# Solving for Concentration Using pseudo-First Order Kinetics

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

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 $$\ce{A}$$.

• Keep in mind that time in first graph is in log scale. – Mathew Mahindaratne Sep 5 '20 at 21:04
• 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. – porphyrin Sep 6 '20 at 7:27