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
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.