Question

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Given the following reaction and thermodynamic data, what is a reccomended temperature for a 1 L pilot scale reactor and how long (time) does it take for the reaction to take place?

CH3OCH2CH2CH2OCH2NH2 -> CH3OCH2CH2CH2OH + CH2=NH

Species	CH3OCH2CH2CH2OCH2NH2	CH3OCH2CH2CH2OH	CH2NH
Hf (Kcal/mol)	-91.15	-92.23	19.75
S (cal/mol/K)	118.08	97.18	54.18
300	39.13	29.37	8.92
400	48.84	36.69	10.39
500	56.65	42.77	11.99
600	63.34	48.03	13.47
800	74.69	56.74	15.95
1000	82.87	62.98	17.87
1500	0	0	21.02

ΔHrxn (Kcal/mol)	ΔSrxn (cal/mol/K)	ΔGrxn (Kcal/mol)	Kp
18.67	33.28	8.69E+03	0.985534268
18.494	32.77367955	5.38E+03	0.993248201
18.292	32.31453957	2.13E+03	0.99785361
18.118	32.00460919	-1.08E+03	1.0009103
17.67	31.31834149	-7.38E+03	1.004656425
17.256	30.84797494	-1.36E+04	1.0068639
43.894	67.11038492	-5.68E+04	1.019230233

Kc	A-forward (sec^-1)	Ea(Kcal/mol)	K-forward (sec^-1)
0.001653304	2.10611E+14	53.09	4.40801E-25
0.001249683	2.80815E+14	52.914	3.42895E-15
0.001004382	3.51018E+14	52.712	3.18437E-09
0.000839549	4.21222E+14	52.538	3.06178E-05
0.000632018	5.61629E+14	52.09	3.295681283
0.000506726	7.02037E+14	51.676	3561.584427
0.000341966	1.05305E+15	78.314	4084.744896

Delta S+c (cal/mol/K)	A-backward (sec^-1)	K-backward (sec^-1)
-24.11856204	45429261.35	9.50818E-32
-23.54895154	59940560.67	7.31917E-22
-23.10712731	74049093.99	6.7176E-16
-22.74613063	87957142.2	6.39344E-12
-22.17652012	114260467	6.70489E-07
-21.73469589	139369642.5	0.000707053
-20.93187498	531011737.2	0.002059767

Answer 1

The data shows that at 1500 degree Kelvin, Kp is the highest ( 1.01923) compared to reaction at all other temperatures. Even the delta G of the recation at 15000 K is most negative ( -5.68E+04) compared to reaction at other temperatures showing that reaction becomes most spontaneous at 1500 K [More negative the delta G - free energy of the reaction is, more spontaneous it is] So 1500 K is the temeprature at which reaction is spontaneous to its maximum

Kp = Kc (RT) ^∆n

1.019230233 = 0.000341966 [(0.0821 L-atm/mole * K) (1500)]^∆n

Upon solving =

2980.501667 = (0.0821 * 1500)^∆n

2980.501667 = (123.15)^∆n

log 2980.501667 = log (123.15)

3.474289 = ∆n * log (123.15)

= ∆n * (2.0904)

∆n = 1.662

So 1.662 moles of product is formed once reaction completes

Rate of forward reaction shows Kf = 4084.744896

meaning 4084.744896 moles of product are formed per second

So, formation of 1.662 moles of product takes (1.662 moles *   4084.744896 moles/sec = 4.0685 * 10^-4 seconds)

So it takes 4.0685 * 10^-4 seconds for the reaction to get completed.