# 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

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

*rate constant* for a given reaction

*pre-exponential factor* (the frequency factor)

*activation energy* of the reaction

**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 *incomplete*.

However, I'll use both values for **specific to a reaction**, i.e. constant for a given reaction, so you should expect the same result for both values of

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 = 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

#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

#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.