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?


1 Answer 1


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? $\endgroup$ Commented Jan 3, 2018 at 19:23
  • $\begingroup$ Don't know that, sorry! $\endgroup$
    – Gert
    Commented Jan 3, 2018 at 20:45

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