For example, for the reaction $\ce{A + 2B -> C}$, we have

$$\text{rate} = -\frac{1}{2}\frac{\mathrm d[\ce{B}]}{\mathrm dt} \quad \text{and} \quad \text{rate} = k[\ce{A}]^x[\ce{B}]^y.$$ Together, this means that

$$-\frac{1}{2}\frac{\mathrm d[\ce{B}]}{\mathrm dt} = k[\ce{A}]^x[\ce{B}]^y$$

However, in many books it is also written simply that

$$-\frac{\mathrm d[\ce{B}]}{\mathrm dt} = k[\ce{A}]^x[\ce{B}]^y.$$

Which is correct?

  • 4
    $\begingroup$ Both are.$\,\!$ $\endgroup$ Sep 20 '18 at 14:22
  • $\begingroup$ @IvanNeretin can you please elaborate? Are we taking $2k$ As $k$ ? $\endgroup$
    – Fawad
    Sep 20 '18 at 14:26
  • 4
    $\begingroup$ Yeah, sort of. What is $k$? Some number. What is $2k$? Some number. You may use either form, as long as you know which one are you using. $\endgroup$ Sep 20 '18 at 14:33
  • $\begingroup$ The rate may be written with both ways. But it is much more frequent to use the rate with the coefficient 1/2 in front of $\pu{\frac{d[B]}{dt}}$ $\endgroup$
    – Maurice
    May 25 '21 at 16:18

Your Answer

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

Browse other questions tagged or ask your own question.