# 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"?

Jan 28, 2017

${t}_{\text{1/2"^((2)) = "85.25 s}}$

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:

${t}_{\text{1/2}} = \ln \frac{2}{k}$

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

${k}_{1} = \ln \frac{2}{{t}_{\text{1/2}}^{\left(1\right)}}$$\text{ "" "" }$${k}_{2} = \ln \frac{2}{{t}_{\text{1/2}}^{\left(2\right)}}$

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

$\boldsymbol{k = A {e}^{- {E}_{a} \text{/} R T}}$

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 $\text{kJ/mol}$.
• $R = \text{0.008314472 kJ/mol"cdot"K}$ is the universal gas constant. Make sure you get the units correct on this!
• $T$ is the temperature in $\text{K}$ (not $\text{^@ "C}$).

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

${k}_{2} = A {e}^{- {E}_{a} \text{/} R {T}_{2}}$$\text{ "" "" }$${k}_{1} = A {e}^{- {E}_{a} \text{/} R {T}_{1}}$

we can divide these:

${k}_{2} / {k}_{1} = {e}^{- {E}_{a} \text{/"RT_2)/e^(-E_a"/} R {T}_{1}}$

Take the $\ln$ of both sides:

$\textcolor{g r e e n}{\ln \left({k}_{2} / {k}_{1}\right)} = \ln \left({e}^{- {E}_{a} \text{/"RT_2)/e^(-E_a"/} R {T}_{1}}\right)$

$= \ln \left({e}^{- {E}_{a} \text{/"RT_2)) - ln(e^(-E_a"/} R {T}_{1}}\right)$

$= - {E}_{a} / \left(R {T}_{2}\right) - \left(- {E}_{a} / \left(R {T}_{1}\right)\right)$

$= \textcolor{g r e e n}{- {E}_{a} / R \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]}$

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

$\ln \left(\left(\cancel{\ln 2} {\text{/"t_"1/2"^((2)))/(cancel(ln2)"/"t_"1/2}}^{\left(1\right)}\right)\right) = - {E}_{a} / R \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]$

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

$\implies \textcolor{g r e e n}{\ln \left(\left({t}_{\text{1/2"^((1)))/(t_"1/2}}^{\left(2\right)}\right)\right) = - {E}_{a} / R \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]}$

Now, we can solve for the new half-life, ${t}_{\text{1/2}}^{\left(2\right)}$, at the new temperature, ${40}^{\circ} \text{C}$. First, convert the temperatures to $\text{K}$:

${T}_{1} = 25 + 273.15 = \text{298.15 K}$

${T}_{2} = 40 + 273.15 = \text{313.15 K}$

Finally, plug in and solve. We should recall that $\ln \left(\frac{a}{b}\right) = - \ln \left(\frac{b}{a}\right)$, so the negative cancels out if we flip the $\ln$ argument.

$\implies \ln \left(\left({t}_{\text{1/2"^((2)))/(t_"1/2}}^{\left(1\right)}\right)\right) = {E}_{a} / R \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]$

$\implies \ln \left(\left({t}_{\text{1/2"^((2)))/("400 s")) = ("80 kJ/mol")/("0.008314472 kJ/mol"cdot"K")[1/("313.15 K") - 1/("298.15 K}}\right)\right]$

$= \left({\text{9621.78 K") (-1.607xx10^(-4) "K}}^{- 1}\right)$

$= - 1.546$

Now, exponentiate both sides to get:

$\left({t}_{\text{1/2"^((2)))/("400 s}}\right) = {e}^{- 1.546}$

$\implies \textcolor{b l u e}{{t}_{\text{1/2"^((2))) = ("400 s}}} \left({e}^{- 1.546}\right)$

$=$ $\textcolor{b l u e}{\text{85.25 s}}$

This should make sense, physically. From the Arrhenius equation, the higher ${T}_{2}$ is, the more negative the $\left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]$ 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 $\ln \left({k}_{2} / {k}_{1}\right)$ is very large, ${k}_{2} \text{>>} {k}_{1}$). Therefore, higher temperatures means larger rate constants.

Furthermore, the rate constant is proportional to the rate of reaction $r \left(t\right)$ in the rate law. Therefore...

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