If the activation energy of a certain reaction is #"125 kJ/mol"#, and the reaction is at #"312.00 K"#, at what new temperature would its rate constant be twice as large?

2 Answers
Oct 30, 2016

Answer:

The rate constant will double at 317 K.

Explanation:

The Arrhenius equation gives the relation between temperature and reaction rates:

#color(blue)(bar(ul(|color(white)(a/a)k = Ae^(-E_"a"/(RT))color(white)(a/a)|)))" "#

where

#k# = the rate constant
#A# = the pre-exponential factor
#E_"a"# = the activation energy
#R# = the Universal Gas Constant
#T# = the temperature

If we take the logarithms of both sides, we get

#lnk = lnA - E_"a"/(RT)#

Finally, if we have the rates at two different temperatures, we can derive the expression

#color(blue)(bar(ul(|color(white)(a/a)ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "#

In your problem,

#k_2 = 2k_1#
#k_1 = "0.200 s"^"-1"#
#E_"a" = "125 kJ/mol" = "125 000 J/mol"#
#R = "8.314 J·K"^"-1""mol"^"-1"#
#T_1 = "312 K"#

Now, let's insert the numbers.

#ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)#

#ln((2color(red)(cancel(color(black)(k_1))))/color(red)(cancel(color(black)(k_1)))) = ("125 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1")))) (1/("312 K") - 1/(T_2 ))#

#ln2 = ("15 035" color(red)(cancel(color(black)("K"))))/(312 color(red)(cancel(color(black)("K")))) - "15 035 K"/T_2 = 48.19 - "15 035 K"/T_2#

#"15 035 K"/T_2 = 48.19 - ln2 = 47.50#

#T_2 = "15 035 K"/47.50 = "317 K"#

Oct 30, 2016

I got #"316.55 K"#.


This is a fairly straightforward problem; the challenge is to interpret what the variables we are given mean, which equation to use, and how to use it.

#E_a = "125 kJ/mol"#
#k_1 = "0.200 s"^(-1)#
#k_2 = 2k_1#
#T_1 = "312.00 K"#
#T_2 = ?#

Each of these variables can be found in the Arrhenius equation

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

which can be written in the form of an initial #(1)# and final state #(2)#:

#k_1 = Ae^(-E_a"/"RT_1)#,
#k_2 = Ae^(-E_a"/"RT_2)#,

noting that for the same reaction at two different temperatures #T_1# and #T_2#, the activation energy #E_a#, frequency factor #A#, and universal gas constant #R# are identical.

Therefore, we can solve for #T_2# by dividing these equations and taking their natural log:

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

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

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

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

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

which we can recognize as the Arrhenius equation for two different temperatures #T_1# and #T_2#:

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

Now, setting #k_2 = 2k_1# as our desired result, we get:

#ln(cancel(k_1)/(2cancel(k_1))) = -(E_a)/(R)[1/(T_1) - 1/(T_2)]#

#-ln(2) = -(E_a)/(R)[1/(T_1) - 1/(T_2)]#

Moving #E_a/R# to the other side:

#(Rln(2))/(E_a) = 1/(T_1) - 1/(T_2)#

Subtracting #1/T_1# to the other side and cross-multiplying:

#(Rln(2)T_1)/(E_aT_1) - (E_a)/(E_aT_1) = -1/(T_2)#

Reciprocating both sides of the equation:

#=> T_2 = -1/[(Rln(2)T_1)/(E_aT_1) - (E_a)/(E_aT_1)]#

#= -1/[(Rln(2)T_1 - E_a)/(E_aT_1)] = 1/[(E_a - Rln(2)T_1)/(E_aT_1)]#

Therefore:

#color(blue)(T_2) = [E_aT_1]/(E_a - Rln(2)T_1)#

#= [("125 kJ/mol")("312.00 K")]/(("125 kJ/mol") - ("0.008314472 kJ/mol"cdot"K")ln(2)("312.00 K"))#

#=# #color(blue)("316.55 K")#

(Note that we needed five sig figs to get the necessary precision. If we had written #T = "317 K"#, as we would for #3# sig figs, we would have gotten that #k_2# is about #2.14# times #k_1# instead, when we wanted #2# times #k_1#.)