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).

1 Answer
Mar 16, 2016

Answer:

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

#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 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

#"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 sig figs

#"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"#.