化工热力学答案 下载本文

解之: nA =0.0348 mol

此时的转化率为:xA= nA /nA0=0.0348/0.4=0.087 又因为等压变容条件下

rA=1?CA0?AdxA?1?xA?2?kCA0?? , 积分得 dtAxA??1?2?(1?A)xA?

1?xA 而

CA0??Aln(1?xA)?ktCA0 (A)

?P0.4A0??7.248×10-3mol/L RT0.082(273?400)所以:

(1?0.4)×0.087-3

?0.4ln(1?0.087)?2×7.248×10t

1?0.087解之 t=6.453min

注:可利用该题的过程来确定气相变摩尔数的反应动力学常数,假设该反应为二级,测定不同时间t下的转化率xA,公式(A) 的左端与t应成直线关系,其斜率为

kCA0,在已知

CA0时,可确定k.若为非直线关系,二级假设不正确。该气相反应也可

在实验室在恒容变压条件下进行测定其动力学参数,由PA=CART,用A的分压或浓度与时间关系求。

5. [等温等容一级可逆过程]

可逆一级液相反应A←→P,已知CA0=0.5mol/L,Cp0=0,当此反应在间歇反应器中进行时,经过8min后,A转化了33.3%,而平衡转化率为66.7%。求此反应的动力学方程式。

dCA?k1CA?k2CP?k1CA?k2(CA0?CA) 解: rA=?dt 初始条件:t=0 ,CA?CA0, Cp0?0,

CA0?CAexAe?ln 积分得:(k1?k2)t?ln

CA?CAexAe?xACAek2? 这里 CA0k1?k2

10.667?0.08645/min (A) 所以 k1?k2?ln80.667?0.333 5

xAek1CA0?CAe0.667???2 (B) 平衡常数K=?k2CAe1?xAe1?0.667联立解(A)(B)得 k1=0.05764min-1 , k2=0.02881min-1 所以,此动力学方程式为:

rA=0.05764CA-0.02881CP=0.05764CA-0.02881(CAO-CA)=0.08645CA-0.01445 (mol/min.L)

注:若为一级不可逆反应,k2=0,rA= k1CA , CAe=0 , 积分式可简化为 k1t?lnCA01?ln ,只要给定一点CA1?xA(t,xA)值,就可确定速率常数k1,当然测定点越多,求得的k1值 就越准确。 6.[等容变温确定活化能过程]

某工厂在间歇反应器中进行液相反应的两次实验,初始浓度相同,达到的转化率相同。第一次20℃,8天;第二次120℃,10min,求活化能E

dC解:rA= ?dtAnkC=A

积分

?CACA0?nCAdCA=??kdt

0t1?n1?n(C1?C)=kt=k0eRTt有 A0A1?n?E

对特定的反应及初始浓度和转化率,左端为一常数, 代入T1=273+20=293K , T2=273+120=393K t1=8×24×60min ,t2=10min

所以:8×24×60k0e?E293R=10k0

e?E393R

解出 E=[8.134ln(48×24)]/(1/293-1/393)=67486J/mol

注:对于具体的反应和级数也可用该方法,至少用2个以上的实验点来测定它的活化能。

7、 【温度对反应速率的影响】

某反应的活化能为7000J/mol,在其它条件不变时,问650℃的反应速度要比500℃时快多少? 解:k=k0e-E/RT ,E=7000J/mol,R=8.314Jmol-1K-1

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k1/k2 = -E/k(1/T1 – 1/T2) = -7000/8.314(1/273+650 – 1/273+500)=1.19 ∴-r650/-r500 = k1/k2 = 1.19 8、 [等温等压变容一级过程]

气相一级分解反应 A → 3P ,在等温活塞流管式反应器中进行,加入原料含A50%,惰性物50%,停留时间为10分钟,系统出口的体积流量为原来的1.5倍,求此时A的转化率及该反应速度方程。

解:A 3P 惰性物

1 0 1 ε0 3 1 ∵ V = V0(1+ε∴ xAf = 1/ε

A =(3+1)-(1+1)/1+1 = 1

τ= 10min

AxAf)

A (V/V0-1) = 1/1(1.5-1)=50%

AxAf)xAf /(1-xAf) = (1+0.5)*0.5/(1-0.5) = 1.5*0.5/0.5 = 1.5

而 kτ= (1+ε

∴ k = 0.15min-1

∴ -rA = 0.15CA(mol/l*min)s

9、 A reaction with stoichiometric equation 1/2A + B → R + 1/2S, has following rate expression -rA = q CA0.5CB

what is the rate expression for this reaction if the stoichiometric equation is written as A + 2B = 2R + S?

Answer : -rA = q CA0.5CB

10、 For the enzyme-substrate reaction of example

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A + E ???1 X X ??3? R + E

The rate of disappearance of substrate is given by -rA = 1760[A][E0]/(6+CA) mol/m3s what are the units of the two constants? Answer :1760m3/mol*s 6mol/m3

、 For a gas reaction at 400K the rate is reported as –dPA/dt = 3.66PA2 atm/hr a) what are the units of the rate constants ?

b) what is the value of the rate constants for this reaction is the rate equation is expressed as -r1dNvAA = -dt = kCA2 mol/m3s

answer : -d(CART)dT = 3.66 CA2(RT)2

= (3.66RT) *CA2 (mol/m3*s) ∴k = 3.66RT

= 3.66*400*0.082* atm*l*k-1atm-1hr-1mol-1 = 3.66*400*0.082*10-3 m-3/3600(s*,mol) =3.66*0.4*0.082/3600(m3/mol*s) = 3.33 m3/mol*s

、 If –rA = -dCA/dt = 0.2mol/l*s when CA = 1mol/l what is the rate of the reaction when CA = 10

mol/l?

Answer : 不确定,级数未知。

、 After 8 minutes in a batch reactor ,reactant (CA0 = 1mol/l) is 80% converted ,after 18

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