# Determining the concentration after half-life period

I was given the reaction

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 $$c_{DNP}(3t_{1/2})=12\mu M$$?

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.
• 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. – xotix Jan 22 at 8:16