7kg 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.

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.

                                  =

Molecular weight of DMT = 198.21 g/mol

Molecular weight of ET = 62.07 g/mol

Hence, calculate the weight average number as follows.

           =          

            = (115.62 + 25.862) g/mol            

             = 141.482 g/mol

Since, efficiency is given as 50% = 0.5

                               0.5 =

                                  =    

                                  = 70.741 g/mol

As,           =

                      =

                                     = 24 g

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

[N2] = [O2] = 0.841M

And [NO] = 0.00172M

Explanation:

The equilibrium constant of this reaction, Kc, is:

Kc = 2400 = [N2] [O2] / [NO]²

Where [] are the equilibrium concentration of each specie.

The initial concentration of [N2] = [O2] = 0.850M. The equilibrium will shift to the left in order to produce NO. The equilibrium concentrations are:

[N2] = [O2] = 0.850M - X

And [NO] = 2X

Replacing:

2400 = [0.850-X]²  / [2X]²

9600X² = 0.7225 - 1.7 X + X²

0 = 0.7225 - 1.7 X - 9599X²

Solving for X:

X = -0.0088M. False solution, there is no negative concentrations.

X = 0.00859M. Right solution.

Replacing:

[N2] = [O2] = 0.850M - 0.00859M

And [NO] = 2*0.00859M

[N2] = [O2] = 0.841M

And [NO] = 0.00172M