Question

7 kg of dimethyl terephthalate is reacted with 5 kg of Ethylene glycol to produce PET. Estimate the amount of polymer that is produced assuming 50% efficiency.

Answer 1

Explanation:

The given data will be as follows.

         7 kg of dimethyl terephthalate + 5 kg of Ethylene glycol = 12 kg of product

Formula for weight average number is as follows.

                       $w_{n} = \sum w_{i} \times M_{i}$

                                  = $\frac{7}{12} \times M_{DMT} + \frac{5}{12} \times M_{EG}$

Molecular weight of DMT = 198.21 g/mol

Molecular weight of ET = 62.07 g/mol

Hence, calculate the weight average number as follows.

          $\frac{7}{12} \times M_{DMT} + \frac{5}{12} \times M_{EG}$

           = $\frac{7}{12} \times 198.21 g/mol + \frac{5}{12} \times 62.07 g/mol$          

            = (115.62 + 25.862) g/mol            

             = 141.482 g/mol

Since, efficiency is given as 50% = 0.5

                               0.5 = $\frac{\bar{M_{avg. PET}}}{w_{n}}$

                $\bar{M_{avg. PET}} = 0.5 \times w_{n}$

                                  = $0.5 \times 141.482 g/mol$    

                                  = 70.741 g/mol

As,          $m_{PET} \times \bar{M_{avg. PET}}$ = $m_{DMT} \times M_{DMT} + m_{EG} \times M_{EG}$

        $m_{PET} \times 70.741 g/mol = 7 \times 198.21 g/mol + 5 \times 62.07 g/mol$  

                      $m_{PET}$ = $\frac{1697.82}{70.741}$

                                     = 24 g

Thus, we can conclude that the amount of given polymer is 24 g.