Calculate Rate constant, k, for a consecutive reaction where k1<<k2

Question

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?

Answer 1

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.]

Answer 2

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.