# A chemist runs a reaction at 65°C and determines its rate to be 0.000364 M/s. If she decreases the temperature to 25°C, what will the rate of the reaction be?

Jun 26, 2018

Here's what I get.

#### Explanation:

There are two ways to approach this problem.

A. Use a rule of thumb

A rule of thumb states that the rate of a reaction changes by a factor of two for every 10 °C change in temperature.

You are decreasing the temperature from 65 °C to 25 °C, a decrease of 40 °C.

Thus, the new rate will be

$\text{rate" = (1/2)^4 × 3.64 × 10^"-4"color(white)(l) "mol·L"^"-1""s"^"-1" = = 1/16 × 3.64 × 10^"-4"color(white)(l) "mol·L"^"-1""s"^"-1" = 2.28 × 10^"-5"color(white)(l)"mol·L"^"-1""s"^"-1}$

B. Use the Arrhenius equation

Ideally, you would know the activation energy ${E}_{\textrm{a}}$ for the reaction.

Then you could use the Arrhenius equation to calculate the rate at the new temperature.

color(blue)(bar(ul(|color(white)(a/a)ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "

where

${k}_{2}$ and ${k}_{1}$ are the rate constants at temperatures ${T}_{2}$ and ${T}_{1}$
${E}_{\text{a}}$ = the activation energy
$R$ = the Universal Gas Constant

Since you are changing only the temperatures, the rates are directly proportional to the rate constants, and we can write:

$\ln \left({r}_{2} / {r}_{1}\right) = {E}_{\text{a}} / R \left(\frac{1}{T} _ 1 - \frac{1}{T} _ 2\right)$

Let's assume that the activation energy is $\text{60.0 kJ·mol"^"-1}$ and set ${T}_{2}$ as the higher temperature. Then

r_2 color(white)(l)= 3.64 × 10^"-4"color(white)(l) "mol·L"^"-1""s"^"-1"
r_1 color(white)(l)= ?
${E}_{\textrm{a}} = \text{60.0 kJ·mol"^"-1}$
$R \textcolor{w h i t e}{l} = \text{8.314 J·K"^"-1""mol"^"-1}$
${T}_{2} = \text{65 °C" = "338.15 K}$
${T}_{1} = \text{25 °C" = "298.15 K}$

ln(r_2/r_1) = ("60 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/("8.314 J"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1")))) (1/(298.15 color(red)(cancel(color(black)("K")))) - 1/(338.15 color(red)(cancel(color(black)("K")))))

= 7217 × 3.967 × 10^"-4" =2.863

"r_2/r_1 = e^2.863 = 17.5

r_1 = r_2/17.5 = (3.64 × 10^"-4"color(white)(l) "mol·L"^"-1""s"^"-1")/17.5 = 2.08 × 10^"-5"color(white)(l) "mol·L"^"-1""s"^"-1""