A second-order reaction has a rate constant of 0.007500/(M · s) at 30°C. At 40°C, the rate constant is 0.05800/(M · s). The activation energy is 161.23 kJ/mol. What is the frequency factor, A?

1 Answer
Apr 1, 2016

Answer:

Here's what I got.

Explanation:

This is a pretty straightforward Arrhenius equation practice problem, so make sure that you're familiar with 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 (the frequency factor)
#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

Notice that you don't really need to know the value of the rate constant at two different temperatures in order to find the value of #A#, so I'm assuming that the question is incomplete.

However, I'll use both values for #k# to double-check the answer. Keep in mind that #A# is specific to a reaction, i.e. constant for a given reaction, so you should expect the same result for both values of #k# and #T#.

Start by converting the temperature 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 have

#T_1 = 30^@"C" + 273.15 = "303.15 K"#

#T_2 = 40^@"C" + 273.15 = "313.15 K"#

Rearrange the Arrhenius equation to solve for #A#

#A = k/("exp"(-E_a/(R * T)))#

Don't forget to convert the activation energy from kilojoules to joules

#"1 kJ" = 10^3"J"#

So, at #T_1# and #k_1 = "0.007500 M"^(-1)"s"^(-1)#, you have

#A = ("0.007500 M"^(-1)"s"^(-1))/("exp"( - (161.23 * 10^3color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("mol"^(-1)))))/(8.314color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("mol"^(-1)))) color(red)(cancel(color(black)("K"^(-1)))) * 303.15color(red)(cancel(color(black)("K")))))#

#A ~~ 4.5 * 10^(25)"M"^(-1)"s"^(-1)#

At #T_2# and #k_2 = "0.05800 M"^(-1)"s"^(-1)#, you have

#A = ("0.05800 M"^(-1)"s"^(-1))/("exp"( - (161.23 * 10^3color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("mol"^(-1)))))/(8.314color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("mol"^(-1)))) color(red)(cancel(color(black)("K"^(-1)))) * 313.15color(red)(cancel(color(black)("K")))))#

#A ~~ 4.5 * 10^(25)"M"^(-1)"s"^(-1)#

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the two temperatures.