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 $"80 kJ/mol"$?

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

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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 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 . Therefore...

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