Question

Decomposition of $N_2{O}_5$ is expressed by the equation, $N_2{O}_5⟶2N{O}_2+\frac{1}{2}{O}_2$.
If during a certain time interval, the rate of decomposition of $N_2{O}_5$ is $1.8\times {10}^{-3}$ $mol$ ${litre}^{-1}$ ${min}^{-1}$, what will be the rates of formation of $N{O}_2$ and $O_2$ during the same interval?

Answer 1

The rate expression for the decomposition of $N_2{O}_5$ is:
$-\frac{\Delta \left[N_2{O}_5\right]}{\Delta t}=\frac{1}{2}\frac{\Delta \left[N{O}_2\right]}{\Delta t}=2\cdot \frac{\Delta \left[O_2\right]}{\Delta t}$
So, $\frac{\Delta \left[N{O}_2\right]}{\Delta t}=2\frac{\Delta \left[N_2{O}_5\right]}{\Delta t}=2\times 1.8\times {10}^{-3}$
$=3.6\times {10}^{-3}$ $mol$ ${litre}^{-1}$ ${min}^{-1}$
and $\frac{\Delta \left[O_2\right]}{\Delta t}=\frac{1}{2}\frac{\Delta \left[N_2{O}_5\right]}{\Delta t}=\frac{1}{2}\times 1.8\times {10}^{-3}$
$=0.9\times {10}^{-3}$ $mol$ ${litre}^{-1}$ ${min}^{-1}$
(Rate is always positive and hence $-\frac{\Delta \left[N_2{O}_5\right]}{\Delta t}$ is taken positive).