1
$\begingroup$

enter image description here

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

http://imgur.com/a/wcoI0u2

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

$\endgroup$
0
4
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.