Question

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?

Answer 1

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

`"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_text(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_"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(r_2/r_1) = E_"a"/R(1/T_1 -1/T_2)`

Let's assume that the activation energy is `"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_text(a) = "60.0 kJ·mol"^"-1"`
`R color(white)(l)= "8.314 J·K"^"-1""mol"^"-1"`
`T_2 = "65 °C" = "338.15 K"`
`T_1 = "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""`