# 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 ${k}_{1}$.)