Chemical Kinetics - Result Question 8
####8. For the reaction $[N_2 O_5(g) \to 2 NO_2(g)+.$ $.1 / 2 O_2(g)]$ the value of rate of disappearance of $N_2 O_5$ is given as $6.25 \times 10^{-3} mol L^{-1} S^{-1}$. The rate of formation of $NO_2$ and $O_2$ is given respectively as :
[2010]
(a) $6.25 \times 10^{-3} mol L^{-1} s^{-1}$ and $6.25 \times 10^{-3}$ $mol L^{-1} s^{-1}$
(b) $1.25 \times 10^{-2} mol L^{-1} s^{-1}$ and $3.125 \times 10^{-3}$ $mol L L^{-1} s^{-1}$
(c) $6.25 \times 10^{-3} mol L^{-1} s^{-1}$ and $3.125 \times 10^{-3}$ $mol L{ }^{-1} s^{-1}$
(d) $1.25 \times 10^{-2} mol L^{-1} s^{-1}$ and $6.25 \times 10^{-3}$ $mol L^{-1} s^{-1}$
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Solution:
- (b) $N_2 O_5(g) \longrightarrow 2 NO_2(g)+1 / 2 O_2(g)$
$-\frac{d}{d t}[N_2 O_5]=+\frac{1}{2} \frac{d}{d t}[NO_2]=2 \frac{d}{d t}[O_2]$
$\frac{d}{d t}[NO_2]=1.25 \times 10^{-2} mol L^{-1} s^{-1}$ and
$\frac{d}{d t}[O_2]=3.125 \times 10^{-3} mol L^{-1} s^{-1}$
