# How the shapes of the plots [dye] vs time, ln[dye] vs time, and 1/[dye] vs time are used to determine the order of the reaction with respect to the dye concentration?

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Feb 7, 2018

#### Answer:

Warning! Long Answer. They come from the integrated rate laws for zero-, first-, and second-order reactions.

#### Explanation:

Chemists like to plot graphs that are straight lines.

Zero-order reactions

The integrated rate law is

${\left[\text{A"] = "-"kt + ["A}\right]}_{0}$

Compare this with the equation for a straight line.

${\left[\text{A"] = "-"kt + ["A}\right]}_{0}$
$\textcolor{w h i t e}{l} y \textcolor{w h i t e}{l l} = m x + \textcolor{w h i t e}{l} b$

If we let $y = \left[\text{A}\right]$ and $x = t$, a plot of $\left[\text{A}\right]$ vs $t$ is a straight line with slope $m = \text{-} k$ and $x$-intercept ${\left[\text{A}\right]}_{0}$

Thus, if a plot of $\left[\text{A}\right]$ vs $t$ is a straight line with a negative slope, the reaction is zero-order.

First-order reactions

The integrated rate law is

$\ln {\left[\text{A"] = -kt + ln["A}\right]}_{0}$

Compare this with the equation for a straight line.

$\ln {\left[\text{A"] = "-"kt + ["A}\right]}_{0}$
$\textcolor{w h i t e}{m} y \textcolor{w h i t e}{l l} = \textcolor{w h i t e}{l} m x + b$

If we let $y = \ln \left[\text{A}\right]$ and $x = t$, a plot of $\ln \left[\text{A}\right]$ vs $t$ is a straight line with slope $m = \text{-} k$ and $x$-intercept ln["A"]_0]

Thus, if a plot of $\ln \left[\text{A}\right]$ vs $t$ is a straight line with a negative slope, the reaction is first-order.

Second-order reactions

The integrated rate law is

$\frac{1}{{\left[\text{A"]) = kt + 1/(["A}\right]}_{0}}$

Compare this with the equation for a straight line.

$\frac{1}{{\left[\text{A"]) = color(white)(l)kt + 1/(["A}\right]}_{0}}$
$\textcolor{w h i t e}{l l} y \textcolor{w h i t e}{l l} = \textcolor{w h i t e}{} m x + \textcolor{w h i t e}{l l} b$

If we let $y = \frac{1}{\left[\text{A}\right]}$ and $x = t$, a plot of $\frac{1}{\left[\text{A}\right]}$ vs $t$ is a straight line with slope $m = k$ and $x$-intercept 1/(["A"]_0.

Thus, if a plot of $\ln \frac{1}{\left[\text{A}\right]}$ vs $t$ is a straight line with a positive slope, the reaction is second-order.

Thus, you plot [dye] vs time, ln[dye] vs time, and 1/[dye] vs time.

The graph that is a straight line gives you the order of the reaction.

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