32.5y = 0.0067x - 0.0276R2 = 0.99882nL1.510.500100200t300400500
Slope?2k11(X?1)CAo?0.0067, Ae?k1?0.0067?6.794?10?4L/(mol?min)2(1
0.5273?1)?5.5When approach to equilibrium, Kk1C2Rec?k?, 2CAeCBeso kk1CAeCBe6.794?10?4?2.622?C2?82?1.364?10?4L/(mol?min) Re5.So the rate equation is ?r(6.794?10?4C2A?ACB?1.364?10?4CR)mol/(L?min)
ii) We postulate it is a 1st order reversible reaction system, so the rate equation is ?rA??dCAdt?k1CA?k2CR After rearranging and integrating, we obtain
Ln(1?XAX)?1Xk'1t eq (2) AeAeDraw Ln(1?XAX)~ t plot, we obtain another straight line: Ae 15
32.52-y = 0.0068x - 0.0156R2 = 0.9986-Ln1.510.500100200x300400500
k1'Slope???0.0068,
XAeSo k1'??0.0068?0.5273??3.586?10?3min?1
k1'CAe?3.586?10?3?2.6k????1.607?10?3min?1
CRe5.8'2So the rate equation is
?rA?(?3.586?10?3CA?1.607?10?3CR)mol/(L?min)
We find that this reaction corresponds to both a 1st and 2nd order reversible reaction system, by comparing eq.(1) and eq.(2), especially when XAe =0.5 , the two equations are identical. This means these two equations would have almost the same fitness of data when the experiment data of the reaction show that XAe =0.5.(The data that we use just have XAe =0.5273 approached to 0.5, so it causes to this.)
3.24 In the presence of a homogeneous catalyst of given concentration, aqueous reactant A is converted to product at the following rates, and CA alone determines this rate:
CA,mol/liter -rA, mol/liter·hr
1 0.06
2 0.1
4 0.25
6 1.0
7 2.0
9 1.0
12 0.5
We plan to run this reaction in a batch reactor at the same catelyst concentration as used in getting the above data. Find the time needed to lower the concentration of A from CA0 = 10 mol/liter to CAf = 2 mol/liter.
Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.
16
20 16 12 -1/Ra
8 4 0 0
2
4
6
Ca
8
10
12
14
3.31 The thermal decomposition of hydrogen iodide 2HI → H2 + I2
is reported by M.Bodenstein [Z.phys.chem.,29,295(1899)] as follows:
T,℃
508
427
393
356
283
k,cm3/mol·0.1059 0.00310 0.000588 80.9×10-6 0.942×10-6 s
Find the complete rate equation for this reaction. Use units of joules, moles, cm3, and seconds.
According to Arrhenius’ Law,
k = k0e-E/R T
transform it,
- In(k) = E/R·(1/T) －In(k0)
Drawing the figure of the relationship between k and T as follows:
16y = 7319.1x - 11.56712R = 0.98792-Ln(k)8400.0010.0021/T0.0030.004From the figure, we get
slope = E/R = 7319.1 intercept = - In(k0) = -11.567
17
E = 60851 J/mol k0 = 105556 cm3/mol·s
From the unit [k] we obtain the thermal decomposition is second-order reaction, so the rate expression is
- rA = 105556e-60851/R T·CA2
18
Chapter 4 Introduction to Reactor Design
4.1 Given a gaseous feed, CA0 = 100, CB0 = 200, A +B→ R + S, XA = 0.8. Find XB,CA,CB. Solution: Given a gaseous feed, CAo?100, CBo?200, A?B?R?S
XA?0, find XB, CA, CB
?A??B?0, CA?CAo(1?XA)?100?0.2?20
XB?bCAoXA1?100?0.8??0.4 CBo200CB?CBo(1?XB)?200?0.6?120
4.2 Given a dilute aqueous feed, CA0 = CB0 =100, A +2B→ R + S, CA = 20. Find XA, XB, CB.
Solution: Given a dilute aqueous feed, CAo?CBo?100,
A?2B?R?S, CA?20, find XA, XB, CB
Aqueous reaction system, so ?A??B?0 When XA?0, V?200 When XA?1, V?100
?C11So ?A??, ?B?ABo??
2bCAo4XA?1?CA20?1??0.8, CAo100XB?bCAoXA2100?0.8????1.6?1, which is impossible. aCBo1100So XB?1, CB?CBo?100
4.3 Given a gaseous feed, CA0 =200, CB0 =100, A +B→ R, CA = 50. Find XA, XB, CB. Solution: Given a gaseous feed, CAo?200, CBo?100,
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