Chapter 13, Problem 13.105QA

Interpretation Introduction

To find:

Therate determining step in the mechanism

Answer to Problem 13.105QA

Solution:

The second step is the rate determining step for the given rate law.

Explanation of Solution

1) Concept:

We need to focus on the mechanism of the reaction. Elementary step 1 involves reversible interconversion of gaseous oxygen molecule to form two oxygen atoms. Elementary step 2 involves collision of one oxygen atom with the nitrogen molecule to form $\mathrm{N}\mathrm{O}\mathrm{ }$ and $\mathrm{N}$ atom. Elementary step 3 involves collision between single atoms of N and O to form $\mathrm{N}\mathrm{O}$ molecule.

Overall reaction involves collision of one nitrogen molecule with one oxygen molecule to form two molecules of $\mathrm{N}\mathrm{O}$. The rate law for any elementary step can be written directly from the balanced reactions using stoichiometric coefficients as exponents in the rate law.

Step 1 ${\mathrm{O}}_2\left(\mathrm{g}\right)\mathrm{ }\leftrightarrow 2\mathrm{O}\mathrm{ }\left(\mathrm{g}\right)\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}={\mathrm{k}}_{\mathrm{f}}\left[{\mathrm{O}}_2\right]={\mathrm{k}}_{\mathrm{r}}{\left[\mathrm{O}\right]}^2$

Step 2 $\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{O}\mathrm{ }\left(\mathrm{g}\right)+{\mathrm{N}}_2\left(\mathrm{g}\right)\to \mathrm{N}\mathrm{O}\mathrm{ }\left(\mathrm{g}\right)+\mathrm{N}\left(\mathrm{g}\right)\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}={\mathrm{k}}_2\mathrm{ }\left[\mathrm{O}\right]\left[{\mathrm{N}}_2\right]$

Step 3: $\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{N}\left(\mathrm{g}\right)+\mathrm{O}\mathrm{ }\left(\mathrm{g}\right)\to \mathrm{N}\mathrm{O}\mathrm{ }\left(\mathrm{g}\right)\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}={\mathrm{k}}_3\mathrm{ }\left[\mathrm{N}\right]\left[\mathrm{O}\right]$

Overall reaction ${\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2\mathrm{N}\mathrm{O}\mathrm{ }\left(\mathrm{g}\right)\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}\mathrm{ }\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]$

From the rate law for the first reversible fast reaction, we can say that the rate of forward and reverse reactions is the same. We can write as:

${\mathrm{k}}_{\mathrm{f}}\left[{\mathrm{O}}_2\right]={\mathrm{k}}_{\mathrm{r}}{\left[\mathrm{O}\right]}^2$

Rearranging the equation for $\left[\mathrm{O}\right]$, we get

${\left[\mathrm{O}\right]}^2=\frac{{\mathrm{k}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{r}}}\mathrm{ }\left[{\mathrm{O}}_2\right]$

$\left[\mathrm{O}\right]={\left(\frac{{\mathrm{k}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{r}}}\left[{\mathrm{O}}_2\right]\right)}^{\frac{1}{2}}$

2) Given:

$\frac{∆\left[\mathrm{N}\mathrm{O}\right]}{∆\mathrm{t}}=\mathrm{k}\mathrm{ }\left[{\mathrm{N}}_2\right]{\left[{\mathrm{O}}_2\right]}^{\frac{1}{2}}$

3) Calculations:

From the rate law for the first reversible fast reaction, we can say that the rate of forward and reverse reactions is the same. We can write as:

${\mathrm{k}}_{\mathrm{f}}\left[{\mathrm{O}}_2\right]={\mathrm{k}}_{\mathrm{r}}{\left[\mathrm{O}\right]}^2$

Arranging the equation for [O], we get

${\left[\mathrm{O}\right]}^2=\frac{{\mathrm{k}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{r}}}\mathrm{ }\left[{\mathrm{O}}_2\right]$

$\left[\mathrm{O}\right]={\left(\frac{{\mathrm{k}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{r}}}\left[{\mathrm{O}}_2\right]\right)}^{\frac{1}{2}}$

In the second step of the given mechanism, NO molecule is formed. So, the rate law for the second step in terms of formation of NO molecule can be written as

$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{ }\mathrm{ }\frac{∆\left[\mathrm{N}\mathrm{O}\right]}{∆\mathrm{t}}\mathrm{ }={\mathrm{k}}_2\mathrm{ }\left[\mathrm{O}\right]\left[{\mathrm{N}}_2\right]$

Inserting the value for concentration of [O] from equation 1, the above equation can be written as

$\frac{∆\left[\mathrm{N}\mathrm{O}\right]}{∆\mathrm{t}}\mathrm{ }={\mathrm{k}}_2\mathrm{ }\left[\mathrm{O}\right]\left[{\mathrm{N}}_2\right]$

$\frac{∆\left[\mathrm{N}\mathrm{O}\right]}{∆\mathrm{t}}\mathrm{ }={\mathrm{k}}_2\mathrm{ }\left[{\left(\frac{{\mathrm{k}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{r}}}\left[{\mathrm{O}}_2\right]\right)}^{\frac{1}{2}}\right]\left[{\mathrm{N}}_2\right]$

Combining all the rate constants as $\mathrm{k}$, we get the final equation as

$\frac{∆\left[\mathrm{N}\mathrm{O}\right]}{∆\mathrm{t}}\mathrm{ }=\mathrm{k}\left[{\mathrm{N}}_2\right]{\left[{\mathrm{O}}_2\right]}^{\frac{1}{2}}$

Thus, it is the second step that determines the given rate of formation of $\mathrm{N}\mathrm{O}$.

O is the intermediate in the reaction, so its concentration remains constant. Hence, it does not appear in the overall reaction.

Conclusion:

Substituting the concentration of the intermediate in the overall rate equation, we can find the step that is the rate determining step for the reaction. In this question, the second step is the rate determining step.