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).

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 .

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

$color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))$

In your case, you will have

$T_1 = 100^@"C" + 273.15 = "373.15 K"$

$T_2 = 120^@"C" + 273.15 = "393.15 K"$

If you take $k_1$ to be the rate constant of the reaction at $T_1$, you can say that

$k_1 = A * "exp"( -E_a/(R * T_1))" " " "color(orange)((1))$

Similarly, if you take $k_2$ to be the rate constant of the reaction at $T_2$, you will have

$k_2 = A * "exp" (-E_a/(R * T_2))" " " "color(orange)((2))$

Now, let's assume that your reaction is $n$ order with respect to a reactant $"A"$

$color(blue)(n"A" -> "products")$

The differential 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

$"rate"_1 = k_1 * ["A"]^n" "$ 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 $color(orange)((2))$ and $color(orange)((1))$ to get

$k_2/k_1 = (color(red)(cancel(color(black)(A))) * "exp"(-E_a/(R * T_2)))/(color(red)(cancel(color(black)(A))) * "exp"(-E_a/(R * T_1)))$

This will be equivalent to

$k_2/k_1 = "exp" [E_a/R * (1/T_1 - 1/T_2)]$

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

$"1 kcal" = 10^3"cal"$

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

$"rate"_2/"rate"_1 = color(green)(|bar(ul(color(white)(a/a)2.8color(white)(a/a)|)))$

Therefore, the reaction will proceed $2.8$ times faster if the temperature is increased by $20^@"C"$.