe T Fe + m water C water T water = 0 C Fe = J g -1 K -1 C water = J g -1 K -1 Memorize Final Temperature (warm) same for Fe & H 2 O Page III-5-3 / Chapter Five Lecture Notes Fe (88.5 g, J/g K, 77.8 o C) and water (244 g, J/g K, 18.8 o C); final temperature? q hot + q cold = 0 m Fe C Fe T Fe + m water C water T water = * * (T f ) * * (T f ) = T f T f = T f = T f = 21.0 o C Efficient method to determine approximate final temperature of mixture

4 Page III-5-4 / Chapter Five Lecture Notes Heat Transfer No Change in State Heat Transfer with Change of State q transferred = (mass)(sp. ht.)( T) As matter absorbs heat, its temperature rises until it undergoes a phase change (change of state) Changes of state involve energy (at constant T) Ice J/g (heat of fusion) -----> Liquid water q = (heat of "something")("mass") Heat & Changes of State Heat & Changes of State What quantity of heat is required to melt 500. g of ice at 0.0 oc and heat the water to steam at oc? Heat of fusion of ice = 333 J/g Specific heat of water = J/g K Heat of vaporization = 2260 J/g What quantity of heat is required to melt 500. g of ice at 0.0 oc and heat the water to steam at oc? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 105 J 2. To raise water from 0.0 oc to oc q = (500. g)(4.184 J/g K)( )K = 2.09 x 105 J J/g J/g To evaporate water at 100. oc q = (500. g)(2260 J/g) = 1.13 x 106 J 4. Total heat energy = 1.51 x 106 J = 1510 kj solid (ice) liquid (0 100 C) gas (steam) CHEMICAL REACTIVITY What drives chemical reactions? How do they occur, and how fast? The first question is answered by THERMODYNAMICS, and the second question is answered by KINETICS. Heat Transfer covered more in Chapter 10 of CH 222 CHEMICAL REACTIVITY THERMODYNAMICS dictates if the reaction will occur or not. Sand will not decompose into silicon and oxygen Paper will combine with oxygen to burn Page III-5-4 / Chapter Five Lecture Notes

5 Page III-5-5 / Chapter Five Lecture Notes CHEMICAL REACTIVITY CHEMICAL REACTIVITY KINETICS dictates how fast the reaction will occur. Example: diamond into graphite is thermodynamically favored, but the kinetics of the reaction is too slow to be of useful We have already seen a number of "driving forces" for reactions that are PRODUCT-FAVORED. formation of a precipitate (precipitation) gas formation (gas forming) H2O formation (acid-base) electron transfer (redox) i.e. in a battery In general, reactions that transfer energy to their surroundings are productfavored. FIRST LAW OF THERMODYNAMICS heat energy transferred E = q + w energy change ENTHALPY Most chemical reactions occur at constant pressure (P), so Heat transferred at constant P = qp work done by the system Energy is conserved! Chemists focus on q more than w qp = H where H = enthalpy E = q + w and E = H + w (P const) w is usually small, so H E ENTHALPY H = Hfinal - Hinitial USING ENTHALPY Consider the formation of water If Hfinal > Hinitial then H is positive Process is ENDOTHERMIC H is positive H2(g) + 1/2 O2(g) --> H2O(g) kj If Hfinal < Hinitial then H is negative Process is EXOTHERMIC H is negative Exothermic reaction - heat is a "product" and H = kj Page III-5-5 / Chapter Five Lecture Notes

6 Page III-5-6 / Chapter Five Lecture Notes H 2 + O 2 gas USING ENTHALPY Making liquid H 2 O from H 2 + O 2 involves two exothermic steps. H 2 O vapor Liquid H 2 O USING ENTHALPY Making H 2 O from H 2 involves two steps. H 2 (g) + 1/2 O 2 (g) ---> H 2 O(g) kj H 2 O(g) ---> H 2 O(liq) + 44 kj H 2 (g) + 1/2 O 2 (g) --> H 2 O(liq) kj Example of HESS'S LAW- If a rxn. is the sum of 2 or more others, the net H is the sum of the H's of the other rxns. Hess's Law & Energy Level Diagrams Forming H 2 O can occur in a single step or in a two steps. H total is the same no matter which path is followed. H is a STATE FUNCTION - depends only on the state of the system and not how it got there. Other state functions: V, T, P, and energy Standard Enthalpy Values Most H values are labeled H o Measured under standard conditions P = 1 bar (approx. 1 atm) Concentration = 1 mol/l T = usually 25 o C with all species in standard states i.e., C = graphite and O 2 = gas Enthalpy Values Depend on how the reaction is written and on phases of reactants and products H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) H = -242 kj 2 H 2 (g) + O 2 (g) --> 2 H 2 O(g) H = -484 kj H 2 O(g) ---> H 2 (g) + 1/2 O 2 (g) H = +242 kj H 2 (g) + 1/2 O 2 (g) --> H 2 O(liquid) H = -286 kj Page III-5-6 / Chapter Five Lecture Notes Standard Enthalpy Values NIST (Nat'l Institute for Standards and Technology) gives values of H f o = standard molar enthalpy of formation - the enthalpy change when 1 mol of compound is formed from elements under standard conditions. These values are available in your text All elements in standard states have Hf = 0 (graphite but not diamond, O2(g) but not O3 or O2(l), etc.)

7 Page III-5-7 / Chapter Five Lecture Notes Use H values to calculate the enthalpy change for: H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) Find enthalpy change for this reaction: H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) From reference books we find: (product is called "water gas") H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) H f of H 2 O vapor = kj/mol C(s) + 1/2 O 2 (g) --> CO(g) H f of CO = kj/mol H 2 O(g) --> H 2 (g) + 1/2 O 2 (g) H o = +242 kj C(s) + 1/2 O 2 (g) --> CO(g) H o = -111 kj H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) H o net = +131 kj Find H o rxn for CaCO 3 (s) --> CaO(s) + CO 2 (g) using Hess's Law and: Ca(s) + 3/2 O 2 (g) + C(s) --> CaCO 3 (s) H o f = kj as well as Ca(s) + 3/2 O 2 (g) + C(s) --> CaO(s) + CO 2 (g) H o rxn = kj To convert 1 mol of water to 1 mol each of H 2 and CO requires 131 kj of energy. The "water gas" reaction is ENDOthermic. CaCO3 CaO Page III-5-7 / Chapter Five Lecture Notes Find H o rxn for CaCO 3 (s) --> CaO(s) + CO 2 (g) CaCO 3 (s) --> Ca(s) + 3/2 O 2 (g) + C(s) H o = -( kj) = kj Ca(s) + 3/2 O 2 (g) + C(s) --> CaO(s) + CO 2 (g) H o rxn = kj CaCO 3 (s) --> CaO(s) + CO 2 (g), and H o rxn = kj + ( kj) = kj

8 Page III-5-8 / Chapter Five Lecture Notes Calculate H of the reaction? In general, when ALL enthalpies of formation are known: H o rxn = Σ H f o (products) - Σ H f o (reactants) H f o, enthalpy of formation H f o = 0 for elements in their standard states. formation = 1 mol product, reactants are elements in standard states Example: Write the H f o equation for Li2CO3(s) 2 Li(s) + C(graph) + 3 /2 O 2 (g) Li2CO3(s) H o rxn = Σ H f o (products) - Σ H f o (reactants), so H o rxn = H f o (Li2CO3(s)) - {2* H f o (Li(s) + H f o (C(graphite)) + 3 /2* H f o (O2(g))} Σ = summation sign, or add up all of the H o rxn = {2* /2*0} = kj/mol H o rxn Example: Find H o rxn for CaCO 3 (s) --> CaO(s) + CO 2 (g) using: H o rxn = Σ H o f (prod) - Σ H o f (react) Answer: H o rxn = { H o f (CaO) + H o f (CO 2 )} - { H o f (CaCO 3 )} H o rxn = { } - { } H o rxn = kj Same answer as before with Hess's Law, but simpler Calculate the heat of combustion of methanol, i.e., H o rxn for CH 3 OH(g) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O(g) H o rxn = Σ H f o (prod) - Σ H f o (react) As before, look up H f o values for reactants and products in your text Elements in standard states have H f o = 0 CH 3 OH(g) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O(g) H o rxn = Σ H f o (prod) - Σ H f o (react) H o rxn = { H f o (CO 2 ) + 2 H f o (H 2 O)} - {3/2 H f o (O 2 ) + H f o (CH 3 OH)} = {( kj) + 2 ( kj)} - {0 + ( kj)} H o rxn = kj per mol of methanol CALORIMETRY Measuring Heats of Reaction Constant Volume "Bomb" Calorimeter Burn combustible sample. Measure heat evolved in a reaction. Derive E for reaction. Page III-5-8 / Chapter Five Lecture Notes

9 Page III-5-9 / Chapter Five Lecture Notes Calorimetry Some heat from reaction warms water qwater = (sp. ht.)(water mass)( T) Some heat from reaction warms "bomb" qbomb = (heat capacity, J/K)( T) Measuring Heats of Reaction CALORIMETRY Calculate heat of combustion of octane: C8H /2 O2 --> 8 CO2 + 9 H2O Burn 1.00 g of octane Temp rises from to oc Calorimeter contains g water Heat capacity of bomb = 837 J/K Total heat evolved = qtotal = qwater + qbomb Measuring Heats of Reaction CALORIMETRY Step 1 Calc. heat transferred from reaction to water. q = (4.184 J/g K)(1200. g)(8.20 K) = 41,200 J Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)( T) = (837 J/K)(8.20 K) = 6860 J Step 3 Total heat evolved 41,200 J J = 48,100 J Heat of combustion of 1.00 g of octane = kj (-5.48*103 kj/mol) End of Chapter Five See also: Chapter Five Study Guide Chapter Five Concept Guide Page III-5-9 / Chapter Five Lecture Notes