At #2^@ "C"#, a certain reaction has a rate constant of #4.60 xx 10^(-6) "s"^(-1)#. If the activation energy is #"108 kJ/mol"#, at what temperature does this reaction have a rate constant of #"0.58 s"^(-1)#?

#T_2 ~~ "366 K"#

This makes sense because the rate constant is directly proportional to the rate of reaction. The rate constant should be much higher only if we are at a much higher temperature.

It is extremely important that you know how to manipulate the Arrhenius equation in general.

If you simply plug in numbers and work through this without units, you will likely forget that #R# must be in the same energy units as #E_a# and that #T# must be in #"K"#.

This is, as we have seen before, using the Arrhenius equation.

#k = Ae^(-E_a//RT)#

• #k# is the rate constant.
• #A# is the frequency factor, which has the same units as the rate constant.
• #E_a# is the activation energy, usually in #"kJ/mol"#.
• #R = "0.008314472 kJ/mol"cdot"K"# is the universal gas constant that ensures the units cancel in the exponential.
• #T# is the temperature in #"K"#.

You should be able to easily declare two states, giving you two equations (holding #A# constant) as follows:

#k_1 = Ae^(-E_a//RT_2)#

#k_2 = Ae^(-E_a//RT_1)#

By dividing these, you can then get an expression for the two rate constants #k_1,k_2# at the corresponding temperatures #T_1,T_2#:

#k_2/k_1 = e^(-E_a/(RT_2))/e^(-E_a/(RT_1))#

Take the #ln# of both sides:

#ln(k_2/k_1) = ln(e^(-E_a//RT_2)/e^(-E_a//RT_1))#

#= ln(e^(-E_a//RT_2)) - ln(e^(-E_a//RT_1))#

#= -E_a/(RT_2) - -E_a/(RT_1)#

#= -E_a/R[1/T_2 - 1/T_1]#

And that would yield the #ln# form of the Arrhenius equation:

#barulbb(|stackrel(" ")(" "ln(k_2/k_1) = -E_a/R[1/T_2 - 1/T_1]" ")|)#

To get the new temperature at which #k_2 = "0.58 s"^(-1)#, you would solve for #T_2#, having:

• #k_1 = 4.60 xx 10^(-6) "s"^(-1)# at #T_1 = "275 K"#
• #E_a = "108 kJ/mol"#
• #R = "8.314472 kJ/mol"cdot"K" = "0.008314472 kJ/mol"cdot"K"#

Solving for #T_2# begins as multiplying by #-R/E_a# on both sides:

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

As a result, the second temperature is obtained by adding #1/T_1# to both sides and taking the reciprocal:

#color(blue)(T_2) = [-R/(E_a) ln(k_2/k_1) + 1/T_1]^(-1)#

#= [-("0.008314472 kJ/mol"cdot"K")/("108 kJ/mol") cdot ln(("0.58 s"^(-1))/(4.60 xx 10^(-6) "s"^(-1))) + 1/"275 K"]^(-1)#

#= ("0.002732 K"^(-1))^(-1)#

#= 1/("0.002732 K"^(-1))#

#=# #color(blue)("366.01 K")#,

which would then be rounded to the appropriate number of significant figures.