1. We combined ethylene glycol (EG) with adipic acid (AA) to make an n-mer, a po

Question

1. We combined ethylene glycol (EG) with adipic acid (AA) to make an n-mer, a polyester whose molecular weight is 1638 g/mol. The product polyester is endcapped with alcohol functional groups (a diol). This diol is then added in a second step to toluene diisocyante (TDI) to produce a polyurethane of molecular weight 15,000 g/mol. The resulting polvurethane is also endcapped with alcohols. a. How much EG, AA, TDI are required to make 100 lb of this polyurethane? b. How much water must be remov 2. If the carboxylic acid (A) is a catalyst in a polyester stepwise polymerization, and "catalysts" don't get used up in a reaction, describe briefly why we do not assume that the concentration of (A) is a constant in the rate law? Find the rate constant, k, for the following data on the reaction of uncatalyzed diethylene glycol with adipic acid at 166 OC. 3. Iime, min 100 200 300 400 0.7844 0.8048 0.8585 0.8820

Explanation / Answer

1.

(A) EG + AA -> PE + H2O

the ratio of EG and AA is 1:1 since both the acid and alcohol are bi-functional

so, EG + AA -> PE + H2O

molar mass 62 + 146-> 1638 + 18   

to find the number of moles required to 1638 g/mol PE, we will add the mol. masses of EG and AA and substract the mol. mass of water from it. since, water is removed in the reaction

so, 2+146= 208; 208-18= 190.

now moles of EG and AA needed will be 1638/190= 8.62

so, 8.62 moles of EG and AA is needed .

in weight, EG= 8.62*62= 534.44g , AA= 8.62*146=1258.52g

for further reaction , PE + TDI -> PU

1638 + 174.2 15000  

SAME WAY, 1638 + 174.2=1812.2

15000/1812.2= 8.28

so, 8.3 moles of TDI is needed or (8.3*174.2) 1440.6g

all the calculation is for 1 mole of PU. for 100lb i.e, 45359.2g lets convert this into moles

moles= 45359.2/ 15000 = 3 moles,

TDI, 8.3 for 1 mol=> 8.3*3= 24.9 for 3 moles

i.e, 24.9* 174.2= 4337.6g TDI for 100lb PU

so, PE needed was 24.9 moles too

so, AA and EG needed was, 8.62 for 1 molPE, so 8.62*24.9 =214.6 moles for 24.9 mol PE.

so, EG 13,305g and AA 31,331g

(b) amount of water removed as seen above is 214.6 * 18= 3862.8g

2. esterification reactions occure only between alcohols and carboxylic acids, so even if you use a carboxylic acid as a catalyst some amount of it will also interefere in the reaction and thus, it's concentration cannot be considered constant in the rate law.

3.it is a first order reaction because if plot a graph of log p vs t, you will get a straight line with negative slope.

taking any two set of values, say first two and putting in rate equation for first order,

ln[A] = -kt + ln[A0]

[A]= 0.8048

[A0]= 0.7844

T= 200-100=100

k= 2.6 * 10-4

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