Question

For the reaction,

$$2\mathrm{NO}(\mathrm{g})+\mathrm{O}{\phantom{A}}_2(\mathrm{g})⟶2\mathrm{NO}{\phantom{A}}_2(\mathrm{g})\underset{\mathrm{dt}}{\overset{\mathrm{d}(\mathrm{NO}{\phantom{A}}_2)}{\to }}=0.052\mathrm{mol}\mathrm{dm}{\phantom{A}}^{-3}\mathrm{s}{\phantom{A}}^{-1}$$

Calculate rate of consumption of NO (g).

Options

• 0.052 mol dm^-3 s^-1

• 0.114 mol dm^-3 s^-1

• 0.026 mol dm^-3 s^-1

• 0.078 mol dm^-3 s^-1

Answer 1

0.052 mol dm^-3 s^-1 

Explanation:

In the given reaction, the rate of consumption of NO = rate of formation of NO₂

So, ${\displaystyle \frac{-\text{d}\left[\text{NO}\right]}{\text{dt}}=\frac{+\text{d}\left[{\text{NO}}_2\right]}{\text{dt}}}$

${\displaystyle \frac{-1}{2}\text{d}\frac{\left[\text{NO}\right]}{\text{dt}}=\frac{1}{2}\frac{\text{d}\left[{\text{NO}}_2\right]}{\text{dt}}}$

${\displaystyle \frac{-\text{d}\left[\text{NO}\right]}{\text{dt}}=\frac{2}{2}\frac{\text{d}\left[{\text{NO}}_2\right]}{\text{dt}}}$

${\displaystyle \frac{-\text{d}\left[\text{NO}\right]}{\text{dt}}=0.052}$ mol dm^-3 s^-1