1
$\begingroup$

I have a consecutive reaction where the first step is the rate determining step so $k_1<<k_2$:

$$\ce{A ->[$k_1$] B ->[$k_2$] C}$$

The rate law can be shown to be:

$$[\ce{C}] = {1 - \exp(-k_1t)} [\ce{A}]_0\tag{eq. 1}$$

My reaction is pseudo first order since the first step is a hydrolysis. If I take the natural log of eq. 1 will I be able to convert it into the following form?:

$$\ln[\ce{C}] = -k_1t + \ln[\ce{A}]_0\tag{eq. 2}$$

... because If I could get it to this form, I will be able to plot $\ln[\ce{C}]$ vs $t$ to get $k$ from the slope.

If I have made a mistake with the natural logarithm (or if it is not possible to convert it to the form of eq. 2), what function do I need to use to find $k$?

My question is specifically: What do I need to plot to determine my rate constant considering I have done a time course NMR experiment?

$\endgroup$
  • $\begingroup$ Your eq. 1 is wrong since 1 is dimensionless an [C] has the dimension of a concentration. Also keep in mind that [A] + [B] + [C] is constant over time and [B] can be neglected. $\endgroup$ – aventurin Sep 22 '18 at 8:59
2
$\begingroup$

In a consecutive reaction A-B-C, A decays as $A=A_0\exp(-k_1t)$ with $A_0$ the initial amount (initial B and C are assumed to be zero) and B rate expression $\displaystyle \frac{dB}{dt}=k_1A-k_2B$ which when substituting for $A$ and integrating gives $\displaystyle B=A_0\frac{k_1}{k_2-k_1}\left(\exp(-k_1t) -\exp(-k_2t \right)$. C is found as $A_0-A-B$

then $\displaystyle C= A_0\left(1+\frac{k_2\exp(-k_1t)-k_1\exp(-k_2t)}{k_1-k_2} \right) $

In your case $k_1<<k_2$ then $\displaystyle C= A_0\left( 1 - \exp(-k_1t) \right) $ which is almost your equation.

Taking logs gives $\displaystyle\ln\left(\frac{C}{A_0}\right)=\ln\left(1-\exp(-k_1t)\right)$ and this is what you should use.

[To go further you will have to expand the series for log and exponential but this will lead to a polynomial and will only be an approximation anyway. The first few terms are $\log(k_1t) - k_1t/2 + (k_1t)^2/24\cdots$; very messy.]

$\endgroup$
  • $\begingroup$ I still don't understand what I need to plot to determine the observed rate constant for my reaction $\endgroup$ – Eleftheria Sep 21 '18 at 16:37
  • $\begingroup$ You should then plot $ C/A_0$ vs $t$ because $ C/A_0=\left( 1 - \exp(-k_1t) \right) $. The curve rises then becomes constant. Then fit the curve by guessing values of $k_1$ until you get a match. You should use a computer to do this. By expanding the exponential, at small times the slope is $+k_1$ so this will give you a starting value. $\endgroup$ – porphyrin Sep 23 '18 at 8:56
0
$\begingroup$

yes, you can plot $\ln[\ce{C}]$ vs $t$ to get $k$ from the slope. // If you have made a mistake with the natural logarithm, it is impossible to predict what happens with the data.

$\endgroup$

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.