# Question #a2fab

##### 1 Answer

#### Explanation:

Your starting point here will be the **Arrhenius equation**, which shows how the **rate constant** of a given reaction depends on the **absolute temperature** at which the reaction takes place

#color(blue)(|bar(ul(k = A * exp(-E_a/(RT))))|)" "# , where

*rate constant*

*pre-exponential factor*, specific to a given reaction

**activation energy** of the reaction

*universal gas constant*, used as

*absolute temperature* of the gas

To find the activation energy of your reaction, take the natural log of both sides of the Arrhenius equation. This will get you

#ln(k) = ln[A * exp(-E_a/(RT))]#

#ln(k) = ln(A) + ln[exp(-E_a/(RT))]#

#ln(k) = ln(A) - E_a/(RT)#

Now, you know the value of the rate constant at two different temperatures

#k_1 = 3.2 * 10^(-6)"M s"^(-1) -> "at T"_1 = "298 K"# #k_2 = 4.2 * 10^(-5)"M s"^(-1) -> "at T"_2 = "398 K"#

This means that you can write

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

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

Divide these two equations to get rid of the

#ln(k_1) - ln(k_2) = color(red)(cancel(color(black)(ln(A)))) - color(red)(cancel(color(black)(ln(A)))) - E_a/(RT_1) + E_a/(RT_2)#

This will be equivalent to

#color(blue)(|bar(ul(ln(k_1/k_2) = E_a/R * (1/T_2 - 1/T_1)))|)#

This equation will allow you to find the value of

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

#E_a = ln((3.2 * 10^(-6)color(red)(cancel(color(black)("M s"^(-1)))))/(4.2 * 10^(-5)color(red)(cancel(color(black)("M s"^(-1)))))) * ("8.314 J mol"^(-1)color(red)(cancel(color(black)("K"^(-1)))))/((1/398 - 1/298)color(red)(cancel(color(black)("K"^(-1)))))#

#E_a = "25387 J mol"^(-1)#

I'll leave the answer rounded to three **sig figs** and expressed in *kilojoules per mole*

#E_a = color(green)(|bar(ul("25.4 kJ mol"^(-1)))|)#

Mind you, you *should* round this off to two sig figs, since that's how many sig figs you have for the two rate constants.