# Given an activation energy of 15 kcal/mol, how do you use the Arrhenius equation to estimate how much faster the reaction will occur if the temperature is increased from 100 degrees Celsius to 120 degrees Celsius? R = 1.987 cal/(mol)(k).

Mar 16, 2016

Here's what I got.

#### Explanation:

The Arrhenius equation looks like this

color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" ", where

$k$ - the rate constant for a given reaction
$A$ - the pre-exponential factor, specific to a given reaction
${E}_{a}$ - the activation energy of the reaction
$T$ - the absolute temperature at which the reaction takes place

In essence, the Arrhenius equation establishes a relationship between the rate constant of a reaction and the absolute temperature at which the reaction takes place.

In other words, this equation allows you to figure out how a change in temperature will ultimately affect the rate of the reaction.

The two temperatures at which the reaction takes place can be calculated using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{T \left[\text{K"] = t[""^@"C}\right] + 273.15} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you will have

${T}_{1} = {100}^{\circ} \text{C" + 273.15 = "373.15 K}$

${T}_{2} = {120}^{\circ} \text{C" + 273.15 = "393.15 K}$

If you take ${k}_{1}$ to be the rate constant of the reaction at ${T}_{1}$, you can say that

${k}_{1} = A \cdot \text{exp"( -E_a/(R * T_1))" " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Similarly, if you take ${k}_{2}$ to be the rate constant of the reaction at ${T}_{2}$, you will have

${k}_{2} = A \cdot \text{exp" (-E_a/(R * T_2))" " " } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

Now, let's assume that your reaction is $n$ order with respect to a reactant $\text{A}$

$\textcolor{b l u e}{n \text{A" -> "products}}$

The differential rate law for this generic reaction would look like this

"rate" = k * ["A"]^n

Assuming that you'll perform the reaction at ${T}_{1}$ and at ${T}_{2}$ using the same concentration for the reactant, you can say that you have

$\text{rate"_1 = k_1 * ["A"]^n" }$ and " " "rate"_2 = k_2 * ["A"]^n

Your goal here will be to find the ratio that exists between the rate of the reaction at ${T}_{2}$ and the rate of the reaction at ${T}_{1}$. This comes down to finding

"rate"_2/"rate"_1 = (k_2 * color(red)(cancel(color(black)(["A"]^n))))/(k_1 * color(red)(cancel(color(black)(["A"]^n)))) = k_2/k_1 = ?

Now, divide equations $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ and $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to get

${k}_{2} / {k}_{1} = \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{A}}} \cdot \text{exp"(-E_a/(R * T_2)))/(color(red)(cancel(color(black)(A))) * "exp} \left(- {E}_{a} / \left(R \cdot {T}_{1}\right)\right)\right)$

This will be equivalent to

${k}_{2} / {k}_{1} = \text{exp} \left[{E}_{a} / R \cdot \left(\frac{1}{T} _ 1 - \frac{1}{T} _ 2\right)\right]$

Before plugging in your values, make sure that you do not forget to convert the activation energy from kcal per mole to cal per mole by using the conversion factor

$\text{1 kcal" = 10^3"cal}$

You will have

k_2/k_1 = "exp"[ (15 * 10^3color(red)(cancel(color(black)("cal"))) color(red)(cancel(color(black)("mol"^(-1)))))/(1.987color(red)(cancel(color(black)("cal")))color(red)(cancel(color(black)("mol"^(-1))))color(red)(cancel(color(black)("K"^(-1))))) *(1/373.15 - 1/393.15)color(red)(cancel(color(black)("K"^(-1))))]

${k}_{2} / {k}_{1} = 2.799$

I'll leave the answer rounded to two sig figs

${\text{rate"_2/"rate}}_{1} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 2.8 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Therefore, the reaction will proceed $2.8$ times faster if the temperature is increased by ${20}^{\circ} \text{C}$.