物化习题答案1

(k1?k?1)?? (k1?k?1)?? (k1?k?1)?? (k1?k?1)?? (k1?k?1)?? (k1?k?1)??k1k?1x?1?x?1211361501651801lnlnlnlnln0.7285?0.13220.72850.7285?0.20460.72850.7285?0.27210.72850.7285?0.33460.72850.7285?0.38840.72850.7285?0.44490.7285?9.535?10?9.158?10?9.38?10?9.46?10?9.52?10?3min?1

?3min?1?1?3min

?1?3minmin?3?1?3?1100ln?9.434?10min

(k1+k-1) (平均)=9.41×10-3min-1

???2.683

1?0.72850.7285 解得：k1 =6.87×10-3min-1，k-1=2.56×10-3min-1 6. 解： cAs=

7034Se???33As???32Ge

??70??70 k1 k2

0.0158n00.0133?0.0158?e?0.0158?10?e?0.0133?10?=0.137n0

P345

2. k1 k-1 k2

A + H+ AH+，AH+ A + H+，AH+ + B P + H+ (1). 用稳态法： cAH??dcAH?dt?k1cAcH??k?1cAH??k2cAH?cB?0

k1cAcH?k?1?k2cB，

dcPdt?k2cAH?cB?k1k2cAcBcH?k?1?k2cB

(2). 当cB很大时，k2cB>> k-1， 当cB很小时，k2cB<< k-1， (3). 用平衡近似法：

dcPdtdcPdtdcPdt?k1cAcH? ?k1k2k?1cAcBcH?

?k2cAH?cB

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K =

k1k?cAH?，

dcPcH?

?1cAcH?dt?k1k2kcAcB?1 与cB很小时，k2cB<< k-1结果一致。 3. 2A+B=2D 历程为：

k1

A + A P Ek1 -1

P 2A E-1 K2

P + B 2D E2 (速率控制步) (1). r = kk2 cPcB ，用平衡近似法：K = 1k?cP2

?1cA r =

k1k222k1k2kcAcB?kcAcB，k?1 + E2 - E-1

?1k，Ea = E?1 (2). k?k1k2k，Ea = E1 + E2 - E-1

?1

4. 2N2O5(g) 4NO2(g) + O2(g)

历程为：

k1

N2O5(g) ? NO2(g) + NO 3(g) k -1

k2

NO 2 (g) + NO 3 (g) NOk2(g) + NO(g) + O2(g) 3

NO(g) + N2O5(g) 3NO2(g) 证明： dcN2O5cN2O5dt??2k1k2k

?1?k2 证明： dcNOdt?k2cNO2cNO3?k3cNOcN2O5?0 dcNO3dt?k1cN2O5?k?1cNO2cNO3?k2cNO2cNO3?0

dcN2O5dt?k?1cNO2cNO3?k1cN2O5?k3cNOcN2O5 (1) 得：k2cNO2cNO3?k3cNOcN2O5

(2) 得：k1cN2O5?k?1cNO2cNO3?k2cNO2cNO3

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(1) (2) (3)

dcN2O5dt??2k2cNO2cNO3 (4)

k1cN2O5k?1?k2 (2) 得：cNOcNO?23，代入(4)

P361

dcN2O5dt??2k1k2cN2O5k?1?k2

1k11?15161k4k3. 解：尝试一级反应：ln1k11?341k11?y?kt，t15/16?ln?ln16?ln2

t3/4?ln?ln4?2kln2， t15/16?2t3/4，因此反应为一级。

4. 解： ClCOOCCl3 2COCl2 p总 t = 0 p0 0

t = t pt 2(p0- pt) 2 p0 - pt t = ∞ 0 p∞ p∞ = 2p0 pt = (p总-2 p0)，尝试一级反应：ln

t/s p/Pa p总-p∞ k?1tlnp0p总?p?p0pt?lnp0p总?p??kt， k?lnt1p0p总?p?

∞ 4000 0 2000 /s-1 500 2520 1480 800 2760 1240 1300 3066 934 1800 3306 694 6.02×10-4 5.98×10-4 5.86×10-4 5.88×10-4 为一级反应，k(平均) = 5.93×10-4 s-1 5. 解： T/K k×104/s-1 lnk 1/T 473.7 0.947 -9.26 0.00211 483.7 2.05 -8.49 0.00207 494.8 4.50 -7.71 0.00202 502.4 9.28 -6.98 0.00199 516.2 27.2 -5.91 0.00194 527.7 70.7 -4.95 0.00190 536.7 116.0 -4.46 0.00186 545.0 186.0 -3.98 0.00183 600.3 564.0 -2.87 0.00166 lnk-1/T作图的一直线(略去)，直线斜率=-(Ea/R)，得：Ea=1.6×105J·mol-1 直线截距=lnA，得：A= 5.2×1013s-1 6. 解： T/K 273.2 293.2 313.2 333.2 k×106/s-1 0.41 7.92 96.0 913 lnk2k1?Ea(T2?T1)R?T2?T1，ln7.920.41?Ea(293.2?273.2)8.314?293.2?273.2，Ea = 98.60×103 kJ· mol-1

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k = Aexp(-Ea/RT)，

A= k / exp(-Ea/RT) = 0.41×10-6/exp(-98600/8.314×273.2) =2.92×1012s-1 ln96.07.9291396.0?Ea(313.2?293.2)8.314?313.2?293.2Ea(333.2?313.2)8.314?333.2?313.2，Ea = 95.24×103 kJ· mol-1

A=96.0×10-6 /exp(-95240/8.314×313.2) =7.358×1011s-1

ln?，Ea = 97.71×103 kJ· mol-1

A= 913×10-6 / exp(-97710/8.314×333.2) =1.90×1012s-1 Ea(平均) = 97.18×103 kJ· mol-1，A(平均) = 1.35×1012s-1 lnk913?97180(373.2?333.2)8.314?373.2?333.2，k = 0.0392 s-1

t1/2 = 0.693/0.0392 = 17.67 s P384

2. 解：n光子=40×60×1.00×10-7=24×10-5mol·dm-3 φ=n产物/ n光子=1 3. 解：n光子=

?光子hc?8.53?106.626?10?3?313?10?98?34?2.9979?1019?23000?1.34?1016?23000

?5.23?101.34?1016 φ=n丙酮/ n光子=P395

?23000?0.17

3. 解：t1/2与CA无关，则：α=1，t1/2=ln2/k’，k’=kCHt1/2,1t1/2,2?CH?,2CH?,1????

，

6.9369.3?(1010?5?4)，β=1，α+β=2

?1?k=

?ln2t1/2CH?A?0.6936.93?10?4?1000mol?dm?min13?1

?4dCdt?k'CA，ln11?y?k't，ln1?0.8?1000?10t，t=1.61min

4. 解：A+B == C+D，二级反应：t1/2?1k'C0dCCdt?kCH?CACB?k'CACB，k’=kCH

?，k’(298K)=1/(0.01×1)=100mol-1·dm3·h-1

k(298K)=100×102=104 mol-2·dm6·h-1

同理：k’(288K)=1/(0.01×2)=50mol-1·dm3·h-1，k(288K)=5×103mol-2·dm6·h-1

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