Chap.5 The First Law of Thermodynamics热力学 The internal energy (E)of an system:the sum of the all kinetic and potential energies of all its components. AEsystem十△Fsounding =0 Closed Heat,q surroundings System Work,w △E =-△E system surrounding =9+w Process Sign Work done by the system on the surroundings Work done on the system by the surroundings Heat absorbed by the system from the surroundings(endothermic process) Heat absorbed by the surroundings from the system(exothermic process)
surroundings Closed System Heat, q Work, w Chap.5 The First Law of Thermodynamics ∆ ∆ = E + E 0 system surrounding The internal energy (E) of an system: the sum of the all kinetic and potential energies of all its components. 热力学 E = - E system surrounding ∆ ∆ = +q w
Transferring Energy:Work (w)&heat In chemistry,two types of work are interesting:electrochemical work P-Vwork: Work done by the system,increases its volume - Work done on the system,decreases its volume No volume change,no expansion work. When a gas system expands against a constant external pressure Pe W=FXd=P×AXd=P.X△V W=-Px·△V g W0: AV
– Work done by the system, increases its volume In chemistry, two types of work are interesting: electrochemical work & P-V work: Transferring Energy: Work (w) & heat When a gas system expands against a constant external pressure P – Work done on the system, decreases its volume – No volume change, no expansion work. W F d P A d P V = × = × × = × ∆ ex ex When a gas system expands against a constant external pressure Pex - W P V = ⋅∆ ex W 0
Transferring Energy:Work (w)&Heat (q) Exercise 1:What is the change in energy when a system does 135 J work and absorbs 45.0 cal of heat from surrounding? w=-135J q=+45.0cal×4.184J/cal=+188J △E=w+q=-135+188=+53J 1 J=1 kg.m2/s2 Traditional: 1 cal 4.184 J calorie (cal) 1 Cal (big Cal)=1000 cal=1 kcal Nutrition/Food=Calorie(big Cal)
Exercise 1: What is the change in energy when a system does 135 J work and absorbs 45.0 cal of heat from surrounding? w = 1- 35 J ∆E = w + q = 135 188 = - + +53 J q = + 45.0 4.1 cal J/ca × 84 = l J +188 E = system ∆ +q w Transferring Energy: Work (w) & Heat (q) Traditional: 1 cal = 4.184 J ‘calorie (cal) 1 Cal (big Cal) = 1000 cal = 1 kcal Nutrition/Food = Calorie (big Cal) ∆E = w + q = 135 188 = - + +53 J 2 2 1 =1 J kg m⋅ / s
Enthalpy焓 H=enthalpy or “Total Heat Content"” Enthalpy (H):used to quantify the heat flow into or out of a system in a process that occurs at constant pressure (qp). H:a state function.No need to calculate H (absolute value)itself, Only difference in H (AH)is necessary. exothermic -H reactants =9p △H0 reactants endothermic Activated Activated complex complex K3ou [enualod E Hreactants K3Jou enusiod Hproducts Hproduets Hreactants Reaction progress Reaction progress
∆ = = H q H - H products rea t c ants p Enthalpy H: a state function. No need to calculate H (absolute value) itself, Only difference in H (∆H) is necessary. Enthalpy (H) : used to quantify the heat flow into or out of a system in a process that occurs at constant pressure (qp). H = enthalpy or “Total Heat Content” exothermic 焓 H H 0 products reactants ∆ > H endothermic
State Function 状态函数 State function:A property of a system,which is independent of the “path”,however,it is dependent only of the“current state” 目前状态 AE-Erinal state-Einitial stateH-H final state-Hinitial state △P=Pfinal state Pinitial state AV-Vrinal state -Vinitial state △T=Tfinal state~ Tintia Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
State function: A property of a system, which is independent of the “path” , however, it is dependent only of the “current state” ∆E= E - E final state initial state ∆P = P - P final state initial state ∆V = V - V final state initial state ∆T = T - T State Function 状态函数 目前状态 ∆H= H - H final state initial state Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. final state initial state ∆T = T - T
Thermochemical Equations 热化学方程式 1.Enthalpy:an extensive property: 广度性质 Given:H2O(s)→H20(0 △H=6.01kJ stoichiometric coefficients 计量系数 2H20(s)→2H20(0△H=？ △H=2×6.01 CH4(g)+2O2(g)=C02(g+2H2O(g)△H=-802k 2CH4(g)+402(g)=2C02(g)+4H2O(g)△H=-1604kJ
H2O (s) H2O (l) ∆H = 6.01 kJ Thermochemical Equations 2H2O (s) 2H2O (l) 1. Enthalpy： an extensive property： ∆H = ? Given: stoichiometric coefficients 热化学方程式 计量系数 广度性质 2H2O (s) 2H2O (l) ∆H = ? ∆H = 2 ×6.01 CH (g)+2O (g) = CO (g)+2H O(g) H = -802 kJ 4 2 2 2 ∆ CH (g)+ O (g) = CO (g) 4 2 2 2 2 4 2 + H O(g) H 4 16 ∆ = - J 04 k
Thermochemical Equations 热化学方程式 2.Change in enthalpy (AH)depends on physical state: State function H2OI)→HO(g) △H=44.0kJ HO(g)→H2O() △H=? △H=-44.0kJ 2H,O(g)→2H,O() △H=-2×44.0kJ Ifyou reverse a reaction,the sign of AH changes
2. Change in enthalpy ( 2. Change in enthalpy (∆H) depends on ) depends on physical state: Thermochemical Equations 热化学方程式 2 2 H O l H O g ( ) ( ) H=44.0 kJ → ∆ State function If you reverse a reaction, the sign of ∆H changes 2 2 H O g H O l ( ) ( ) H= → ∆ ? ∆H= 4- 4.0 kJ 2 2 2 2 - 2 H O g H O l ( ) ( ) H= → ∆ × 44.0 kJ
Exercise 2:Given the changes of enthalpy of reaction (1)&(2) Calculate how much heat is involyed in the 3rd reaction? (1)2H2(g)+O2(8)=2H0(g)△H=-483.9kJ/mo1 (2)2HO(g)→2H,O()△H=-88kJ1mol (3)2H2(g)+O2(g)→2HOI)△H=?kJ1mol (1) 2H2(g)+O2(g)=2HO(g) +(2) 2H,O(g)=2H,O() Hess's Law (3) 2H2(g)+O2(g)=2HO() △H3=△H1+△H2=-438.9kJ-88kJ=-572kJ AH of overall reaction is the sum of the AH for each individual step. △H=△H1+△H2+△H
2 2 (2) 2 ( ) 2 ( H O H O H kJ mo g l → ∆ = ) 88 / − l 2 2 2 (1) 2 ( ) ( ) = 2H O(g) H 483.9 / H O mol g + g ∆ = − kJ 2 2 2 (1) 2 ( ) ( ) = 2H O( ) H O g g + g = 2 2 2 (3) 2 ( ) ( ) 2 ( ) / H O H g + = g → O mo l ? ∆H kJ l Exercise 2: Given the changes of enthalpy of reaction (1) & (2). Calculate how much heat is involved in the 3rd reaction? 2 2 2 2 2 +(2) 2 ( ) 2 ( ) (3) 2 ( ) ( ) = 2 ( ) H O H O H O H O g l g g l = + 3 1 2 ∆H = ∆ + ∆ H H = − = − 438.9 88 572 kJ kJ − kJ ∆H of overall reaction is the sum of the ∆H for each individual step. ∆ ∆ ∆ ∆ H = H + H + H 1 2 3 Hess’s Law
Exercise 3:Given the changes of enthalpy of reaction (1)&(2). Calculate how much heat is involyed in the 3rd reaction? N2g+202g→2N02g H1=66.4kJ 2N0(g+02g)→2N028 H2=-114.2kJ N2(g+O2g→2N0(g H3=？ N2(g)+22g)→2e2(g △H=66.4kJ +2O,(g)→2NO(g)+Dg) △H,=+114.2kJ N2(g)+O2(g)→2NO(g) △H,=(66.4+114.2)kJ Hss'sLaw:△H=△H1+△H2+△H
2 2 2 N (g) + 2O (g) 2NO (g) → → N2(g) + 2O 2(g) → 2NO 2(g) ∆H1 = 66.4 kJ 2 NO (g) + O 2(g) → 2NO 2(g) ∆H2 = - 114.2 kJ N2(g) + O 2(g) → 2NO(g) ∆H3 = ? 1 H = 66.4 kJ ∆ ∆ Exercise 3: Given the changes of enthalpy of reaction (1) & (2). Calculate how much heat is involved in the 3rd reaction? 2 2 + 2NO (g) 2 NO 2 2 N (g) + O (g) 2 (g) + O (g) N ) O(g → → ' 2 3 +114.2 kJ (66.4+114.2)kJ H = H = ∆ ∆ 1 2 3 Hess s Law ' : H = H + H + H ∆ ∆ ∆ ∆
Exercise 4:Given the changes of enthalpy of reaction (1),(2)&(3) Calculate how much heat is involved in the 4th reaction? N0g+O3g→NO2g+02g 4H1=-198.9kJ 203(8→3028g H2=-284.6kJ 028g→208g) H3=495.0kJ NOg+O(g→N02,g H=? NO(g)+(g)NO,(g)+e.(g) △H=-198.9kJ ,(g)→g) △H2=-(-284.6)/2kJ 0g)→6，(g) AH,=-(495.0)/2kJ N0(g)+O(g)→NO2(g)△H4=-198.9kJ-(-284.6)/2kJ-(495.0)/2kJ =-304.1kJ Hess'sLaw:△H=△H1+△H2+AH3
NO(g) + O (g) NO (g) + O (g) H = -198.9 kJ → ∆ NO(g) + O 3(g) → NO 2(g) + O2(g) ∆H1 = -198.9 kJ 198.9 kJ 2 O3(g) → 3 O 2(g) ∆H2 = -284.6 kJ 284.6 kJ O 2(g) → 2O(g) ∆H3 = 495.0 kJ NO(g) + O(g) → NO 2(g) ∆H4= ? Exercise 4: Given the changes of enthalpy of reaction (1), (2) & (3) Calculate how much heat is involved in the 4th reaction? 3 ' 3 2 2 1 2 ' 2 3 2 NO(g) + O (g) NO (g) + O (g) H = -198.9 kJ O (g) O (g) = (-284.6) kJ O(g) O (g) = (495.0) kJ 3 H - 2 2 1 H - 2 2 → ∆ → → ∆ ∆ NO(g) + O(g) NO (g) H = -198.9 kJ (-284. 2 4 6) kJ (495.0) kJ = - 2 - 2 -304.1kJ → ∆ 1 2 3 Hess s Law ' : H = H + H + H ∆ ∆ ∆ ∆