# 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).

Here’s what I got.

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”##.

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