化学反应工程习题答案PPT幻灯片课件.ppt

<p>单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,习题,2.1,一个,75kg,的人大约每天消耗,6000kJ,热量的食物,假设食物为葡萄糖,其反应方程式为,:,C,6,H,12,O,6,+6O,2,=6CO,2,+6H,2,O -,H=2816kJ,如果人吸入的空气中含,CO,2,0.04%,、,O,2,20.96%,，呼出的气体中,CO,2,含量上升到,4.1%,、,O,2,含量降低到,16.1%,，请问人每千克体重代谢所消耗的空气的速率。,1,习题,2.1,C,6,H,12,O,6,+6O,2,+inert=6CO,2,+6H,2,O+inert,mol 20.96 79 0.04 79 t=0,mol 20.96(1-x)79 0.04+20.96x 20.96x 79 t=t,反应前后,O,2,与,CO,2,摩尔数之和应不变,则反应后总摩尔数为,:,21/20.2,100,=103.96(mol),O,2,浓度,:20.96(1-x)/103.96=16.1%,则转化率为,:x=0.201,反应需,O,2,:6000/2816/0.201,6,=63.6(mol),空气消耗速率,:63.6/0.2096/75=4.05 mol/kg,2,习题,2.1,如果该题目改为,:,人对反应的水份不产生影响,.,且不给出,CO,2,的出口浓度,则,:,重要知识点,:,关键组分的转化率,3,习题,2.2,在,210,等温条件下,进行亚硝酸乙酯的气相分解反应,该反应为一级不可逆反应,反应速率常数与温度的关系为,:,活化能,E,的单位为,cal/mol,若反应在恒容下进行,系统起始状态为,1at,的纯亚硝酸乙酯,试计算亚硝酸乙酯分解率为,80%,时,亚硝酸分解速率及乙醇的生成速率,.,4,习题,2.2,t (1-x)x 0.5x 0.5x,反应总压力,:P=1+x=1.8(at),知识点,:,反应速率,:,Arrhenius law:,5,习题,2.3,对于不可逆基元反应：,aA+bBcC+dD,其速率方程式可表示为：,r,A,=kC,A,a,C,B,b,如果以,A,为基准物，反应方程式两边除以,A,有计量系数,a,，方程式可写为,A+(b/a)B(c/a)C+(d/a)D,则其动力学方程按此方程式可写为,r,A,=kC,A,C,B,b/a,请问后一种速率方程式是否正确，为什么？,6,习题,2.3,通常对基元反应而言,可以说,:,单分子反应、双分子反应、三分子反应,什么是基元反应？,反应物分子在碰撞中一步直接转化为生成物分子的反应,.,Guldberg and WaageJ.Prckt.Chem.19,71(1879),指出：化学反应的速率与反应物的有效质量成正比,只有基元反应,在反应物浓度不太大的情况下才符合质量作用定律,7,习题,2.4,工业上以氮氧化物生产硝酸,氮氧化物由氨和空气气相氧化得到,4NH,3,+5O,2,4NO+6H,2,O,如果进气中含氨,15%,进气状态为,8.2at,227,.,试求,:,(1),进气总浓度和氨浓度,;,(2),恒压间歇反应器中,分别写出反应转化率对,pi,Ci,和体积,V,的函数,;,(3),恒容间歇反应器中,分别写出反应转化率对,pi,Ci,和总压,P,的函数,;,8,习题,2.4,进气总浓度,:,氨浓度,:C,A,=15%C=29.59(mol/m,3,),4NH,3,+5O,2,+inert 4NO+6H,2,O+inert,n,A0,n,B0,n,i0,n,P0,n,S0,n,i0,t=0,n,A,n,B,n,i0,n,P,n,S,n,i0,t=t,n,A,=n,A0,(1-x)n,B,=n,B0,-(b/a)n,A0,x,n,P,=n,P0,+(p/a)n,A0,x n,s,=n,s0,+(s/a)n,A0,x,膨胀因子,:,当每消耗,(,或生成,),1mol,反应物,A(,或产物,P),时,整,个物系总摩尔数的变化值,.,9,习题,2.4,n,t,=n,0,+,A,n,A0,x,恒容反应,:P=P,0,(1+0.0375x),恒压反应,:P=P,0,膨胀率,:,当反应物,A,全部反应后,系统体积的变化率,:,以膨胀率表征变容程度时,不但要考虑反应的计量关系,还要考虑系统内是否含有惰性气体,而以膨胀因子表达时,与惰性物料是否存在无关,.,10,习题,2.5,NO,和,O,2,氧化为,NO,2,的反应是一个三级反应,2NO+O,2,=2NO,2,在,30,及,1kg/cm,2,下测得其反应速率常数为,k,c,=2.65,10,4,L,2,/(mol,2,s).,如果将速率方程表示为分压的函数,:r,A,=k,p,p,NO,2,p,O,2,请问反应速率常数的,k,p,值和单位是什么,?,11,习题,2.5,反应速率常数的,k,p,值,:,知识点,:,化学反应速率常数,12,习题,2.6,有一反应,已知下列速率常数,:,温度,/,58.1,77.9,k/(1/h),0.117,0.296,求该反应的活化能和指前因子,13,习题,2.6,知识点,:,活化能、指前因子、,Arrhenius,方程,K,0,=1641236.4(1/h),14,习题,2.7,乙醇同乙酸在盐酸水溶液中的可逆酯化反应,CH,3,COOH+C,2,H,5,OH=H,2,O+CH,3,COOC,2,H,5,实验测得,100,时的反应速率常数为,:,r,A,=k,1,C,A,C,B,-k,2,C,p,C,s,k,1,=4.7610,-4,m,3,/(minkmol),k,2,=1.6310,-4,m,3,/(minkmol),今有一反应器,充满,0.3785m3,水溶液,其中含,CH,3,COOH90.8kg,含,C,2,H,5,OH181.6kg,所用盐酸浓度相同,假定在反应器中的水分不蒸发,物料密度恒定为,1043kg/m,3,求,:,(1),反应,120min,后,CH,3,COOH,转化酯的转化率,;,(2),忽略逆反应影响,反应,120min,后,乙醇的转化率,;,(3),平衡转化率,15,习题,2.7,CH,3,COOH+C,2,H,5,OH=H,2,O+CH,3,COOC,2,H,5,C,A0,C,B0,C,p0,t=0,C,A0,(1-x)C,B0,-C,A0,x C,p0,+C,A0,x C,A0,x t=t,16,习题,2.7,C,A0,=4.066mol/m,3,C,B0,=10.43mol/m,3,C,P0,=17.96mol/m,3,x=0.356,令,17,习题,2.7,(2):,忽略逆反应的影响,x=0.418,乙醇转化率,:(C,B0,-C,A0,x)/C,B0,=0.839,(3),平衡转化率,xe=0.5465,18,等温恒容不可逆反应的动力学方程,反应,动力学方程,动力学方程积分式,0,级,-r,A,=k,1,级,-r,A,=kC,A,2,级,-r,A,=kC,A,2,-r,A,=kC,A,C,B,(C,A0,=C,B0,),2,级,-r,A,=kC,A,C,B,(C,A0,C,B0,),3,级,-r,A,=kC,A,2,C,B,19,等温恒容可逆反应的动力学方程,反应,反应式,动力学方程积分式,1,级,A,P,2,级,A+B,P+S,(C,A0,=C,B0,),2,级,2A,2P,同上式,m=2,2,级,2A,P+S,同上式,m=1,2,级,A+B,2P,同上式,m=4,20,习题,2.10,乙炔与氯化氢在,HgCl,2,/,活性炭催化剂上合成氯乙烯的反应,C,2,H,2,+HCl=C,2,H,3,Cl,其动力学方程有以下几种可能的形式,:,(1)r=k(p,A,p,B,-p,c,/K)/(1+K,A,p,A,+K,B,p,B,+K,c,P,c,),2,(2)r=kK,A,K,B,p,A,p,B,/(1+K,A,p,A,)(1+k,B,p,B,+k,c,P,c,),(3)r=kK,A,p,A,p,B,/(1+K,A,p,A,+k,B,p,B,),(4)r=kK,B,p,A,p,B,/(1+k,B,p,B,+k,c,P,c,),试根据理想表面假设说明各式对应的机理假设及其控制步骤,21,习题,2.10,(1),机理,:A+,A B+B A+BC+,(,控制步骤,)CC+,k,aA,p,A,v,-k,dA,A,=0 ,A,=K,A,p,A,V,B,=K,B,p,B,V,C,=K,C,p,C,V,A,+,B,+,C,+,V,=1 ,v=1/(1+K,A,p,A,+K,B,p,B,+K,C,p,C,),r=k,+,A,B,-k-,C,V,r=k(p,A,p,B,-p,c,/K)/(1+K,A,p,A,+K,B,p,B,+K,c,P,c,),2,k=k,+,K,A,K,B,K=k,+,K,A,K,B,/(k,-,K,C),22,习题,2.10,(2),机理,:A+,1,A B+,2,B,2,A,1,+B,2,C,2,+,2,(,控制步骤,)C,2,C+,2,(2),机理,:A+,1,A B+,2,B,2,A,1,+B,2,C,2,+,2,(,控制步骤,)C,2,C+,2,k,aA,p,A,v1,-k,dA,A,=0 ,A,=K,A,p,A,V1,A,+,V1,=1 ,V1,=1/(1+K,A,p,A,),B,=K,B,p,B,V2,C,=K,C,p,C,V 2,B,+,C,+,V2,=1 ,v2,=1/(1+K,B,p,B,+K,C,p,C,),r=k,+,A,B,r=kp,A,p,B,/(1+K,A,p,A,)(1+K,B,p,B,+K,c,P,c,),k=k,+,K,A,K,B,23,习题,2.10,(3),机理,:A+,A B+B,A+BC+,(,控制步骤,),k,aA,p,A,v,-k,dA,A,=0 ,A,=K,A,p,A,V,B,=K,B,p,B,V,B,+,A,+,V,=1 ,v,=1/(1+K,B,p,B,+K,A,p,A,),r=k,+,A,p,B,r=k,+,K,A,p,A,p,B,/(1+K,B,p,B,+K,A,P,A,),24,习题,2.10,(4),机理,:,B+B,CC+,A+B C,(,控制步骤,),k,aB,p,B,v,-k,dB,B,=0 ,B,=K,B,p,B,V,C,=K,C,p,C,V,C,+,B,+,V,=1 ,v,=1/(1+K,B,p,B,+K,C,p,C,),r=k,+,p,A,B,r=k,+,K,B,p,A,p,B,/(1+K,B,p,B,+K,C,P,C,),25,习题,2.10,解题要点,:,根据吸附项的幂可以确定有几个活性点参加此控制反应,根据吸附项中是否出现开根号,可以判断是否有物质解离,根据推动力项中是否出现减号,可以判断是否是可逆反应,根据推动力项中反应物出现的幂数可以确定参加反应的物质的数量,26,习题,2.11,水蒸汽和一氧化碳变换反应,CO+H,2,O=CO,2,+H2,在以,Fe,3,O,4,为主剂的催化剂上进行的,Redox,反应,.,该反应的机理为,:,氧化过程,:H,2,O+,=O,+H,2,还原过程,:CO+O,=CO,2,+,在不同的温区,反应的控制步骤可能发生变化,.,试分别推导,:,(1),过程为氧化步骤控制时的动力学方程,(2),过程为还原步骤控制时的动力学方程,27,习题,2.11,(1),氧化步骤控制,还原反应采用拟稳态,:,k,2+,p,CO,O,-k,2-,p,CO2,V,=0,V,=Kp,CO,/(p,CO2,+Kp,CO,)K=k,2+,/k,2-,O,=p,CO2,/(p,CO2,+Kp,CO,),r=k,1+,p,H2O,V,-k,1-,p,H2,O,=(k,1+,Kp,H2O,p,CO,-k,1-,p,H2,p,CO2,)/(p,CO2,+Kp,CO,),28,习题,2.11,(2),还原步骤控制,氧化反应采用拟稳态,:,k,1+,p,H2O,V,-k,1-,p,H2,O,=0,V,=p,H2,/(K,1,p,H2O,+p,H2,)K,1,=k,1+,/k,1-,O,=K,1,p,H2O,/(K,1,p,H2O,+p,H2,),r=k,2+,p,CO,O,-k,2-,p,CO2,v,=(k,2+,K,1,p,CO,p,H2O,-k,2-,p,CO2,p,H2,)/(K,1,p,H2O,+p,H2,),29,习题,2.12,在氧化钽催化剂上进行乙醇氧化反应：,C,2,H,5,OH(A)+0.5O,2,(B),CH,3,CHO(L)+H,2,O(M),乙醇和氧分别在两类活性中心,1,和,2,上离解吸附，反应机理为：,C,2,H,5,OH+2,1,=C,2,H,5,O,1,+H,1,O,2,+2,2,=2O,2,C,2,H,5,O,1,+O,2,-,C,2,H,4,O+OH,2,+,1,OH,2,+H,1,=H,2,O,2,+,1,H,2,O,2,=H,2,O+,2,试推导该反应的动力学方程式。,30,习题,2.8,用纯组分,A,在一恒容间歇反应器中进行可逆反应,A=2.5P,实验测得反应体系的压力数据为,:,时间,/min,0,2,4,6,8,10,12,14,p,A,1,0.8,0.625,0.51,0.42,0.36,0.32,0.28,0.2,试确定该反应的速率方程式,31,习题,2.8,设反应为一级可逆反应,:,-r,A,=k,1,C,A,-k,2,(C,A0,-C,A,),积分得,:,作图,K=k,1,/k,2,=(C,A0,-C,Ae,)/C,Ae,=4,k,1,=0.129 k,2,=0.0322,-r,A,=0.129C,A,-0.0322(C,A0,-C,A,),32,习题,2.9,有一复杂反应,如果几个反应的指前因子相差不大,而活化能,E,1,E,2,E,1,E,3,E,4,E,3,.,应如何控制操作温度才能使产物,S,的收率增大,?,33,习题,2.9,r,s,=k,3,C,B,-r,A,=(k,1,+k,2,)C,A,34,物料流率,:,反应时间,:,所需反应器体积,:,习题,3.1,35,习题,3.2,等温操作的间歇反应器中进行一级液相反应,13min,后反应物转化率达到,70%.,如果分别在平推流和全混流反应器中进行此反应,达到相同转化率,所需空时多少,?,36,习题,3.2,解,:,一级反应,间歇过程,:,ln(1/(1-x)=kt,得,:k=0.0926,平推流与间歇釜相同,为,13min,全混流,:,t=x/k(1-x)=25.19(min),37,习题,3.3,38,习题,3.3,50%A,时,y,A0,=0.5,C,A0,=244.47 mol/m,3,=1,k=0.05676,单管流量,:4.7328m,3,/h,故需管数,:68,根,并联,39,习题,3.4,一级反应,停留时间,t=8.5S V,sp,=1/t=0.117,反应器体积,:V=2.8m,3,进料物流,:,40,3.5,时间,t=507.6min,平推流,V,0,=171.23ml/h,全混流,时间,t=2538.1min,反应器体积,:V=1448.67mL,反应器体积,:V=7243.4mL,41,习题,3.7,乙酸酐在,25,等温水解,加料速度为,500ml/min,反应速率为,rA=0.158CA,(1),以一个,5L,全混流或两个,2.5L,串联全混流,那一种情况的转化率大,(2),两个,2.5L,全混流并联,每个加料速率,250ml/min,转化率为多少,?,(3)2.5L,平推流和,2.5L,全混流串联,求转化率,(4)5L,平推流反应器操作,求转化率,?,42,习题,3.7,(1)5L,全混流,:,得,x,A,=0.6124,2.5L,两釜串联,:x,A1,=0.4413,x,A2,=0.6879,(2):,停留时间相同,故转化率相同,x,A,=0.6124,(3):kt=-ln(1-x,A1,),得转化率为,:x,A1,=0.546,x,A2,=0.7465,(4):kt=-ln(1-x,A,),得转化率为,:x,A,=0.794,43,习题,3.9,(1),平推流,:,全混流,:,(2),二级反应,:,平推流,:,全混流,:,44,习题,3.9,(2),级反应,平推流,:,全混流,:,(2)-1,级反应,:,平推流,:,全混流,:,45,习题,3.9,(3),一级反应为例,平推流,:,全混流,:,x,A,=0.6,k,m,=1.637k,p,T,m,=158.9,x,A,=0.9 km=3.908k,p,T,m,=175.6,46,习题,3.10,平推流,:,即,ABC,组分,B,的收率,:C,B,/C,A0,=47.73%,47,习题,3.10,全混流,:,即,ABC,组分,B,的收率,:C,B,/C,A0,=33.33%,48,3.11,A,与,B,停留时间相同,则,B,的转化率,X,B,=1/6,因而产品中,R,为,:75%,49,习题,3.12,(1)c(t)t,曲线,=5min,（,2,）,V=V,0,=2840L,t,0,1,2,3,4,5,6,7,8,9,10,C(t),0,0.1,0.2,0.3,0.4,0.5,0.4,0.3,0.2,0.1,0,E(t),0,0.04,0.08,0.12,0.16,0.2,0.16,0.12,0.08,0.04,0,F(t),0,0.02,0.08,0.18,0.32,0.5,0.68,0.82,0.92,0.98,1,50,习题,3.12,(3),平推流：,x,A,=0.857,(4),全混流：,x,A,=0.667,(5),完全离析流：,x,A,=0.8305,51,习题,3.12,(6),最大混合模型,Euler,式,转化率：,0.79,t,0.00,0.25,0.50,0.75,1.00,1.25,1.50,1.75,2.00,2.25,2.50,2.75,3.00,3.25,3.50,3.75,4.00,4.25,4.50,4.75,5.00,C(t),0.00,0.03,0.05,0.08,0.10,0.13,0.15,0.18,0.20,0.23,0.25,0.28,0.30,0.33,0.35,0.38,0.40,0.43,0.45,0.48,0.50,E(t),0.00,0.01,0.02,0.03,0.04,0.05,0.06,0.07,0.08,0.09,0.10,0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18,0.19,0.20,F(t),0.00,0.00,0.01,0.01,0.02,0.03,0.05,0.06,0.08,0.10,0.13,0.15,0.18,0.21,0.25,0.28,0.32,0.36,0.41,0.45,0.50,xi,0.79,0.78,0.77,0.76,0.75,0.73,0.72,0.71,0.70,0.69,0.68,0.66,0.65,0.64,0.62,0.61,0.59,0.58,0.56,0.55,0.54,5.25,5.50,5.75,6.00,6.25,6.50,6.75,7.00,7.25,7.50,7.75,8.00,8.25,8.50,8.75,9.00,9.25,9.50,9.75,10.00,0.48,0.45,0.43,0.40,0.38,0.35,0.33,0.30,0.28,0.25,0.23,0.20,0.18,0.15,0.13,0.10,0.08,0.05,0.02,0.00,0.19,0.18,0.17,0.16,0.15,0.14,0.13,0.12,0.11,0.10,0.09,0.08,0.07,0.06,0.05,0.04,0.03,0.02,0.01,0.00,0.55,0.60,0.64,0.68,0.72,0.76,0.79,0.82,0.85,0.88,0.90,0.92,0.94,0.96,0.97,0.98,0.99,1.00,1.00,1.00,0.53,0.52,0.51,0.50,0.49,0.48,0.47,0.45,0.44,0.42,0.40,0.38,0.35,0.33,0.29,0.24,0.40,-0.15,0.30,0.00,52,习题,3.14,53,习题,3.14,转化率：,54,习题,3.15,解,:(1):,55,习题,3.15,56,习题,3.15,(3)F(t)|,t=230-270,=0.37,(4):F(t)|,t=250,=0.42,(5):,(7),方差,:,(8):,平推流,:,2,=0,全混流,:,2,=,2,实际反应器,0,2,2,2,=6844234,57,习题,4.9,在直径,6mm,的球形催化剂上进行一级不可逆反应,.C,As,=9.510,-5,mol/cm,3,T,s,=623K,k(T,s,)=11.8s,-1,.(-,H,r,)=3.1510,4,cal/mol,E=2.4810,4,cal/mol.,催化剂有效导热系数,4.810,-4,cal/(cmsK),De=1.510,-2,cm,2,/s.,(1),颗粒中心与催化剂外表面的最大温差,(2),等温处理的效率因子,(3),非等温处理的效率因子,58,习题,4.9,(1):,T-T,s,=(-,D,H,r,)D,e,(c,As,-c,A,)/,e,=93.52K,(2):,=(K,v,/D,e,),0.5,=2804.8,=,R/3=2.8,=0.314,(3):,查图,4.6,=20.=0.15,=2.8,=1.3,非等温,效率因子可大于,1.,差异非常大,59,习题,4.13,900,1atm,含,O,2,8%,的气体焙烧球形锌矿,.,2ZnS+3O,2,2ZnO+2SO,2,反应按收缩未反应芯模型,.k=2cm/s,B,=4.13g/cm,3,=0.0425mol/cm,3,.,De=0.08cm,2,/s.,若气膜阻力可以忽略,试分别计算,r,s,为,1mm,和,0.05mm,时颗粒完全反应时间及反应过程中产物层内扩散阻力的相对大小,60,习题,4.13,考虑表面化学反应和内扩散阻力,:,结合,:,可得,:,61,习题,4.13,分别得到完全反应的时间为,:,5432.7s,和,195.7s,阻力比,:,取积分平均,:,因而可得,:,阻力比分别为,0.417,和,0.021,62,习题,4.14,1atmH,2,还原铁矿,4H,2,+Fe,3,O,4,4H,2,O+3Fe,反应可近似用收缩未反应芯模型,.,反应速率近似正比于气相中,H,2,的浓度,.,k=1.9310,5,exp(-2400/RgT)cm/s.,r,S,=5mm,B,=4.6g/cm,3,De=0.03cm,2,/s,忽略气膜阻力,有无其它阻力控制,计算,500,时的完全反应时间,63,习题,4.14,k=40516,得总反应时间为,6986.7s,按内扩散计算为,:6986.67s,阻力比为,:112545,内扩散控制,64,习题,5.2,在连续流动反应器中,基元反应,:A+B,2C,A,、,B,等物质进料，体积流量为,2L/s,27,进料。,H,A273K,=-20kcal/mol,H,B273K,=-15kcal/mol,H,c273K,=-41kcal/mol,C,A0,=0.1mol/L,C,p,A,=C,p,B,=15cal/(molK),C,p,c,=30,300K,时，,k=0.01L/(molS)E=10000cal/mol.,(1)85%,转化率，平推流、全混流绝热反应器体积,（,2,）全部转化，反温度不超过,550K,，进料温度,（,3,）平推流的转化率、温度变化曲线,（,4,）,500L,绝热全混流反应器的转化率,65,习题,5.2,解：,(1),全混流反应器：,H,r,273K,=2,Hc,273,K-H,A,273K,-H,B,273K,=-47kcal/mol=-196.46kJ/mol,C,p,=2C,p,c,-C,p,B,-C,p,A,=30cal/(molK)=125.4 J/(molK,),H,r,(T,)=,H,r,(T,r,)+,C,p,(T-T,r,)=-196.4610,3,+125.4(T-273),F,0,C,p0,=F,A0,C,p,A,+F,B0,C,p,B,+F,C0,C,p,C,=20.1215=6cal/(sK)=25.08 J/(sK),66,习题,5.2,FC,pt,=F,A0,(1-x,A,)C,p,A,+F,B0,(1-x,A,)C,p,B,+2F,A0,x,A,C,p,C,=20.1215+(220.130-220.115)x,A,=6+6x,A,cal/(sK)=25.08(1+x,A,)J/(sK),全混流热衡算：,FC,Pt,T-F,0,CP,0,T,0,=-,H,r,(T)F,A0,x,A,T=690.3K,K=k,300,exp(-E/RT)/exp(-E/RTr)=130.33,L/(mol,s),r,A,=kC,A,C,B,=kC,A,2,=kC,A0,2,(1-x,A,),2,=0.029(mol/Ls),V=v,0,(c,A0,-c,A,)/r,A,=v,0,C,A0,x,A,/r,A,=5.8L,67,习题,5.2,管式反应器,K=k,300,exp(-E/RT)/exp(-E/RTr),FC,Pt,T-F,0,CP,0,T,0,=-,H,r,(T)F,A0,x,A,积分可得反应器体积：,62L,为什么全混流体积比平推流小？,68,习题,5.2,(2),进料最高温度,FC,Pt,T-F,0,C,P0,T0=-,Hr(T)F,A0,x,A,其中,:T=550K,x,Af,=1,则求得,T,0,=-189.7,69,习题,5.2,70,习题,5.2,(4)FC,Pt,T-F,0,CP,0,T,0,=-,H,r,(T)F,A0,x,A,K=k,300,exp(-E/RT)/exp(-E/RTr)=130.33,L/(mol,s),r,A,=kC,A,C,B,=kC,A,2,=kC,A0,2,(1-x,A,),2,V=v,0,(c,A0,-c,A,)/r,A,=v,0,C,A0,x,A,/r,A,=500L,设一转化率,求温度,T,得到,r,A,再求,V,试差计算,得到最终转化率为,:0.984,71,习题,5.9,在平推流、全混流反应器中进行基元气相反应，进料温度,27,，含,A80%,，其余为惰性组分，进料速率,100L/min,，,A,进料浓度,0.5mol/L,，达到,80%,的绝热平衡转化率。,（,1,）反应在平推流反应器中绝热进行，计算反应器体积。,（,2,）如果平推流反应器直径,5cm,，做出反应转化率和温度沿反应管长度的变化关系。,（,3,）反应在全混流反应器中绝热进行，计算反应器体积。,（,4,）如果反应在平推流反应器中进行，不绝热，反应器直径,5cm,，环境温度,27,，总传热系数,h=10w/(m2K),，计算反应转化率、温度沿反应管长度方向的变化关系。,物化数据为：,C,p,A,=12J/(molK),，,C,p,B,=10J/(molK),，,C,p,l,=12J/(molK),，,300K,时反应热,H,r,=-75000J/molA,，,300K,时,k,1,=0.217min,-1,K,e,=70000mol/L,，,k,1,随温度变化，,340K,时，,k,1,=0.324min,-1,。,72,习题,5.9,反应,A,2B,初始反应物量：,n,A0,C,A0,惰性组分：,0.2F,0,=0.25n,A0,反应达到平衡时：,n,Ae,n,B,=2(n,A0,-n,Ae,),平衡时的总摩尔数,:,n,Ae,+2(n,A0,-n,Ae,)+0.25n,A0,=2.25n,A0,-n,Ae,平衡常数,73,习题,5.9,而,:,由,Vant Hoff,方程有,:,平衡时的体积流量：,平衡浓度：,74,习题,5.9,平衡常数,热衡算,:,设定平衡转化率,热衡算得温度,再比较平衡常数,试差得,:0.0774,75,习题,5.9,(1),平推流反应器体积的计算,活化能的计算,:,k,1,=k,2,exp(-E/RT,1,)/exp(-E/RT,2,)=k,2,exp(E/R(1/T,2,-1/T,1,),E=Rln(k,1,/k,2,)/(1/T,2,-1/T,1,)=8498 J/mol,温度与反应速率常数的关系,:,k=k,1,exp(-E/RT)/exp(-E/RT,1,),反应速率,:,r,A,=kC,A,=1.25kC,A0,(1-x,A,)/(1.25+x,A,)T,0,/T,利用数值积分可求得,V=2.66L,76,习题,5.9,全混流反应器,:,V=v,0,(c,A0,-c,A,)/r,A,=v,0,C,A0,x,A,/r,A,=1.75L,77,</p>