# The rate constant of a first order reaction is 4.60 xx 10^(-4) "s"^(-1) at 350^@ "C". If the activation energy is "104 kJ/mol", what is the temperature at which the rate constant is 8.80 xx 10^(-4) "s"^(-1)?

Apr 28, 2016

I get ${371}^{\circ} \text{C}$.

Let's look at what we have:

${k}_{1} = 4.60 \times {10}^{- 4} {\text{s}}^{- 1}$ at ${T}_{1} = {350}^{\circ} \text{C}$, the rate constant for the reaction at temperature $1$

${k}_{2} = 8.80 \times {10}^{- 4} {\text{s}}^{- 1}$ at T_2 = ?, the rate constant for the same reaction at temperature $2$

${E}_{a} = \text{104 kJ/mol}$

THE ARRHENIUS EQUATION RELATES RATE CONSTANTS TO TEMPERATURE AND THE ACTIVATION ENERGY

These are related via the Arrhenius equation, which, for a given rate constant ${k}_{i}$ of reaction $i$, is

$\setminus m a t h b f \left({k}_{i} = A {e}^{- {E}_{a} \text{/} R {T}_{i}}\right)$

where:

• $A$ is the pre-exponential factor, and it is basically an experimentally-acquired constant correlating with collision frequency.
• $R$ is the universal gas constant. With energy units in $\text{J}$, we have $R = \text{8.314472 J/mol"cdot"K}$.
• $T$ is temperature in $\text{K}$.
• ${E}_{a}$, as you can tell, is the activation energy, or the energy needed to help the reactants make it over the transition state "hill" and promote a successful reaction.

To determine the temperature associated with ${k}_{2}$, what we should do is divide the equations for each ${k}_{i}$, cancelling out the pre-exponential factor (given that for the same reaction, despite being at different temperatures, $A$ is the same).

$\left[{k}_{2} = \cancel{A} {e}^{- {E}_{a} \text{/"RT_2)]/[k_1 = cancel(A)e^(-E_a"/} R {T}_{1}}\right]$

Also note that for the same reaction, the activation energy does not change with temperature. Instead, the average number of particles in an ensemble that have enough energy is higher.

DETERMINING THE Y = MX + B FORM OF THE ARRHENIUS EQUATION

We proceed to derive what would be the graphable form of this equation.

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

Taking the natural logarithm allows a separation of terms, and a cancellation of exponentials via the fact that $\ln x$ is the inverse of ${e}^{x}$:

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

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

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

$= - \frac{{E}_{a}}{R {T}_{2}} + \frac{{E}_{a}}{R {T}_{1}}$

$= \frac{{E}_{a}}{R} \left[- \frac{1}{{T}_{2}} + \frac{1}{{T}_{1}}\right]$

So that gives us:

$\textcolor{b l u e}{\ln \setminus \frac{{k}_{2}}{{k}_{1}} = - \frac{{E}_{a}}{R} \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]}$

This is the graphable form of the Arrhenius equation, and would normally be used with known temperatures to determine the activation energy.

DETERMINING THE NEW TEMPERATURE

However, we have been solving for ${T}_{2}$. So, let's get the final expression.

$- \frac{R \ln \setminus \frac{{k}_{2}}{{k}_{1}}}{{E}_{a}} = \frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}$

$- \frac{R \ln \setminus \frac{{k}_{2}}{{k}_{1}}}{{E}_{a}} + \frac{1}{{T}_{1}} = \frac{1}{{T}_{2}}$

${T}_{2} = \frac{1}{\frac{1}{{T}_{1}} - \frac{R \ln \setminus \frac{{k}_{2}}{{k}_{1}}}{{E}_{a}}}$

Yikes, kind of messy. So let's simplify that some more... Cross-multiply:

${T}_{2} = \frac{1}{\frac{{E}_{a}}{{E}_{a} {T}_{1}} - \frac{{T}_{1} R \ln \setminus \frac{{k}_{2}}{{k}_{1}}}{{E}_{a} {T}_{1}}}$

Combine in the denominator:

${T}_{2} = \frac{1}{\frac{{E}_{a} - {T}_{1} R \ln \setminus \frac{{k}_{2}}{{k}_{1}}}{{E}_{a} {T}_{1}}}$

Now flip! Much nicer.

$\textcolor{b l u e}{{T}_{2} = \frac{{E}_{a} {T}_{1}}{{E}_{a} - {T}_{1} R \ln \setminus \frac{{k}_{2}}{{k}_{1}}}}$

So, finally, let's get ${T}_{2}$:

color(blue)(T_2) = ["104000 J/mol"cdot("350+273.15 K")]/("104000 J/mol" - ("350+273.15 K")cdot"8.314472 J/mol"cdot"K"cdotln\frac(8.80xx10^(-4) "s"^(-1))(4.60xx10^(-4) "s"^(-1)) )

$= \left(64807600 \cancel{\text{J/mol")cdot"K")/([104000-(5181.1632268*0.648695)] cancel("J/mol}}\right)$

$=$ $\text{643.96 K}$

$\to \textcolor{b l u e}{\approx {371}^{\circ} \text{C}}$

So, this is saying that a higher temperature increases the rate constant, which makes sense.

The higher the rate constant, the faster the reaction occurs (given that it is in reciprocal time units, where less time is faster), and the reaction should be faster because the average kinetic energy of the system is higher, promoting more successful collisions more quickly.

And to make sure we didn't botch our calculations, let's check to see we get the right ${k}_{2}$ back:

$\ln \setminus \frac{{k}_{2}}{{k}_{1}} = - \frac{{E}_{a}}{R} \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]$

${k}_{2} = {k}_{1} {e}^{- \frac{{E}_{a}}{R} \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]}$

So, we have:

color(green)(k_2) = 4.60xx10^(-4) "s"^(-1)cdote^(-"104000 J/mol"/("8.314472 J/mol"cdot"K")(1/("643.96 K") - 1/("623.15 K"))

= 8.7997xx10^(-4) "s"^(-1) ~~ color(green)(8.80xx10^(-4) "s"^(-1))

It works!