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I was given the reaction enter image description here

In diluted solution and with neutral pH, the reaction reacts according to a reaction velocity law of pseudo-1. order:

$$-\frac{dc_{DNPA}}{dt}=k_{obs},\quad k_{obs}=k\cdot c_{H_2O}$$

At $t=0$ we have $96\mu M$ DNPA.

How big is the concentration of DNPA and DNP after three half-time periods ($3\cdot t_{1/2}$)?

Of course, for DNPA we would just have $1/8$th of $96\mu M$ which is $12\mu M$. But what about DNP?

For me it's clear that it correlates directly with the concentration of DNPA since we have a heavy diluted solution, so the concentration of the water can be assumed as constant.

Further I know that $v_c(t)=\frac{1}{\nu_i}\frac{dd_i(t)}{dt}$ whereas $\nu_i$ are the stoichiometric coefficients. So basically, the concentration od DNPA, DNP and HOAc are the same since all have a stoichiometric coefficient of 1. Right? So they all change the same. Meaning: if DNPA changes to $1/8$-th, so does DNP and HOAc.

I am still a bit unsure, but can I now conclude, that $c_{DNP}(3t_{1/2})=12\mu M$?

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I'm afraid you are mistaken. The DNPA essentially turns into DNP in the course of the reaction, so the answer is $84\ \mu M$, assuming the initial DNP concentration is zero. Formally, the stoichometric coefficient of DNPA is -1, which illustrates nicely that the sum of the amounts of DNPA and DNP is constant in this system.

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  • $\begingroup$ S0 $c_{DNPA}(3t_{1/2})=12\mu M$ and $c_{DNP}(3t_{1/2})=84\mu M$? I'f so, do I also have $c_{HOA_c}(3t_{1/2})=84\mu M$? And of course you are right about the coefficient. $\endgroup$ – xotix Jan 22 at 8:16
  • $\begingroup$ @xotix Full agreement. $\endgroup$ – TAR86 Jan 22 at 9:48
  • $\begingroup$ Thanks! I actually just saw that I missed a little bit of theory here and of course everything makes sense! THanks. $\endgroup$ – xotix Jan 22 at 10:13

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