The rate constant, #k#, for a particular reaction is #1.3xx10^(-4)"M"^(-1)"s"^(-1)# at #100^@"C"# and #1.5xx10^(-3)"M"^(-1)"s"^(-1)# at #150^@"C"#. What is the #E_a# (in kJ) for this reaction?

1 Answer
Mar 17, 2016

Answer:

#"64 kJ mol"^(-1)#

Explanation:

Your tool of choice here will be the Arrhenius equation, which 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
#R# - the universal gas constant, useful here as #8.314"J mol"^(-1)"K"^(-1)#
#T# - the absolute temperature at which the reaction takes place

As you can see, the Arrhenius equation establishes a relationship between the rate constant and the absolute temperature at which the reaction takes place.

In other words, this equation allows you to determine how a change in temperature affects the rate of the reaction.

Now, convert the two temperatures from degrees Celsius to Kelvin by using the conversion factor

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

You will thus have

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

#T_2 = 150^@"C" + 273.15 = "423.15 K"#

So, you know that at #T_1#, the rate constant for your reaction takes the form

#k_1 = A * "exp"( -E_a/(R * T_1))#

Likewise, at #T_2# the rate constant takes the form

#k_2 = A * "exp"( -E_a/(R * T_2))#

Divide these two equations to get rid of the pre-exponential factor, #A#

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

This will be equivalent to - using the properties of exponents

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

Take the natural log of both sides of the equation to get

#ln(k_1/k_2) = E_a/R * (1/T_2 - 1/T_1)#

Rearrange to solve for #E_a#, the activation energy of the reaction

#color(purple)(|bar(ul(color(white)(a/a)color(black)(E_a = R * ln(k_1/k_2)/(1/T_2 - 1/T_1))color(white)(a/a)|)))#

Finally, plug in your values to get

#E_a = "8.314 J mol"^(-1) color(red)(cancel(color(black)("K"^(-1)))) * ln( (1.3 * 10^(-4)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1)))))/(1.5 * 10^(-3)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1))))))/((1/423.15 - 1/373.15)color(red)(cancel(color(black)("K"^(-1)))))#

#E_a = "64,212.3 J mol"^(-1)#

Expressed in kilojoules per mole, the answer will be

#E_a = color(green)(|bar(ul(color(white)(a/a)"64 kJ mol"^(-1)color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs.