Question

3.3 (a) Write down a balanced chemical equation for polymerization of a ?-hydroxy carboxylic acid (HORCO2H). (b) The following data were obtained during the condensation of 12-hydroxystearic acid at 160.5 °C in the molten state by sampling the reaction mixture at various times. The [COOH] was determined for each sample by titrating with ethanolic sodium hydroxide. t / h [COOH] / mol dm–3 0 3.10 0.5 1.30 1.0 0.83 1.5 0.61 2.0 0.48 2.5 0.40 3.0 0.34 Determine the rate coefficient for the reaction under these conditions and the order of the reaction. Also suggest whether or not a catalyst was used. (c) What would be the extent of the reaction after (i) 1 h and (ii) 5 h?

          3.3 (a) Write down a balanced chemical equation for polymerization of a ?-hydroxy
carboxylic acid (HORCO2H).
(b) The following data were obtained during the condensation of 12-hydroxystearic acid
at 160.5 °C in the molten state by sampling the reaction mixture at various times.
The [COOH] was determined for each sample by titrating with ethanolic sodium
hydroxide.
t / h [COOH] / mol dm–3
0 3.10
0.5 1.30
1.0 0.83
1.5 0.61
2.0 0.48
2.5 0.40
3.0 0.34
Determine the rate coefficient for the reaction under these conditions and the order
of the reaction. Also suggest whether or not a catalyst was used.
(c) What would be the extent of the reaction after (i) 1 h and (ii) 5 h?

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3.3 (a) Write down a balanced chemical equation for polymerization of a ?-hydroxy
carboxylic acid (HORCO2H).
(b) The following data were obtained during the condensation of 12-hydroxystearic acid
at 160.5 °C in the molten state by sampling the reaction mixture at various times.
The [COOH] was determined for each sample by titrating with ethanolic sodium
hydroxide.
t / h [COOH] / mol dm–3
0 3.10
0.5 1.30
1.0 0.83
1.5 0.61
2.0 0.48
2.5 0.40
3.0 0.34
Determine the rate coefficient for the reaction under these conditions and the order
of the reaction. Also suggest whether or not a catalyst was used.
(c) What would be the extent of the reaction after (i) 1 h and (ii) 5 h?

Added by Jose Ignacio W.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Sri K

05/08/2022

Step 1

The polymerization of ω-hydroxy carboxylic acid involves the condensation reaction where the hydroxyl group (-OH) of one molecule reacts with the carboxyl group (-COOH) of another molecule, releasing water (H₂O) and forming an ester bond. The general equation Show more…

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Write down a balanced chemical equation for polymerization of a ̉ω-hydroxy carboxylic acid (HORCO2H). The following data were obtained during the condensation of 12-hydroxystearic acid at 160.5 °C in the molten state by sampling the reaction mixture at various times. The [COOH] was determined for each sample by titrating with ethanolic sodium hydroxide. Determine the rate coefficient for the reaction under these conditions and the order of the reaction. Also suggest whether or not a catalyst was used. (c) What would be the extent of the reaction after (i) 1 h and (ii) 5 h?

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00:01 Let's discuss this question here we have three questions.

00:03 Firstly, we need to write down the balanced chemical equation for polymerization of omega -hydroxy carboxic acid.

00:10 In second question, we need to determine the rate coefficient.

00:14 And in third question, we need to find what would be the extent of the reaction after one hour and five hour.

00:20 So here we can say for question number a here, when n -h -o, r -c -double -bond o, o, s.

00:33 Gives us o r c double bond o n times plus n minus 1 h2o for question number b here we have time our chart and here we have c double o h per one mole decimeter u inverse and here we have 1 divided by c double oh h so here we have 0 when time is 0 0 .5 1 .0, 1 .5, 2 .0, 2 .5, and 3 .0.

01:18 So here, the concentration of cw will be 3 .10.

01:24 Here it will be 1 .30.

01:26 Here it will be 0 .83, 0 .61.

01:34 Then it will be 0 .48, 0 .40.

01:39 And finally, it will be 0 .34.

01:41 And here when we do 1 divided by cwh, so it will be 0 .322585, 0645, here 0 .769230769.

02:00 Here, 1 .204819277.

02:05 Here 1 .6393, 4262.

02:10 Here, 2 .08 3333, 333.

02:15 3 3 and here the value will be 2 .5 and finally here the value will be 2 .94117 6471 so here we can say while we draw the graph so here we have 1 2 3 4 5 and so on and here we have 0 0 .5 1 1 .5 2 2 2 .5 2 .5 and so on and so on so on.

02:52 So here if we plot the graph it will be one...

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