发布时间 : 星期六 文章化工热力学答案更新完毕开始阅读b8c4b68029ea81c758f5f61fb7360b4c2e3f2ac4
minutes ,conversion is 90% Find a rate equation to represent this reaction.. Answer : n = 2级。
14、 Aqueous A at a concentration CA0 = 1mol/l is introduced into a batch reactor where it reacts
away to form product R according to stoichiometriy A → R. The concentration of A in the reactor is monitored at various time ,as shown below: T min 0 100 200 300 400 CA mol/l 1000 500 333 250 200
For CA0 = 500mol/m3 find the conversion of reactant after 5 hours in the batch reactor Answer :CA-1 10-3 2*10-3 3*10-3 4*10-3 5*10-3 二级。 1/ CA = 1/ CA0+ kt 斜率 k = (5-2)* 10-2/300 = 10-5m3/mol min 1/ CA = 2*10-3+10-5*5*6= 5*10-3 ∴ CA = 0.2*103 = 200mol/m3.
xA = CA0 - CA /CA0 = 500 – 200 /500 = 60%.
15. Find the first order rate constant for the disappearance of A in the gas reaction on holding the pressure constant ,the volume of the reaction ,starting with 80%A, decreases by 20% in 3min .
answer :2A → R 惰性物 nt
0.8 0 0.2 1 0 0.2 0.2 0.6 ε
A = (0.6-1)/1 = -0.4
V = V0 (1+εAxA) = V0 (1-0.4xA)
V/ V0 – 1 =
V?V0V0 = -0.4 xA = -0.2
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xA = 0.5 ㏑1?1x = kt
A㏑1?10.5= ㏑2 = 3k k = 1/3㏑2 min-1
1. 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？（5分）
解：k=k0e-E/RT ，E=7000J/mol，R=8.314Jmol-1K-1
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
2. 可逆一级液相反应A?R ，已知A的初始浓度为0.5mol/l，初始产物浓度为零。此反应在间歇反应器中，经过8分钟后A转化了33.3％，而平衡转化率为66.7％，试求正逆反应的速度常数和该反应的平衡常数。（10分）
解： A?R CA0=0.5mol/l, t=8min, xA=33.3%. xAe=66.7%
C㏑(k?K?k1(1?k1)t??(K1)C?CCa0cAA0CKCC?1)(1?xA)?1
KC =
xae1?xAek?1?00.667..667?2?k
12∴ ㏑(2?1)(1?20.333)?1=k1(1+1/2)*8 解得k1=0.05776min-1 ∴ k2= 1/2k1=0.02888min-1
3. 某气相一级分解反应 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
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及该反应速度方程。（10分） 解：A 3P 惰性物
1 0 1 εA =（3+1）-（1+1）/1+1 = 1
0 3 1 τ= 10min
∵ V = V0(1+εAxAf)
∴ xAf = 1/ε
A (V/V0-1) = 1/1(1.5-1)=50%
而 kτ= (1+ε
AxAf)xAf /(1-xAf) = (1+0.5)*0.5/(1-0.5) = 1.5*0.5/0.5 = 1.5
∴ k = 0.15min-1
∴ -rA = 0.15CA(mol/l*min)s
4. 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
5. For the enzyme-substrate reaction of example
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
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6. 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 -rA = -vanswer : -d(CART)dT1dNAdt = kCA2 mol/m3s
= 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
7. 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。
8. After 8 minutes in a batch reactor ,reactant (CA0 = 1mol/l) is 80% converted ,after 18 minutes ,conversion is 90% Find a rate equation to represent this reaction.. Answer : n = 2级。
9. Aqueous A at a concentration CA0 = 1mol/l is introduced into a batch reactor where it reacts away to form product R according to stoichiometriy A → R. The concentration of A in the reactor is monitored at various time ,as shown below:
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T min 0 100 200 300 400 CA mol/l 1000 500 333 250 200
For CA0 = 500mol/m3 find the conversion of reactant after 5 hours in the batch reactor Answer :CA-1 10-3 2*10-3 3*10-3 4*10-3 5*10-3 二级。 1/ CA = 1/ CA0+ kt 斜率 k = (5-2)* 10-2/300 = 10-5m3/mol min 1/ CA = 2*10-3+10-5*5*6= 5*10-3 ∴ CA = 0.2*103 = 200mol/m3.
xA = CA0 - CA /CA0 = 500 – 200 /500 = 60%.
10. Find the first order rate constant for the disappearance of A in the gas reaction on holding the pressure constant ,the volume of the reaction ,starting with 80%A, decreases by 20% in 3min .
answer :2A → R 惰性物 nt
0.8 0 0.2 1 0 0.2 0.2 0.6 ε
A = (0.6-1)/1 = -0.4
V = V0 (1+εAxA) = V0 (1-0.4xA)
V/ V0 – 1 = V?V0V0 = -0.4 xA = -0.2
xA = 0.5 ㏑11?xA = kt
㏑11?0.5= ㏑2 = 3k k = 1/3㏑2 min-1
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