# What is the rate law equation for the following reaction (given the experimental data below)?

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Mar 9, 2018

The rate law is $r = k {\left[\text{A}\right]}^{2}$.

#### Explanation:

The rate law is

$r = k {\text{[A]}}^{m}$

where

$r \textcolor{w h i t e}{l} =$ the reaction rate
$k \textcolor{w h i t e}{l} =$ the rate constant
$m =$ an integer whose value we must determine

Consider Trial Runs 1 and 2

r_2/r_1 = (color(red)(cancel(color(black)(k)))(0.8 color(red)(cancel(color(black)("mol/L")))) ^m)/(color(red)(cancel(color(black)(k)))(0.4 color(red)(cancel(color(black)("mol/L"))))^m) = (0.8/0.4)^m = (8.0 color(red)(cancel(color(black)("mol·L"^"-1""s"^"-1"))))/ (2.0 color(red)(cancel(color(black)("mol·L"^"-1""s"^"-1"))))

${2}^{m} = 4.0$

Thus, doubling the concentration quadruples the rate.

The reaction is second order in $\left[\text{A}\right]$.

If you use logarithms, you get

$\textcolor{b l u e}{m \log 2 = \log 4.0}$

color(blue)(m = log4.0/log2 = 2.0 ≈ 2)

The reaction is second order in $\text{[A]}$.

The rate law is $r = k {\left[\text{A}\right]}^{2}$.

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