Ethane is chlorinated in a continuous reactor c2h6 cl2 a...

Ethane is chlorinated in a continuous reactor: C2H6 + Cl2 → C2H5Cl + HCl Some of the product monochloroethane is further chlorinated in an undesired side reaction: C2H5Cl + Cl2 → C2H4Cl2 + HCl Take a basis of 100 mol C2H5Cl produced. Assume that the feed contains only ethane and chlorine and that all of the chlorine is consumed and carry out a degree-of-freedom analysis based on atomic species balances. The reactor is designed to yield a 15% conversion of ethane and a selectivity of 14 mol C2H5Cl/mol C2H4Cl2. Calculate the feed ratio (mol Cl2/mol C2H6) and the fractional yield of monochloroethane.

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00:01 Hello students in the given question we have to calculate the feed ratio and the fractional yield of monochloroethane which is c2 h5 cl given the chlorination of ethane that is c2 h6 plus cl2 it gives c2 h5 cl plus hcl and under undesired side reaction which is c2h5 cl plus cl plus cl to give c2 c2 c2 h4 cl2 plus h .c .l.

01:08 Selectivity of c2h5cl divided by c2h4 cl2, it is equal to 10 and yield is equal to 15 % conversion of ethane.

01:29 Extent of reaction for.

01:32 Extent of reaction for reaction 1 is e1 and for reaction 2 it is e2, n1 for c2 h6 and n2 for cl2.

01:53 Given basis, 100 moles, 100 moles of c2h5, cl produces produced e1 minus e2 that is equal to 100 and cl2 is outlet and cl2 in outlet is 0 that is equal to n2 minus e1 minus e2 is equal to 0 the selectivity selectivity it is equal to e1 minus of minus e2 divided by e2 and that is equal to 10 which further can be written as e1 is equal to 11 times of e2 now e1 minus e2 it is equal to 100 and from this we can write that e 1 it is equal to e 2 plus 100 moving ahead with the solution from here e 2 plus 100 is equal to 11 e2 from the these two we get to know therefore from this we get to know finally that e2 is equal to 10 and e1 is equal to 110 this is on solution now 15 % conversion of ethane n1 minus n1 minus e1 within brackets divided by n1 this is equal to e1 divided by n1 and that is equal to e1 divided by n1 and that is equal to 0 .15, it is in percentage.

03:48 Therefore, n1 will be equal to 110 divided by 0 .15.

03:54 And on solving, we get n1 to be equal to 733 .33 moles.

04:00 This is n1...

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