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I am not able to solve this question. Here's my working


Have I done this in the right way? Because my textbook gives answer as 4 moles


Ok, there are two reactions for the 2 moles of A

$$\ce{A ->[rate\space =\space k_1] 2B}$$

$$\ce{A ->[rate\space =\space k_2] 2C}$$

$$\dfrac{k_1}{k_2} = \dfrac{1}{2}$$

The question asks for the total number of moles of A+B+C when the reaction is 75% complete.

  • At that point 25% of A remains or 0.5 moles of A
  • It does matter what the ratio of B and C produced is since either reaction of A creates 2 moles of product. So there is 2*1.5 = 3 moles of product (B+C).

Thus there is 0.5 + 3 = 3.5 moles of A+B+C in the solution when the reaction is 75% complete.

There are 4 moles when the reaction is 100% complete.

So the book answer is wrong.

  • $\begingroup$ I don't understand how either reaction produces 2 moles of product ......how did you conclude this.Is my solution wrong because I am getting different concentrations of C and B $\endgroup$
    – Aladdin
    Aug 4 '18 at 15:50
  • $\begingroup$ @GENESECT - It is just the stoichiometry of the reaction. 1 molecule of A makes either 2 molecules of B or 2 molecules of C. So 1 mole of A produces 2 moles of B+C. $\endgroup$
    – MaxW
    Aug 4 '18 at 15:53
  • $\begingroup$ So some part of A makes B and remaining part makes C .The sum of which is 2 $\endgroup$
    – Aladdin
    Aug 4 '18 at 16:03
  • $\begingroup$ But what have I done wrong in my solution.......I am also getting same answer but it's wrong as It seems from your comments $\endgroup$
    – Aladdin
    Aug 4 '18 at 16:05
  • $\begingroup$ Yes. I could figure out how much of B and how much of C, but that is extra work and I am lazy. // Really -- the less stuff you calculate the less you're likely to do wrong. So solve the problem the simplest way possible. $\endgroup$
    – MaxW
    Aug 4 '18 at 16:06

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