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

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

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

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?

• 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. – aventurin Sep 22 '18 at 8:59

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 $$k_1< then $$\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.]

• I still don't understand what I need to plot to determine the observed rate constant for my reaction – Eleftheria Sep 21 '18 at 16:37
• 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. – porphyrin Sep 23 '18 at 8:56

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.