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Thermochemistry Q p H ( product)- H (reacant) r H r H m B H m ( B) B H △rHm (298.15K) B f m ( B) (298.15K) B r H m (298.15K) B c H m (B, 298.15K) B H2(g,p ) + I2(g,p )=2HI(g,p ) △rHm(298.15K) = -51.8kJ·mol Dependence of standard molar enthalpy of Kirchhoff equation reaction on temperature r H m (T1 ) + bB aA H1 H 2 + bB aA r H m (T2 ) yY + zZ H 4 H 3 yY + zZ r H m T1 H1 H 2 r H m T2 H3 H 4 since T2 H a C p,m (A)d T, H b C p,m (B)d T, 1 T1 H y 3 then T2 T1 2 T2 T1 C p,m (Y)d T, H z 4 T2 T2 T1 C p,m (Z)d T, r H m (T2 ) r H m (T1 ) BC p ,m ( B)dT T 1 Heat of phase change Q p H H H ( ) H ( ) From a liquid to a gas (vaporization) From a solid to a liquid (fusion) From a solid to a gas (sublimation) Crystal form transition W, Q, ΔU,ΔH of phase transformation ΔH = ΔU + Δ(PV)=Q+W+△(PV) For melting and crystal transition process at constant pressure and constant temperature Qp H W pV 0 U H For vaporization and sublimation processes Qp H W pV pV ( g ) nRT U H ( pV ) H pV ( g ) H nRT Temperature dependence of enthalpy of phase change H (T2 ) H m ( ) H (T1 ) H m ( ) T1 H m ( ) C p ,m ( )dT T2 T2 H m ( ) C p ,m ( )dT T1 Cp,m =C p ,m ( )-C p ,m ( ) T2 H m ( )+H m ( ) C p ,m dT T1 T2 H m (T2 ) H m (T1 )+ C p ,m dT T1 1 373 K , P , r H m θ θ H2O(g, Pθ) H2(g, P ) + O2(g, P ) 2 ΔH4 H2O(l, Pθ,373K) ΔH3 ΔH2 1 298 K , P , H m ,1 θ θ r H2O(l, Pθ,298K) H2(g, P ) + O2(g, P ) 2 θ Δ rH θ m(373K) = Δ rH =ΔrHθm(298K) + 373 298 m1 +ΔH2 +ΔH3 +ΔH4 ΔCPdT +ΔH4 Practice: How high could the temperature get if methane (CH4, g) deflagrate in air at 25℃ (CH4:O2=1:4molar ratio),p≈100kPa? T0=298.15K CH4(g)1mol,O2 (g) 4mol N2 (g) 15.05mol Qp=H=0 T CO2(g)1mol, H2O (g) 2mol O2 (g) 2mol,N2 (g) 15.05mol ( )p, adiabatic rHm H2 T0=298.15K CO2(g)1mol, H2O (g) 2mol O2 (g) 2mol,N2 (g) 15.05mol Qp= H = rHm + H2 =0 H 2 C p ,m CO2 , g 2C p ,m H 2 O, g 2C p ,m O 2 , g 15.05C p ,m N 2 , g dT T T0 T =1477K Home work A: P89: 3.12, 3.14, 3.24 Y: p45 : 48, 53 Preview: The second law of thermodynamics A: 4.1-4.4 Y:2.1-2.6 Practice 1: Calculation of work 1. A water bubble is expanded from a radius of 1.00cm to 3.25cm. The surface tension of water is 71.99N.m-1. 104 m 2 w d 2 4 (r ri ) 8 71.99 N .m (3.25 cm 1.00 cm ) cm 2 1.73J 2 f 2 1 2 2 2 2 2. A current of 3.20 A is passed through a heating coil for 30.0s.The electrical potential across the resistor is 14.5 V. The work done on the coil. w dQ Q It 14.5V 3.20 A 30.0s 1.39kJ 3. If the force to stretch a fiber a distance x is given by F=-kx with k=100 N.cm-1, how much work does it take to stretch the fiber 0.15cm? kx2 xf 100.0 N .m1 x 2 0.15 w F .dl kxdx [ ]x 0 [ ]0 1.1J 2 2 x0 xf Practice 2: Calculate △U and △H (1) 1.00 mol of H2O goes from 25.0℃ and 1.00 atm to 30.0 ℃and 1.00 atm. (density 0.9956 g.cm-3 and 0.9970 g.cm-3) △U=Q+W W=-P. △V=-0.0024J Q=75.295J. mol-1.K-1×5×1.0=376.475J T2 U C dT V T1 (2) 1.0 mol CaCO3 change in the form of calcite converts to aragonite. The density of the solids are 2.71 g.cm-3 and 2.93 g.cm-3. ΔH=ΔU+Δ(PV) PΔV=-0.3J ΔU=210J ΔU≈ΔH Discussions next class: Application of the first law Group 1: Understanding about the atmosphere and climate phenomenon (1) Marine climate/ Continental climate (2) Altitude/temperature Group 2: Is it possible water being used as fuel? Group 3: Food and energy reserves. Energy content and availability of the major food components percent availabi lity 92 type of food food ΔH°, kJ g–1 Protein meat 22.4 egg 23.4 butter 38.2 animal fat 39.2 starch 17.2 glucose (sugar) 15.5 99 ethanol 29.7 100 Fat Carbohydrate 95 Energy densities of some common fuels fuel MJ kg–1 wood (dry) 15 coal (poor) 15 coal (premium) 27 ethanola 30 petroleum-derived products 45 methane, liquified natural gas 54 hydrogenb