# How to calculate the exponents in the differential rate law?

## GIVEN:

Nov 18, 2017

You use the formulas from the integrated rate laws.

#### Explanation:

Shen you have a list of concentrations as a function of time, you want to plot something that gives a straight line.

The equation for a straight line is:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} y = m x + b \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can write the integrated rate laws for zero-, first, and second-order reactions all as straight-line functions.

$\underline{\boldsymbol{\text{Order} \textcolor{w h i t e}{m l l} y = \textcolor{w h i t e}{l} m \textcolor{w h i t e}{l} x + b}}$
$\textcolor{w h i t e}{m l l} 0 \textcolor{w h i t e}{m m m l} c = \textcolor{w h i t e}{l} \text{-} k \textcolor{w h i t e}{l l} t + {c}_{0}$
$\textcolor{w h i t e}{m l l} 1 \textcolor{w h i t e}{m m} \ln c = \textcolor{w h i t e}{l} \text{-} k \textcolor{w h i t e}{l l} t + \ln {c}_{0}$
color(white)(mll)2color(white)(mmm)1/c = color(white)(ll )kcolor(white)(ll)t + 1/c_0

Thus,

• For a 0-order reaction, a plot of $c$ vs $t$ should give a straight line with slope $\text{-} k$
• For a 1st-order reaction, a plot of $\ln \left(c\right)$ vs $t$ should give a straight line with
slope $\text{-} k$
• For a 2nd-order reaction, a plot of $\frac{1}{c}$ vs $t$ should give a straight line with slope $k$

So, we plot each of these functions and see which gives the best straight line.

I used Excel to convert your data into the following table.

Then I plotted each of the three functions.

Zero-Order Plot

The graph appears to be a curve.

First-Order Plot

The graph appears to be a straight line.

Second-Order Plot

The graph appears to be a curve.

Conclusion

Your data give the best fit to a first-order rate law of the form

$r a t e = k {\left[{\text{SO"_2"Cl}}_{2}\right]}^{1}$

Nov 23, 2017

Jose suggested a different way to approach this, so I figured I'd share it.

The starting rate law is:

$r \left(t\right) = k {\left[{\text{SO"_2"Cl}}_{2}\right]}^{m}$

Jose suggested that one could take the $\ln$ of both sides like this:

$\ln r \left(t\right) = m \ln \left[{\text{SO"_2"Cl}}_{2}\right] + \ln k$

But this requires that we generate a new column of data:

$r \left(t\right) = - \frac{\Delta \left[{\text{SO"_2"Cl}}_{2}\right]}{\Delta t}$

and for $n$ data points, we generate $n - 1$ new data points to use here. We use the midpoint concentrations of ${\text{SO"_2"Cl}}_{2}$ (call them ["SO"_2"Cl"_2]_"mp") and generate each $r \left(t\right)$ as described above.

ul(ln[r(t) xx 10^4] ("M/s")" "" "" "ln["SO"_2"Cl"_2]_"mp" ("M")" ")

$\ln 1.24 \text{ "" "" "" "" "" "" "" "" } \ln 0.0938$
$\ln 1.08 \text{ "" "" "" "" "" "" "" "" } \ln 0.0822$
$\ln 0.95 \text{ "" "" "" "" "" "" "" "" } \ln 0.0721$
$\ln 0.83 \text{ "" "" "" "" "" "" "" "" } \ln 0.0632$
$\ln 0.73 \text{ "" "" "" "" "" "" "" "" } \ln 0.0554$
$\ln 0.64 \text{ "" "" "" "" "" "" "" "" } \ln 0.0485$

You get the idea. I'll leave out the remaining $5$ data points because six is quite sufficient to establish a trend.

Plotting $\ln r \left(t\right)$ vs. $\ln \left[{\text{SO"_2"Cl}}_{2}\right]$, the y-intercept should be $\ln k$, and the slope should be whatever the order of reaction is.

The regression equation was found as follows using my trusty TI-83 calculator.

1. $\text{STAT}$ $\to$ $\text{Edit}$, enter all the data in there. Have the $x$ variable in $\text{L1}$ and $y$ variable in $\text{L2}$.
2. "STAT" -> "CALC" -> "LinReg"("ax+b")

This gave:

lnr(t) = 1.000765854 cdot ln["SO"_2"Cl"_2]_"mp" - 6.630082087

This implies that the order of reaction is approx. $\textcolor{b l u e}{1}$ and that

$\ln k = - 6.630082087$

and

$\textcolor{b l u e}{k} = {e}^{- 6.630082087} = \textcolor{b l u e}{{\text{0.0013 s}}^{- 1}}$

So this method works too. However, you can see it is a big hassle, when you have this many data points, to rewrite all of them into $\ln r \left(t\right)$ and ln["SO"_2"Cl"_2]_"mp". I would say Ernest's method is faster.