Question

For the reaction NO$_2$Cl $\rightarrow$ NO$_2$ + $\frac{1}{2}$Cl$_2$, the proposed mechanism involves the following two elementary steps:
NO$_2$Cl $\xrightarrow{k_1}$ NO$_2$ + Cl
NO$_2$Cl + Cl $\xrightarrow{k_2}$ NO$_2$ + Cl$_2$
Assuming steady state approximation for Cl, the correct expression for the rate of formation of NO$_2$ is

• A. $\frac{d[\text{NO}_2]}{dt} = 2k_1[\text{NO}_2\text{Cl}]$
• B. $\frac{d[\text{NO}_2]}{dt} = \left(\frac{k_1}{k_2}\right)[\text{NO}_2\text{Cl}]$
• C. $\frac{d[\text{NO}_2]}{dt} = (k_1 + k_2)[\text{NO}_2\text{Cl}]$
• D. $\frac{d[\text{NO}_2]}{dt} = (k_1k_2)[\text{NO}_2\text{Cl}]$

Hint:

In steady state approximation, rate of formation of intermediate = rate of consumptionUse this to eliminate intermediate concentration

Answer 1

The Correct Option is A

Step 1: Write rate of formation of NO$_2$.
From mechanism, NO$_2$ is formed in both steps:
\[ \text{Rate} = k_1[\text{NO}_2\text{Cl}] + k_2[\text{NO}_2\text{Cl}][\text{Cl}] \]

Step 2: Apply steady state approximation for Cl.
\[ \frac{d[\text{Cl}]}{dt} = 0 \]
\[ k_1[\text{NO}_2\text{Cl}] = k_2[\text{NO}_2\text{Cl}][\text{Cl}] \]

Step 3: Solve for [Cl].
\[ [\text{Cl}] = \frac{k_1}{k_2} \]

Step 4: Substitute in rate expression.
\[ \text{Rate} = k_1[\text{NO}_2\text{Cl}] + k_2[\text{NO}_2\text{Cl}] \cdot \frac{k_1}{k_2} \]

Step 5: Simplify.
\[ \text{Rate} = k_1[\text{NO}_2\text{Cl}] + k_1[\text{NO}_2\text{Cl}] \]
\[ \text{Rate} = 2k_1[\text{NO}_2\text{Cl}] \]

Step 6: Final rate expression.
\[ \frac{d[\text{NO}_2]}{dt} = 2k_1[\text{NO}_2\text{Cl}] \]

Step 7: Conclusion.
\[ \boxed{2k_1[\text{NO}_2\text{Cl}]} \]