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?

Oct 30, 2016

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
${E}_{\text{a}}$ = the activation energy
$R$ = the Universal Gas Constant
$T$ = the temperature

If we take the logarithms of both sides, we get

$\ln k = \ln A - {E}_{\text{a}} / \left(R T\right)$

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)|)))" "

${k}_{2} = 2 {k}_{1}$
${k}_{1} = \text{0.200 s"^"-1}$
${E}_{\text{a" = "125 kJ/mol" = "125 000 J/mol}}$
$R = \text{8.314 J·K"^"-1""mol"^"-1}$
${T}_{1} = \text{312 K}$

Now, let's insert the numbers.

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

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

$\frac{\text{15 035 K}}{T} _ 2 = 48.19 - \ln 2 = 47.50$

${T}_{2} = \text{15 035 K"/47.50 = "317 K}$

Oct 30, 2016

I got $\text{316.55 K}$.

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.

${E}_{a} = \text{125 kJ/mol}$
${k}_{1} = {\text{0.200 s}}^{- 1}$
${k}_{2} = 2 {k}_{1}$
${T}_{1} = \text{312.00 K}$
T_2 = ?

Each of these variables can be found in the Arrhenius equation

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

which can be written in the form of an initial $\left(1\right)$ and final state $\left(2\right)$:

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

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:

${k}_{1} / {k}_{2} = {e}^{- {E}_{a} \text{/"RT_1)/e^(-E_a"/} R {T}_{2}}$

$\ln \left({k}_{1} / {k}_{2}\right) = \ln \left({e}^{- {E}_{a} \text{/"RT_1)/e^(-E_a"/} R {T}_{2}}\right)$

$= \ln \left({e}^{- {E}_{a} \text{/"RT_1)) - ln(e^(-E_a"/} R {T}_{2}}\right)$

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

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

which we can recognize as the Arrhenius equation for two different temperatures ${T}_{1}$ and ${T}_{2}$:

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

Now, setting ${k}_{2} = 2 {k}_{1}$ as our desired result, we get:

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

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

Moving ${E}_{a} / R$ to the other side:

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

Subtracting $\frac{1}{T} _ 1$ to the other side and cross-multiplying:

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

Reciprocating both sides of the equation:

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

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

Therefore:

$\textcolor{b l u e}{{T}_{2}} = \frac{{E}_{a} {T}_{1}}{{E}_{a} - R \ln \left(2\right) {T}_{1}}$

$= \left[\left(\text{125 kJ/mol")("312.00 K")]/(("125 kJ/mol") - ("0.008314472 kJ/mol"cdot"K")ln(2)("312.00 K}\right)\right)$

$=$ $\textcolor{b l u e}{\text{316.55 K}}$

(Note that we needed five sig figs to get the necessary precision. If we had written $T = \text{317 K}$, 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 ${k}_{1}$.)