A chemist runs a chemical reaction at 15°C and decides that it proceeds far too slowly. As a result, he decides that the reaction rate must be increased by a factor of 16. At what temperature should the chemist run the reaction to achieve this goal?

Jun 26, 2018

Use the Arrhenius Equation to define the change in temperature to the desired rate.
$k = A {e}^{- {E}_{a} / \left(R T\right)}$

Explanation:

For thermodynamics the ratios will NOT WORK unless the temperatures are all in ABSOLUTE DEGREES!! (Degrees Kelvin). Thus, you must convert the Celcius degree value to Kelvins before doing the multiplication.

Other ratios would require different temperatures.
https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Kinetics/Modeling_Reaction_Kinetics/Temperature_Dependence_of_Reaction_Rates/The_Arrhenius_Law/The_Arrhenius_Law%3A_Arrhenius_Plots

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 want to increase the rate by a factor of 16.

$16 = {2}^{4}$

So, you will increase the temperature by 4 × 10 °C = 40 °C.

The new temperature will be 55 °C.

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 $\text{80.0 kJ·mol"^"-1}$ and set ${T}_{2}$ as the higher temperature. Then

${r}_{2} / {r}_{1} = 16$
${E}_{\textrm{a}} = \text{55.0 kJ·mol"^"-1}$
$R \textcolor{w h i t e}{l} = \text{8.314 J·K"^"-1""mol"^"-1}$
T_2 = ?
${T}_{1} = \text{15 °C" = "288.15 K}$

$\ln 16 = \left(\text{55 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J")))"·K"^"-1"color(red)(cancel(color(black)("mol"^"-1")))) (1/"288.15 K} - \frac{1}{T} _ 2\right)$

= 6615 × 3.470 × 10^"-3" -"6615 K"/T_2 = 22.96 -"6615 K"/T_2#

$2.773 = 22.96 - \frac{\text{6615 K}}{T} _ 2$

$\frac{\text{6615 K}}{T} _ 2 = 22.96 - 2.773 = 20.19$

${T}_{2} = \text{6615 K"/2019 = "328 K = 55 °C}$