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For the reaction $N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g)$ the value of rate of disappearance of $N_2O_5$ is given as $6.25\times10^{-3}$ mol $L^{-1}s^{-1}$. The rate of formation of $NO_2$ and $O_2$ is given respectively as
A
$1.25\times10^{-2}$ mol $L^{-1}s^{-1}$ and $6.25\times10^{-3}$ mol $L^{-1}s^{-1}$
B
$6.25\times10^{-3}$ mol $L^{-1}s^{-1}$ and $6.25\times10^{-3}$ mol $L^{-1}s^{-1}$
C
$1.25\times10^{-2}$ mol $L^{-1}s^{-1}$ and $3.125\times10^{-3}$ mol $L^{-1}s^{-1}$
D
$6.25\times10^{-3}$ mol $L^{-1}s^{-1}$ and $3.125\times10^{-3}$ mol $L^{-1}s^{-1}$
Detailed Solution
For $N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$: $-\frac{d[N_2O_5]}{dt} = \frac{1}{2}\frac{d[NO_2]}{dt} = 2\frac{d[O_2]}{dt}$
$\frac{d[NO_2]}{dt} = 2\times\left(-\frac{d[N_2O_5]}{dt}\right) = 2\times6.25\times10^{-3}$
$\frac{d[NO_2]}{dt} = 1.25\times10^{-2}$ mol $L^{-1}s^{-1}$
$\frac{d[O_2]}{dt} = \frac{1}{2}\times\left(-\frac{d[N_2O_5]}{dt}\right) = \frac{1}{2}\times6.25\times10^{-3}$
$\frac{d[O_2]}{dt} = 3.125\times10^{-3}$ mol $L^{-1}s^{-1}$
$\frac{d[NO_2]}{dt} = 2\times\left(-\frac{d[N_2O_5]}{dt}\right) = 2\times6.25\times10^{-3}$
$\frac{d[NO_2]}{dt} = 1.25\times10^{-2}$ mol $L^{-1}s^{-1}$
$\frac{d[O_2]}{dt} = \frac{1}{2}\times\left(-\frac{d[N_2O_5]}{dt}\right) = \frac{1}{2}\times6.25\times10^{-3}$
$\frac{d[O_2]}{dt} = 3.125\times10^{-3}$ mol $L^{-1}s^{-1}$