Question

Consider the hypothetical reaction

$\mathit{B}\mathbf{\to }\mathit{E}\mathbf{+}\mathit{F}$

which is assumed to occur by the mechanism

$$\begin{gathered} \mathit{B}\mathbf{+}\mathit{B}\underset{{\mathbf{K}}_{\mathbf{-}\mathbf{1}}}{\overset{{\mathbf{K}}_{\mathbf{1}}}{\mathbf{⟷}}}{\mathit{B}}^{\mathbf{*}}\mathbf{+}\mathit{B} \\ {\mathbf{B}}^{\mathbf{*}}\stackrel{{\mathbf{K}}_{\mathbf{2}}}{\mathbf{\to }}\mathit{E}\mathbf{+}\mathit{F} \end{gathered}$$

where ${\mathit{B}}^{\mathbf{*}}$represents a $\mathit{B}$molecule with enough energy to surmount the reaction energy barrier.

(a) Derive the rate law for the production of $\mathit{E}$using the steady-state approximation.

(b) Assume that this condition is known to be first order. Under what conditions does your derived rate law (from part a) agree with this observation.

(c) Explain how a chemical reaction can be first order, since even in a simple case $\mathbf{(}\mathit{B}\mathbf{\to }\mathit{E}\mathbf{+}\mathit{F}\mathbf{)}$molecules must collide to build up enough energy to get over the energy barrier. Why aren’t all reactions at least second order ? In other words, explain the physical significance of the result from part b.

Short answer

(a) Rate= $\frac{{\text{K}}_1{\text{K}}_1{\left[\text{B}\right]}^2}{{\text{K}}^{-1}\left[\text{O}\right]+{\text{K}}_2}$

(b) When rate :${\text{K}}_2>>{\text{K}}_{-1}\left[\text{B}\right]$

$=\frac{{\text{K}}_1{\text{K}}_2}{{\text{K}}_{-1}}\times \left[\text{B}\right]$

(c) Collisions between $B$molecules are occurring more faster than decomposition of $B^{*}$.

Step-by-step solution

Deriving the rate law equation to demonstrate production of  .

$\text{rate}={\text{K}}_2\left[{\text{B}}^{*}\right]$

Now we can introduce steady-state approximation of $B^{*}$

${\text{K}}_1{\left[\text{B}\right]}^2={\text{K}}_{-1}\left[{\text{B}}^{*}\right]\left[\text{B}\right]+{\text{K}}_2\left[{\text{B}}^{*}\right]$

When we solve for $B^{*}$:

$\left[{\text{B}}^{*}\right]=\frac{{\text{K}}_1{\left[\text{B}\right]}^2}{{\text{K}}_{-1}\left[\text{B}\right]+{\text{K}}_2}$

Last thing left to do is to substitute that into the initial rate law equation:

Rate= $\frac{{\text{K}}_1{\text{K}}_2{\left[\text{B}\right]}^2}{{\text{K}}_{-1}\left[\text{B}\right]+{\text{K}}_2}$

Understanding how the first rate law equation agrees with this observation and how collision needs energy barrier to proceed a reaction.

When :${\text{K}}_2>>{\text{K}}_{-1}\left[\text{B}\right]$

Rate= $\frac{{\text{K}}_1{\text{K}}_2}{{\text{K}}_{-1}}\times \left[\text{B}\right]$

Also collision between $B$molecules are occurring much faster than decomposition of $B^{*}$ . This is the physical significance of the result from part B.