Burkus and Eckert (14) studied the kinetics of the triethylamine-catalyzed reaction of 2,6-toluene diisocyanate (A) with 1-butanol (B) in toluene solution. The reactions may be represented as A titrimetric method was used to follow the progress of the reaction at 39.69∘C On the basis of the data

Question

Burkus and Eckert (14) studied the kinetics of the triethylamine-catalyzed reaction of 2,6-toluene diisocyanate (A) with 1-butanol (B) in toluene solution. The reactions may be represented as A titrimetric method was used to follow the progress of the reaction at $39.69^{\circ}$ C On the basis of the data given below, what are the corresponding values of the reaction rate constants, $k_{1}$ and $k_{2}$ ?

Initial concentration of 53.2 mol/ $m^{3}$ 2,6-tolylene diisocyanate

Initial concentration of 1-butanol 106.4 mol/ $m^{3}$

Catalyst concentration (triethylamine) 31.3 mol/ $m^{3}$

Time	1-Butanol reacted
(K)	(%)
0.36	20
0.657	30
1.118	40
1.866	50
3.282	60

Initial concentration of 53.2 mol/[latex]m^{3}[/latex] 2,6-tolylene diisocyanate

Initial concentration of 1-butanol 106.4 mol/[latex]m^{3}[/latex]

Catalyst concentration (triethylamine) 31.3 mol/[latex]m^{3}[/latex]

Time	1-Butanol reacted
(K)	(%)
0.36	20
0.657	30
1.118	40
1.866	50
3.282	60

Accepted Answer

Because equivalent amounts of reactants were employed [latex](A_{0}=2B_{0})[/latex], the time-ratio method of Frost and Schwemer (11–13) may be used in the solution of this problem. From Table 5.3 the following values of 1/k may be determined at the time ratios indicated.

[latex]\frac{t_{6}}{t_{50}}=\frac{3.282}{1.866}=1.759[/latex] [latex]\frac{1}{k}=6.21[/latex]

[latex]\frac{t_{60}}{t_{40}}=\frac{3.282}{1.116}=2.941[/latex] [latex]\frac{1}{k}=6.08[/latex]

[latex]\frac{t_{60}}{t_{30}}=\frac{3.282}{0.657}=4.995[/latex] [latex]\frac{1}{k}=6.16[/latex]

[latex]\frac{t_{60}}{t_{20}}=\frac{3.282}{0.360}=9.117[/latex] [latex]\frac{1}{k}=6.16[/latex]

[latex]\frac{t_{50}}{t_{30}}=\frac{1.866}{0.657}=2.840[/latex] [latex]\frac{1}{k}=6.09[/latex]

[latex]\frac{t_{50}}{t_{20}}=\frac{1.866}{0.360}5.183[/latex] [latex]\frac{1}{k}=6.11[/latex]

Average: [latex]\frac{1}{k}=6.13[/latex]

The values of k that correspond to the various conversion levels and 1/k =6.13 may be found in Table 5.2. The rate constant [latex]k_{1}[/latex] may then be calculated using equation (5.4.18). Hence,

Conversion		[latex]k_{1}= au ^{\ast }/(B_{0}t)[/latex]
(%)	[latex] au ^{\ast }[/latex]	[latex][m^{3}/(Kmol\cdot Ks)][/latex]
20	0.2722	14.21
30	0.4966	14.21
40	0.8397	14.14
50	1.413	14.23
60	2.476	14.18
		Average = 14.19

Thus, [latex]k_{2}=\kappa k_{1}=14.19/6.13=2.31 \: m^{3}/(Kmol\cdot Ks)[/latex].

Question : Burkus and Eckert (14) studied the kinetics of the triethyla...