Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# 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".

LEARN MORE EFFECTIVELY AND GET BETTER GRADES!