Question

Following reaction takes place by mechanism:
$O _{2}+ CO \longrightarrow NO + CO _{2}$
Step - $I: NO _{2}+ NO _{2}\xrightarrow{k_{1}, \text { Slow }}NO _{3}+ NO$
Step $- II : NO _{3}+ CO \xrightarrow{ k _{2}, \text{Fast} } NO _{2}+ CO _{2}$
Hence, $\left(\frac{d x}{d t}\right)$ for the given reaction is

• A. $k _{1}\left[ NO _{2}\right]^{2}- k _{2}\left[ NO _{3}\right][ CO ]$
• B. $k _{1}\left[ NO _{2}\right]^{2}$
• C. $k _{1}\left[ NO _{2}\right]^{2}+ k _{2}\left[ NO _{3}\right][ CO ]$
• D. $k _{1}\left[ NO _{2}\right]+ k _{2}[ CO ]$

Answer 1

Answer: (B)

Slow step I is the rate-determining step.

$\therefore \left(\frac{d x}{d t}\right)=k_{1}\left[ NO _{2}\right]^{2}$