How does the stoichiometric coefficient affect the time until a given percentage of the reactant is consumed (if at all)?

### Problem from Serway's College Physics test bank:




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The reaction $2NO\_2\rightarrow2NO+O\_2$ obeys the rate law:


$\large \frac{\Delta[O\_2]}{\Delta t} = (1.40\times10^{-2})[NO\_2]^2 \space at \space 500^\circ \space K $


If the initial concentration of $NO\_2$ is 1.00 M, how long will it take for the [$NO\_2$] to decrease to 35.8% of its initial value?


A) 45.9 s


B) 73 s


C) 128 s


D) 1.40 x 10-2 s


E) cannot be determined from this data


**ANSWER: C**




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### My work


We are given the reaction rate for $O\_2$. Since for a given reaction $aA + bB \rightarrow cC + dD$, the rate of the reaction $r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$, I conclude that $\frac{\Delta[O\_2]}{\Delta t} = -\frac{1}{2}\frac{\Delta[NO\_2]}{\Delta t} \Rightarrow -\frac{\Delta[NO\_2]}{\Delta t} = 2\frac{\Delta[O\_2]}{\Delta t}$. Therefore:


$-\frac{\Delta[NO\_2]}{\Delta t} = 2 \times (1.40\times10^{-2})[NO\_2]^2 = (2.8\times10^{-2})[NO\_2]^2 \Rightarrow$


$\textbf k = 2.8\times10^{-2}$


Knowning that the reaction is second-order in $NO\_2$, I plug in $[NO\_2] = 0.358[NO\_2]\_0$ into the integrated rate law and solve for t as so:


$\frac{1}{0.358[NO\_2]\_0} = kt + \frac{1}{[NO\_2]\_0}$


$\frac{1}{0.358\times1M} - \frac{1}{1M} = 2.8\times10^{-2}\times t$


$t \approx 64 s$




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### Answer given in the test bank


However, the textbook gives 128s as the answer, indicating that the reaction rate was NOT multiplied by 2 (from $\frac{\Delta[O\_2]}{\Delta t}$ to $-\frac{\Delta[NO\_2]}{\Delta t}$).


Is there a gap in my understanding of conversion between reaction rates according to the stoichiometric coefficient of the reactant/product being handled or did the test bank simply give an incorrect answer?




You should not multiply the rate constant by an stoichiometric coefficient.


The rate of this chemical reactionis given by


$$ v = -\frac{1}{2}\frac{d|NO2\_2|}{{dt}} = +\frac{1}{2}\frac{d|NO|}{{dt}} = +\frac{1}{1}\frac{d|O2\_2|}{{dt}} = k |NO\_2|^2 $$


where k is the reaction rate constant. In your exercise, $k = 1,4\times10^{-2} \;mol^{-1}\,dm^3\,s^{-1}$.


Notice that the stoichimetric coefficients appear in the reaction rate definition, thus the reaction rate does not depend on which reactant or product used to obtain it.


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