1、Energy & Chemistry,2H2(g) + O2(g) 2H2O(g) + heat and light This can be set up to provide ELECTRIC ENERGY in a fuel cell. Oxidation: 2 H2 4 H+ + 4 e- Reduction: 4 e- + O2 + 2 H2O 4 OH-,H2/O2 Fuel Cell Energy, page 288,Energy & Chemistry,ENERGY is the capacity to do work or transfer heat. HEAT is the
2、form of energy that flows between 2 objects because of their difference in temperature. Other forms of energy light electrical kinetic and potential,Positive and negative particles (ions) attract one another. Two atoms can bond As the particles attract they have a lower potential energy,NaCl compose
3、d of Na+ and Cl- ions.,Potential & Kinetic Energy,Kinetic energy energy of motion.,Internal Energy (E),PE + KE = Internal energy (E or U) Internal Energy of a chemical system depends on number of particles type of particles temperature,The higher the T the higher the internal energy So, use changes
4、in T (T) to monitor changes in E (E).,Thermodynamics,Thermodynamics is the science of heat (energy) transfer.,Heat transfers until thermal equilibrium is established. T measures energy transferred.,SYSTEM The object under study SURROUNDINGS Everything outside the system,Directionality of Heat Transf
5、er,Heat always transfer from hotter object to cooler one. EXOthermic: heat transfers from SYSTEM to SURROUNDINGS.,Directionality of Heat Transfer,Heat always transfers from hotter object to cooler one. ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM.,Energy & Chemistry,All of thermodynam
6、ics depends on the law of CONSERVATION OF ENERGY. The total energy is unchanged in a chemical reaction. If PE of products is less than reactants, the difference must be released as KE.,Energy Change in Chemical Processes,Potential Energy of system dropped. Kinetic energy increased. Therefore, you of
7、ten feel a Temperature increase.,HEAT CAPACITY,The heat required to raise an objects T by 1 C.,Which has the larger heat capacity?,Specific Heat Capacity,How much energy is transferred due to Temperature difference? The heat (q) “lost” or “gained” is related to a) sample mass b) change in T and c) s
8、pecific heat capacity,Table of specific heat capacities,A Assuming an altitude of 194 meters above mean sea level (the worldwide median altitude of human habitation), an indoor temperature of 23 C, a dewpoint of 9 C (40.85% relative humidity), and 760 mmHg sea levelcorrected barometric pressure (mol
9、ar water vapor content = 1.16%).,Aluminum,Specific Heat Capacity,If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?,Heat/Energy Transfer No Change in State,q transferred = (sp. ht.)(mass)(T),Heat Transfer,Use heat transfer as a way to find specific heat cap
10、acity, Cp 55.0 g Fe at 99.8 C Drop into 225 g water at 21.0 C Water and metal come to 23.1 C What is the specific heat capacity of the metal?,Heating/Cooling Curve for Water,Note that T is constant as ice melts or water boils,Chemical Reactivity,But energy transfer also allows us to predict reactivi
11、ty. In general, reactions that transfer energy to their surroundings are product-favored.,So, let us consider heat transfer in chemical processes.,FIRST LAW OF THERMODYNAMICS,E = q + w,Energy is conserved!,The First Law of Thermodynamics,Exothermic reactions generate specific amounts of heat. This i
12、s because the potential energies of the products are lower than the potential energies of the reactants.,The First Law of Thermodynamics,There are two basic ideas of importance for thermodynamic systems. Chemical systems tend toward a state of minimum potential energy.Chemical systems tend toward a
13、state of maximum disorder.The first law is also known as the Law of Conservation of Energy. Energy is neither created nor destroyed in chemical reactions and physical changes.,SYSTEM,E = q + w,ENTHALPY,Most chemical reactions occur at constant P, so,and so E = H + w (and w is usually small) H = heat
14、 transferred at constant P E H = change in heat content of the system H = Hfinal - Hinitial,Heat transferred at constant P = qp qp = H where H = enthalpy,If Hfinal Hinitial then H is negative Process is EXOTHERMIC,If Hfinal Hinitial then H is positive Process is ENDOTHERMIC,ENTHALPY,H = Hfinal - Hin
15、itial,Consider the formation of water H2(g) + 1/2 O2(g) H2O(g) + 241.8 kJ,USING ENTHALPY,Exothermic reaction heat is a “product” and H = 241.8 kJ,Making liquid H2O from H2 + O2 involves two exothermic steps.,USING ENTHALPY,H2 + O2 gas,Liquid H2O,H2O vapor,Making H2O from H2 involves two steps. H2(g)
16、 + 1/2 O2(g) H2O(g) + 242 kJH2O(g) H2O(l) + 44 kJ H2(g) + 1/2 O2(g) H2O(l) + 286 kJ Example of HESSS LAW If a rxn. is the sum of 2 or more others, the net H is the sum of the Hs of the other rxns.,Hesss Law & Energy Level Diagrams,Forming H2O can occur in a single step or in a two steps. Htotal is t
17、he same no matter which path is followed.,Active Figure 6.18,Hesss Law,Hesss Law of Heat Summation, Hrxn = H1 +H2 +H3 + ., states that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps. Hesss Law is true because H is a state functio
18、n. If we know the following Hos,Hesss Law,For example, we can calculate the Ho for reaction 1 by properly adding (or subtracting) the Hos for reactions 2 and 3. Notice that reaction 1 has FeO and O2 as reactants and Fe2O3 as a product. Arrange reactions 2 and 3 so that they also have FeO and O2 as r
19、eactants and Fe2O3 as a product. Each reaction can be doubled, tripled, or multiplied by half, etc. The Ho values are also doubled, tripled, etc. If a reaction is reversed the sign of the Ho is changed.,Hesss Law,Hesss Law in a more useful form. For any chemical reaction at standard conditions, the
20、standard enthalpy change is the sum of the standard molar enthalpies of formation of the products (each multiplied by its coefficient in the balanced chemical equation) minus the corresponding sum for the reactants.,Hesss Law,Hesss Law,Given the following equations and Hovaluescalculate Ho for the r
21、eaction below.,Hesss Law,Use a little algebra and Hesss Law to get the appropriate Hovalues,Some Thermodynamic Terms,Notice that the energy change in moving from the top to the bottom is independent of pathway but the work required may not be! Some examples of state functions are: T (temperature), P
22、 (pressure), V (volume), E (change in energy), H (change in enthalpy the transfer of heat), and S (entropy) Examples of non-state functions are: n (moles), q (heat), w (work),H along one path = H along another path,This equation is valid because H is a STATE FUNCTION These depend only on the state o
23、f the system and not how it got there. V, T, P, energy and your bank account! Unlike V, T, and P, one cannot measure absolute H. Can only measure H.,Some Thermodynamic Terms,The properties of a system that depend only on the state of the system are called state functions. State functions are always
24、written using capital letters. The value of a state function is independent of pathway. An analog to a state function is the energy required to climb a mountain taking two different paths. E1 = energy at the bottom of the mountain E1 = mgh1 E2 = energy at the top of the mountain E2 = mgh2 E = E2-E1
25、= mgh2 mgh1 = mg(h),Standard States and Standard Enthalpy Changes,Thermochemical standard state conditions The thermochemical standard T = 298.15 K. The thermochemical standard P = 1.0000 atm. Be careful not to confuse these values with STP. Thermochemical standard states of matter For pure substanc
26、es in their liquid or solid phase the standard state is the pure liquid or solid. For gases the standard state is the gas at 1.00 atm of pressure. For gaseous mixtures the partial pressure must be 1.00 atm. For aqueous solutions the standard state is 1.00 M concentration.,Hfo = standard molar enthal
27、py of formationthe enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 6.2 and Appendix L,Enthalpy Values,H2(g) + 1/2 O2(g) H2O(g) H = -242 kJ2H2(g) + O2(g) 2H2O(g) H = -484 kJH2O(g) H2(g) + 1/2 O2(g) H = +242 kJ H2(g) + 1/2 O2(g) H2O(l) H = -286 kJ,De
28、pend on how the reaction is written and on phases of reactants and products,Hfo, standard molar enthalpy of formation,H2(g) + O2(g) H2O(g) Hf (H2O, g)= -241.8 kJ/mol C(s) + O2(g) CO(g) Hf of CO = - 111 kJ/mol By definition, Hfo = 0 for elements in their standard states.,Use Hs to calculate enthalpy
29、change for H2O(g) + C(graphite) H2(g) + CO(g),Using Standard Enthalpy Values,In general, when ALL enthalpies of formation are known,Horxn = Hfo (products) - Hfo (reactants),Remember that always = final initial,Using Standard Enthalpy Values,Calculate the heat of combustion of methanol, i.e., Horxn for CH3OH(g) + 3/2 O2(g) CO2(g) + 2 H2O(g) Horxn = Hfo (prod) - Hfo (react),
