Question

Following reaction takes place by mechanism
$N{O}_2+CO\to NO+C{O}_2$
$StepIN{O}_2+N{O}_2\stackrel{k_1,slow}{\to }N{O}_3+NO$
$StepIIN{O}_3+CO\stackrel{k_2,fast}{\to }N{O}_2+C{O}_2$
Hence, $\left(\frac{dx}{dt}\right)$ for the given reaction is:

• A. $k_1{\left[N{O}_2\right]}^2-k_2\left[N{O}_3\right]\left[CO\right]$
• B. $k_1{\left[N{O}_2\right]}^2+k_2\left[N{O}_3\right]\left[CO\right]$
• C. $k_1{\left[N{O}_2\right]}^2$
• D. $k_1\left[N{O}_2\right]+k_2\left[CO\right]$

Answer 1

Answer: (C)

$k_1{\left[N{O}_2\right]}^2$

${NO}_2+CO\rightleftharpoons NO+{CO}_2$

Step I ${NO}_2+{NO}_2\underrightarrow{K_1,slow}{NO}_3+NO$

Step II ${NO}_3+CO\underrightarrow{K_2,fast}{NO}_2+{CO}_2$

$\left(\frac{dx}{dt}\right)$ i.e. rate of reaction is only dependent on slow step of mechanism

Hence $\frac{dx}{dt}=K_1{\left[{NO}_2\right]}^2$