中国高校课件下载中心 》 教学资源 》 大学文库

上海交通大学:《大学化学 Chemistry》教学资源(课件讲稿)Chapter 19 Chemical Thermodynamics

文档信息
资源类别:文库
文档格式:PDF
文档页数:57
文件大小:1.8MB
团购合买:点击进入团购
内容简介
• Spontaneous Processes • Entropy • 2nd Law of Thermodynamics • Calculation of ∆S • Gibbs Free Energy ∆G • Free Energy and Temperature • ∆G & the Equilibrium Constant K
刷新页面文档预览

Chapter 19 Chemical Thermodynamics .Spontaneous Processes ·Entropy 2nd Law of Thermodynamics ·Calculation of△S ·Gibbs Free Energy△G Free Energy and Temperature 。△G&the Equilibrium Constant K

Chapter 19 Chemical Thermodynamics • Spontaneous Processes • Entropy • 2nd Law of Thermodynamics • Calculation of ∆S • Gibbs Free Energy ∆G • Free Energy and Temperature • ∆G & the Equilibrium Constant K

Spontaneous vs.Non-spontaneous Spontaneous process A process,once started,proceeds on its own without outside influence. .Non-spontaneous process: A process requires external help for it to proceed. Spontaneous Nonspontaneous

Spontaneous vs. Non-spontaneous • Spontaneous process : A process, once started, proceeds on its own without outside influence. • Non-spontaneous process spontaneous process : A process requires external help for it to proceed

Is this a spontaneous process? Other Factor Transferring heat from a hot object to a cold object. Yes,always. Spontaneous for T>0C Spontaneous for T0? HOC Yes,ifTOC

Is this a spontaneous process? • Transferring heat from a hot object to a cold object. Yes, always. Other Factor ? • Ice melting. Water freezing. Yes, if T > 0oC No, if T 0oC if T = 0 oC ? ∆H > 0 ? H > 0 ? ∆H < 0 ? H < 0 ?

Thermodynamics Laws of Thermodynamics show that spontaneity is predictable by considering two factors: ENTHALPY&ENTROPY S Enthalpy Change (AH)total Entropy (S)randomness energy difference between factor accounts for energy reactants products lost in the process Gibbs Free Energy (G):net amount available Energy for useful work after considering the two factors Thermodynamics:a study of energy flow as well as how much energy can be converted to useful work

Thermodynamics Laws of Thermodynamics Laws of Thermodynamics show that show that spontaneity is predictable by considering two factors: ENTHALPY& ENTROPY H S Enthalpy Change (∆H) : total Entropy (S) : randomness factor accounts for energy Thermodynamics: a study of energy flow as well as how much energy can be converted to useful work Enthalpy Change (∆H) : total energy difference between reactants & products factor accounts for energy lost in the process Gibbs Free Energy (G) : net amount available Energy for useful work after considering the two factors

Why consider spontaneity of reactions? Spontaneity:the tendency in nature for certain things to take place >Can make use of this tendency to work for us. To know the amount of useful work to expect from a reaction. What is the maximum amount of Energy I can get out ofone tank of gasoline? Useful work obtained through a spontaneous combustion reaction of Thermodynamics one tank of gasoline

Why consider spontaneity of reactions? • Spontaneity: the tendency in nature for certain things to take place  Can make use of this tendency to work for us.  To know the amount of useful work to expect from a reaction. What is the maximum amount of Useful work obtained through a spontaneous combustion reaction of one tank of gasoline. What is the maximum amount of Energy I can get out of one tank of gasoline ? Thermodynamics

Entropy Entropy (S):a measure of the randomness or disorder of a system order S △S>0 If a change results in an increase in randomness initial S inal

Entropy (S) : a measure of the randomness or disorder of a system order S disorder S If a change results in an increase in randomness Entropy - final initial ∆ = S S S ∆S>0 If a change results in an increase in randomness initial final S < S

Entropy,S Entropy a thermodynamic quantity,similar to enthalpy ·A state.function: AS=Sfinat -Stmimal the change of entropy (AS)depends only on the initial and final states of the system not on the path it takes to change states An extensive property its value depends on the quantity (mass,mole)of a substance. S(of 1mole O2(g))<S(of 2 mole O2(g))

Entropy, S Entropy : a thermodynamic quantity, similar to enthalpy • A state function: – the change of entropy (∆S) depends • only on the initial and final states of the system - final initial ∆ = S S S • not on the path it takes to change states • An extensive property : its value depends on the quantity (mass, mole) of a substance. 2 2 S(of 1mole O (g)) < S(of 2 mole O (g))

Entropy Boltzman the entropy (S)of a system is related to the natural log of the number of microstates (W). S=k.InW AS=S final-Similial AS=S inal -S'initial =k In Winal -k In Wininal =kIn W itial WFmal >Winital,AS> WEmal <Wial.AS<

Boltzman : the entropy (S) of a system is related to the natural log of the number of microstates (W). Entropy final initial S k = ⋅lnW ∆ = − S S S ln ln ln final initial final final initial initial S S S W k W k W k W ∆ = − = − = , 0 W W S Final initial > ∆ > , 0 W W S Final initial < ∆ <

Ludwig Boltzmann s-k fog w An Austrian physicist famous for his founding contribution in the fields of statistical mechanics and statistical thermodynamics EVDIG BOLIZMANN DPILFAULA BOLTZMANN 5 鞋 1 S=k·lnW BOLEZNNN OCTZMANK BOUZMANN Among his students in Graz were Svante Arrhenius and Walther Nernst

An Austrian physicist famous for his founding contribution in the Ludwig Boltzmann Nov. 2008 S k = ⋅ln W his founding contribution in the fields of statistical mechanics and statistical thermodynamics Among his students in Graz were Svante Arrhenius and Walther Nernst

Entropy ep.Four Molecules in a Two-Bulbed Flask S=k.InW W:the possible arrangements (states) Arrangement I Microstates three Possible 0 Arrangements W=3 Arrangement II k=R/N=8.314J/mol.K/6.02×1023mol-1 =1.38×10-23J/K S=k.InW=1.38x1023J/KxIn3 Arrangement III =1.516×10-23J/K Probability and the locations of gas molecules

Microstates : three Possible Arrangements S k = ⋅lnW Entropy W=3 ep. :Four Molecules in a Two in a Two-Bulbed Flask Bulbed Flask W: the possible arrangements (states) Arrangements 23 S k J K ln 1.38 10 ln W / 3 − = ⋅ = × × 23 1 23 8.314 / 6.02 10 = 1.38 10 / k R N J mol K mol J K − − = = ⋅ × × W=3 23 1.516 10 / J K − = × Probability and the locations of gas molecules

刷新页面下载完整文档
VIP每日下载上限内不扣除下载券和下载次数;
按次数下载不扣除下载券;
注册用户24小时内重复下载只扣除一次;
顺序:VIP每日次数-->可用次数-->下载券;
相关文档