# Find the rate constant for parallel or competitive reactions?

## a) ${k}_{\text{obs}} = {k}_{1} + {k}_{2} + \ldots + {k}_{n}$ b) ${k}_{\text{obs}} = n {k}_{i}$ c) ${k}_{\text{obs}} = {k}^{n}$

Jun 11, 2016

I get $a$, which matches my source.

Okay, so for the 1st order reaction, we have:

$A \stackrel{{k}_{1} \text{ }}{\to} {A}_{1}$
$A \stackrel{{k}_{2} \text{ }}{\to} {A}_{2}$
$\text{ } \vdots$
$A \stackrel{{k}_{n} \text{ }}{\to} {A}_{n}$

We can say that the same starting $A$ becomes the $n$ products.

Also, we are partitioning the same reactant $A$ across $n$ reactions (we know that the same species behaves the same way across all $n$ identical reaction types).

Next, we can write the rate law for each parallel competitive subreaction:

${r}_{1} \left(t\right) = {k}_{1} \left[A\right] = \frac{d \left[{A}_{1}\right]}{\mathrm{dt}}$ (Eq. 1.1)
${r}_{2} \left(t\right) = {k}_{2} \left[A\right] = \frac{d \left[{A}_{2}\right]}{\mathrm{dt}}$ (Eq. 1.2)
$\text{ } \vdots$
${r}_{n} \left(t\right) = {k}_{n} \left[A\right] = \frac{d \left[{A}_{n}\right]}{\mathrm{dt}}$ (Eq. 1.n)

where:

• The reaction rate is ${r}_{i}$. These may or may not be different across subreactions.
• The rate constant is ${k}_{i}$. These may or may not be different across subreactions.
• The partitioned concentration is ${\left[A\right]}_{i}$. These are presumed to be the same across each subreaction.
• The rate of appearance of ${A}_{i}$ with respect to time is $\frac{d \left[{A}_{i}\right]}{\mathrm{dt}}$. These may or may not be different across subreactions.

The full rate law would be

$\setminus m a t h b f \left(r \left(t\right) = k \left[A\right] = - \frac{d \left[A\right]}{\mathrm{dt}}\right) ,$ (Eq. 2)

and $\left[A\right]$ is the total initial concentration.

Now, ultimately, since each subreaction is presumed to start at the same time, and the sum of all the rates of appearance of products is equal to the rate of disappearance of $A$, the rates of change in concentration all add up:

$- \frac{d \left[A\right]}{\mathrm{dt}} = \frac{d \left[{A}_{1}\right]}{\mathrm{dt}} + \frac{d \left[{A}_{2}\right]}{\mathrm{dt}} + \cdots + \frac{d \left[{A}_{n}\right]}{\mathrm{dt}} ,$ (Eq. 3)

and consequently, from Eqs. 1.1 - 1.n and Eq. 3, we get:

$r \left(t\right) = - \frac{d \left[A\right]}{\mathrm{dt}} = {r}_{1} \left(t\right) + {r}_{2} \left(t\right) + \cdots + {r}_{n} \left(t\right)$. (Eq. 4)

Now, adding up Eqs. 1.1 - 1.n and utilizing Eq. 4:

$r \left(t\right) = {r}_{1} \left(t\right) + {r}_{2} \left(t\right) + \cdots + {r}_{n} \left(t\right)$

$= {k}_{1} \left[A\right] + {k}_{2} \left[A\right] + \cdots + {k}_{n} \left[A\right]$

$= \left({k}_{1} + {k}_{2} + \cdots + {k}_{n}\right) \left[A\right]$

Therefore, comparing back to Eq. 2:

$\textcolor{b l u e}{r \left(t\right)} = k \left[A\right]$

$= \textcolor{b l u e}{\left({k}_{1} + {k}_{2} + \cdots + {k}_{n}\right) \left[A\right]}$

Comparing back to the options:

• $\setminus m a t h b f \left(a\right)$ matches this result.
• $b$ would imply that $n \left[A\right] = \left[A\right]$, but we already assumed that $n \ne 1$.
• $c$ would change the units of $k$ compared to each ${k}_{i}$, which doesn't make sense. It would falsely imply the units of $k$ to be ${\text{s}}^{- n}$ instead of ${\text{s}}^{- 1}$, for a 1st-order reaction.