# Question 68ca2

Feb 7, 2016

Here's what's going on here.

#### Explanation:

That equation represents a variation of the integrated rate law for a first-order reaction.

For a generic first-order reaction

$\textcolor{b l u e}{\text{A " -> " products}}$

you know that the rate of the reaction depends linearly on the concentration of the reactant

"rate" prop ["A"]

Now, you can express the rate of the reaction in terms of the rate of change in the concentration of the reactant with time. If we take ${A}_{0}$ to be the concentration of $\text{A}$ at the beginning of the reaction, we can say that after a time $t$, the concentration of $\text{A}$ will be

["A"]_t = A_0 - x" ", where

$x$ - how much of the reactant is consumed in a period of time $t$

This means that you can write

$\frac{\mathrm{dx}}{\mathrm{dt}} \propto {A}_{0} - x$

The rate of formation of the product, $\frac{\mathrm{dx}}{\mathrm{dt}}$, is proportional to how much reactant remains unconsumed.

You can write the rate law for this reaction, which establishes a relationship between the concentration of the reactant and the rate of the reaction, by introducing a rate constant $k$

$\frac{\mathrm{dx}}{\mathrm{dt}} = k \cdot \left({A}_{0} - x\right)$

To find a relationship between the concentration of the reactant and the time of the reaction, you need to integrate this equation. Rearrange to get

$\frac{\mathrm{dx}}{{A}_{0} - x} = k \cdot \mathrm{dt}$

$\int \frac{1}{{A}_{0} - x} \mathrm{dx} = k \cdot \int \left(\mathrm{dt}\right)$

This will be equal to

-ln(A_0 -x ) = k * t + c" " " "color(red)("(*)")#

Here $c$ is an integration constant.

To get rid of $c$, use the fact that at $t = 0$, i.e. before the reaction takes place, you have $\left[\text{A}\right] = {A}_{0}$ and $t = 0$.

This means that you have

$- \ln \left({A}_{0} - 0\right) = k \cdot 0 + c \implies c = - \ln \left({A}_{0}\right)$

Plug this into equation $\textcolor{red}{\text{(*)}}$ to get

$- \ln \left({A}_{0} - x\right) = k \cdot t - \ln \left({A}_{0}\right)$

Rearrange and use the property of logs to find

$k \cdot t = \ln \left({A}_{0}\right) - \ln \left({A}_{0} - x\right)$

$k \cdot t = \ln \left({A}_{0} / \left({A}_{0} - x\right)\right)$

Now, use the conversion factor between the natural log, $\text{ln}$, and the common log, $\text{log}$

$\ln \left(x\right) = 2.303 \cdot \log \left(x\right)$

to write

$\textcolor{b l u e}{k = \frac{2.303}{t} \cdot \log \left({A}_{0} / \left({A}_{0} - x\right)\right)}$

The idea now is that you can express concentration as volume.

For a reaction that features two gases, you can say that the volume at infinity, ${V}_{\infty}$, which is simply the volume of the gaseous product collected at a time after the reaction is complete, will depend on the initial amount of reactant present.

${V}_{\infty} \propto {A}_{0}$

Likewise, the volume at time $t$, ${V}_{t}$, will depend on how much product was produced as a given time time $t$

${V}_{t} \propto x$

And so you can thus say that

$\textcolor{g r e e n}{k = \frac{2.303}{t} \cdot \log \left({V}_{\infty} / \left({V}_{\infty} - {V}_{t}\right)\right)}$