物理化學(xué)最新版本

1、整理課件1例題 例3 . 在一抽成真空的容器中單獨放有NH4Cl(s)，在340oC時分解反應(yīng)NH4Cl(s)=NH3(g) + HCl (g)的平衡壓力P1=104.6kPa；同樣條件下，若單獨放有NH4I(s), 分解反應(yīng)NH4I(s)=NH3(g) + HI (g) 的平衡壓力P2 =18.85kPa。求NH4Cl(s)與NH4I(s)兩種純固態(tài)物質(zhì)同時存在，加熱到340oC ，當(dāng)上述兩個反應(yīng)都達到平衡時，系統(tǒng)的總壓為多大? 設(shè)HI(g)不分解，兩個鹽類不形成固態(tài)溶液。整理課件2Answer 1NH4Cl(s) alone: NH4Cl(s)=NH3(g)+HCl(g) P(NH3)=P

2、(HCl)=P1/2NH4I(s) alone: NH4I(s)=NH3(g)+HI(g) P(NH3)=P(HI)=P2/2 23112()()1()0.2738()4P NHP HClPKPP23222()()1()0.0089()4P NHP HIPKPP整理課件3 When NH4Cl (ammonium chloride)and NH4I (ammonium iodide) coexist in the system, the partial pressure(分壓) of ammonia(NH3) gas must accord with both two equilibrium

3、mentioned above. If the partial pressure is P(NH3) for NH3(g), P(HCl) for HCl(g) and P(HI) for HI, So (1) (2) and P(NH3) = P(HCl)+P(HI) (3) The answer is P(total) = P(NH3)+P(HCl)+P(HI) =106.3kPa312()()0.2738()P NHP HClKP322()()0.0089()P NHP HIKP整理課件4Answer 2 Let x and y denote the amounts of HCl and

4、 HI when the two equilibrium coexisted in the reaction system. So the amount of NH3 in the system must be x+y and the total amounts of three gas is 2(x+y).Thus, the partial pressure of NH3 is and 3()()/22()totaltotalxyP NHPPxy()2()totalxP HClPxy()2()totalyP HIPxy整理課件5(we can see the partial pressure

5、 of P(NH3) = P(HCl) + P(HI) ) X=30.83y P(total)=106.3kPa12(/2)2()0.2738()totaltotalxPPxyKP22(/2)2()0.0089()totaltotalyPPxyKP整理課件6Key problem in this example 1. The partial pressure is a intensive variable(強度變量）, and the amount of species is extensive variable(廣度變量） 2. Equilibrium constant for hetero

6、geneous reactions(復(fù)相反應(yīng)的平衡常數(shù)) 3. The calculate rule for the equilibria coexisted in a system(同時平衡的計算原則） 整理課件7Review 1 The criterion of a spontaneous chemical reaction 2 The description of an equilibrium of reaction 3 The isothermal(constant temperature) equation 整理課件8Review 1 Several symbols1The Gibb

7、s energy of a reaction system, G 2The reaction Gibbs energy: rGm = (G/)P,T3 The standard reaction Gibbs Energy4 is the Gibbs energy of mixing A and B5 Standard Gibbs energy of formation rmGmixGfmG整理課件92 Several relations: 1 relation of reaction Gibbs energy (rGm) to the chemical potentials of the sp

8、ecies, rGm =products - reactants2 The standard equilibrium constant (K) and the standard reaction Gibbs energy = -RT In Ko3 general expression for rGm at an arbitrary stage of the reaction, rGm = + RT In QrmGrmG整理課件10Review5thermodynamic equilibrium constant, (熱力學(xué)平衡常數(shù))can expressed in terms of activ

9、ities (or fugacities): K for Perfect gas reaction, solution reaction and heterogeneous reaction has different form. 6 in terms of standard Gibbs energies of formation, tanoormfmfmproductsreactsGvGvG()JvJequilibriumJKarmG整理課件11The content for this class1The standard Gibbs energy of reaction in soluti

10、on 2 The response of equilibria to the conditions 1 How equilibria respond to pressure? 2The effect of inert gas on equilibria 3 Le Chateliers principle 4The response of equilibria to temperature (van t Hoff equation)3 The coexist equilibrium整理課件121 The standard Gibbs energy of reaction in solution

11、(1) Because the value of standard Gibbs energy formation( 標(biāo)準(zhǔn)生成吉布斯自由能)is for a pure species, so the value of the standard reaction Gibbs energy(反應(yīng)的標(biāo)準(zhǔn)吉布斯自由能變) indicated that the standard state of each reactant and product is in the pure state. rmifmiGvGrmG整理課件13 The standard state for the species in s

12、olution reaction maybe not the pure state. (1) The solution need not distinguish solvent (溶劑)and solute(溶質(zhì)) Equation (1) can be used to calculate the (2) The solvent and solute need distinguish and the solution is dilute(稀溶液). The standard state of each reactant and product is ci = 1mol.dm-3 and the

13、 solution submit Henrys law hypothetical.rmG整理課件14Example Calculate the ionization constant （電離常數(shù)）for benzoic acid C6H5COOH in solution at 298 K. C6H5COOH(aq.)=C6H5COO-(aq.) + H+(aq.) Given that for C6H5COOH in the solid state, and for C6H5COO-(C = 1.0 mol.dm-3). The solubility of C6H5COOH in water

14、is 0.02787 mol.dm-3 at 298K.1(298)245.27.fmKGkJ mol 1(298)223.84.fmKGkJ mol 整理課件152 The response of equilibria to the conditions1How equilibria respond to pressure (P234): The standard equilibrium constant is independent of the pressure, but the equilibrium composition depends on the pressure.(壓力對化學(xué)

15、平衡的影響，實際上并非是對平衡常數(shù)的影響，而是指對平衡移動的影響）()0TKP1/21()14/P K整理課件16 Le Chateliers principle: A system at equilibrium, when subjected to a disturbance, responds in a way that tends to minimize the effect of the disturbance. The principle implies that if a system at equilibrium is compressed, then the reaction

16、will adjust so as to minimize the increase in pressure.ln()iximTvKVPPRT 整理課件172 effect of inert gas(惰性氣體) on equilibrium 1. If an inert gas is added to an equilibrium mixture of gases at constant temperature and volume, there is no effect on the equilibrium. Because the addition of an inert gas leav

17、es the molar concentrations of the original gases unchanged, as they continue to occupy the same volume.整理課件18 2. If adding an inert gas at constant temperature and pressure which has the same effect as lowering the original gas partial pressure. If , the addition of an inert gas at constant pressur

18、e reduces the sum of the partial pressure of the reactants and products, and the reaction shifts to the left to compensate for this. (/)()iiifgfgvvvFGFGPxdedeDEDEiinertsix xn nPKK PP Px xn nPnn0iv 整理課件19Example At 1273 K and a total pressure of 30.4 atm the equilibrium in the reaction CO2(g) + C(s)

19、= 2CO(g) is such that 17 mol % of the gas is CO2. (a) What would be CO2 if the total pressure were 20.3 atm? (b) What would be the effect on the equilibrium of adding N2 to the reaction mixture in a closed vessel until the partial pressure of N2 is 10 atm? If the reactant gases contain a mole of N2

20、in addition to 1 mole of CO2, what is the equilibrium extent of reaction at 30.4 atm? (d) At what pressure of the reactants will 25% of the gas be CO2?整理課件204 The response of equilibria to temperature According to Le Chateliers principle, a system at equilibrium will tend to shift in the endothermic

21、 direction if the temperature is raised, while the equilibrium tend to shift in the exothermic direction if the temperature is lowered.整理課件21The van t Hoff equation The van t Hoff equation is an expression for the effect of temperature on the standard equilibrium constant Ko. Here is the standard reaction enthapy at the temperature T.2ln()rmPHKTRTrmH整理課件22the value of K at different temperature General, the integral(積分) of Vant Hoff Equation is used for gas reaction which the pressure is low. (1) If varies only slightly with temperature over the temperature range of interest,