Lets say rate = k[A][B]n
(A is in excess, lets say at a concentration of 50mM)

My data includes the initial concentration of B, and the initial rate of reaction at that concentration

I can rewrite the rate equation as as:
rate = kapp[B]n

I then take the natural logarithm of this equation and plot a graph of ln(rate) against ln[B], the gradient will give me the value of n.

Am I right in thinking that the y intercept will give a value of kapp? And if it does, to determine the true rate constant do I just divide kapp by 50? If not, how do you determine the true rate constant?


There is no real need to invoke a "$k^{app}$".

Let the rate be $r$, so:


Since as $\ce{A}$ is in large excess:

$$[\ce{A}]=C(A)\gg[\ce{B}] \implies r=kC(A)[\ce{B}]^n$$

So $[\ce{A}]$ doesn't vary perceptibly during the rate measurement interval.

Plot the value of $\ln r$ versus $\ln [\ce{B}]$ in the knowledge that:

$$\ln r=\ln k + \ln C(A) + n\ln [\ce{B}]$$

So the gradient of that graph is $n$.

And the intercept of that graph is $\ln k + \ln C(A)$. Since as $C(A)$ is known, $\ln k$, and thus $k$, is also known.

  • $\begingroup$ Similar to the pseudo-first order approximation in case of Fischer esterification? $\endgroup$ Jan 3 '18 at 19:23
  • $\begingroup$ Don't know that, sorry! $\endgroup$
    – Gert
    Jan 3 '18 at 20:45

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.