Question

The following reaction was carried out at ${44}^{o}C$:
$N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$
The concentration of $N{O}_2$ is $6.0\times {10}^{-3}M$ after $10$ minutes of the start of the reaction. Calculate the rate of production of $N{O}_2$ over the first ten minutes of the reaction.

Answer 1

$N_2{O}_5⟶2{NO}_2+\frac{1}{2}{O}_2$

$\frac{d\left[N_2{O}_5\right]}{dt}=\frac{1}{2}\frac{d\left[{NO}_2\right]}{dt}=\frac{2d\left[O_2\right]}{dt}$

$\frac{d\left[{NO}_2\right]}{dt}=\frac{6\times {10}^{-3}}{10}=6\times {10}^{-4}mol{L}^{-1}{min}^{-1}$