2
$\begingroup$

The following graphs represent a pseudo-first order bi-molecular reversible reaction with the formula $\ce{A + B <=> C}$:

Pseudo-First Order Kinetics

The reaction product has an extinction coefficient of $\pu{50000 M-1 cm-1}$. Solve for the concentration of $\ce{A}$.

I think you are supposed to use Beer-Lambert's equation ($A=\epsilon cl$) with the assumption that cellular path, $l=\pu{1.00cm}$, and get a value for $A$ (absorption) from the graphs to then solve for $c$. I just do not know how to get a value for absorption from the graphs. Furthermore, after getting a concentration for $\ce{C}$, I do not know where to go from there to solve for the concentration of $\ce{A}$. I think it has something to do with the equation $y=2.0172+0.498x$ (the slope is equal to $k_1$ and the y-intercept is equal to $k_{-1}$) and messing around with the rate constants to solve for $\ce{A}$.

$\endgroup$
2
  • $\begingroup$ Keep in mind that time in first graph is in log scale. $\endgroup$ – Mathew Mahindaratne Sep 5 '20 at 21:04
  • $\begingroup$ In a reversible reaction $k_{obs}=k_1(A_e+B_e)+k_{-1}$. You know $B_e$, the equilibrium conc'n of B from the right hand graph, calculate the value $A_e$ from the absorbance. Species B is in excess. $\endgroup$ – porphyrin Sep 6 '20 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.