Question

What is the half-life of a first-order reaction at `40^@ "C"` if its half-life at `25^@ "C"` was `"400 s"` and the activation energy is `"80 kJ/mol"`?

Answer 1

`t_"1/2"^((2)) = "85.25 s"`

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Notice how you're given the half-life (for one temperature), a second temperature, and the activation energy. The key to doing this problem is recognizing that:

• the half-life for a first-order reaction is related to its rate constant.
• the rate constant changes at different temperatures.

Go here for a derivation of the half-life of a first-order reaction. You should find that:

`t_"1/2" = ln2/k`

Therefore, if we label each rate constant, we have:

`k_1 = ln2/(t_"1/2"^((1)))` `" "" "" "` `k_2 = ln2/(t_"1/2"^((2)))`

Recall that the activation energy can be found in the Arrhenius equation:

`bb(k = Ae^(-E_a"/"RT))`

where:

• `A` is the frequency factor, i.e. it is proportional to the number of collisions occurring over time.
• `E_a` is the activation energy in `"kJ/mol"`.
• `R = "0.008314472 kJ/mol"cdot"K"` is the universal gas constant. Make sure you get the units correct on this!
• `T` is the temperature in `"K"` (not `""^@ "C"`).

Now, we can derive the Arrhenius equation in its two-point form. Given:

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

we can divide these:

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

Take the `ln` of both sides:

`color(green)(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))`

`= color(green)(-E_a/R[1/(T_2) - 1/(T_1)])`

Now if we plug in the rate constants in terms of the half-lives, we have:

`ln((cancel(ln2)"/"t_"1/2"^((2)))/(cancel(ln2)"/"t_"1/2"^((1)))) = -E_a/R[1/(T_2) - 1/(T_1)]`

This gives us a new expression relating the half-lives to the temperature:

`=> color(green)(ln((t_"1/2"^((1)))/(t_"1/2"^((2)))) = -E_a/R[1/(T_2) - 1/(T_1)])`

Now, we can solve for the new half-life, `t_"1/2"^((2))`, at the new temperature, `40^@ "C"`. First, convert the temperatures to `"K"`:

`T_1 = 25 + 273.15 = "298.15 K"`

`T_2 = 40 + 273.15 = "313.15 K"`

Finally, plug in and solve. We should recall that `ln(a/b) = -ln(b/a)`, so the negative cancels out if we flip the `ln` argument.

`=> ln((t_"1/2"^((2)))/(t_"1/2"^((1)))) = E_a/R[1/(T_2) - 1/(T_1)]`

`=> ln((t_"1/2"^((2)))/("400 s")) = ("80 kJ/mol")/("0.008314472 kJ/mol"cdot"K")[1/("313.15 K") - 1/("298.15 K")]`

`= ("9621.78 K") (-1.607xx10^(-4) "K"^(-1))`

`= -1.546`

Now, exponentiate both sides to get:

`(t_"1/2"^((2)))/("400 s") = e^(-1.546)`

`=> color(blue)(t_"1/2"^((2))) = ("400 s")(e^(-1.546))`

`=` `color(blue)("85.25 s")`

This should make sense, physically. From the Arrhenius equation, the higher `T_2` is, the more negative the `[1/T_2 - 1/T_1]` term, which means the larger the right hand side of the equation is.

The larger the right hand side gets, the larger `k_2` is, relative to `k_1` (i.e. if `ln(k_2/k_1)` is very large, `k_2 ">>" k_1`). Therefore, higher temperatures means larger rate constants.

Furthermore, the rate constant is proportional to the rate of reaction `r(t)` in the rate law. Therefore...

The higher the rate constant, the faster the reaction, and thus the shorter its half-life should be.