Question

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?

Answer 1

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"`

Answer 2

I got `"316.55 K"`.

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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`.)