5
$\begingroup$

Consider the reactions

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

$$\ce{C ->[$k_2$] E + F}$$

with reaction rate constants $k_1$ and $k_2$. I know that the two reactions can be written as follows

$$\ce{A +B->E +F}$$

with a rate constant $k_3$. How does $k_1$ and $k_2$ relate to $k_3$? Does it matter how many species are in the reactants (such as if the first reaction was simply $\ce{A->C}$)?

$\endgroup$
4
$\begingroup$

It is not clear that they are related. It is simpler to consider the equivalent scheme $A \rightarrow C \rightarrow E$ with rate constnats $k_1,\, k_2$ then the decay of A is $A_t/A_0= \exp(-k_1t)$ the amount of C rises and falls; it is $$\displaystyle C_t/A_0 = \frac{k_1}{k_2-k_1}(\exp(-k_1t)-\exp(-k_2t))$$ and E is $$E/A_0 = 1-A_t-C_t=1-\exp(-k_1t)-\frac{k_1}{k_2-k_1}(\exp(-k_1t)-\exp(-k_2t))$$

When $k_2 $ is very large and much bigger than $k_1$ then $\exp(-k_2t) \rightarrow 0 $ and $\displaystyle E/A_0= 1-(1+\frac{k_1}{k_2})\exp(-k_1t)$ and in the opposite limit $k_1 \gg k_2$ then $E/A_0= 1-\exp(-k_2t)$.

So you can choose which limit is important. In either case the amount of E rises as time proceeds but with different rate constants. In the intermediate case there seems to be no rate constant but only a slope to the curve which will depend on the time at which the measurement in made. You could for example find the half rise time and try to relate that to rate constants, but it looks messy.

$\endgroup$
  • $\begingroup$ Can you explain how do you come with the first equation which describes the concentration of $[C_t]$ in terms of $k_1$ and $k_2$ as a function of time?(it will be a complete answer you describe that too.) $\endgroup$ – pranjal verma Apr 28 at 15:25
  • 1
    $\begingroup$ Write down $dC_t/dt =k_1A_t-k_2C_t$ and substitute for $A_t$ as given and solve the equation. At $t=0,\,C_0=0$. $\endgroup$ – porphyrin Apr 29 at 6:49

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.