# Question 4cb27

Apr 13, 2015

To solve this problem you have to use the Arhennius equation, which looks like this

$k = A \cdot {e}^{- {E}_{A} / \left(R T\right)}$, where

$k$ - the rate constant;
$A$ - the Arhennius factor;
${E}_{A}$ - the activation energy for the reaction;
$R$ - the gas constant, equal to $8.314 {\text{J/mol"^(-1)"K}}^{- 1}$
$T$ - the tmperature at which the reaction takes place - expressed in Kelvin.

A more useful form of the Arhennius equation looks like this

$\ln \left(k\right) = \ln \left(A \cdot {e}^{- {E}_{A} / \left(R T\right)}\right) = \ln \left(A\right) + \ln \left({e}^{- {E}_{A} / \left(R T\right)}\right)$

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

The idea is very simple - as temperature increases, more molecules will accumulate enough kinetic energy to overcome the energy barrier needed for the reaction to take place.

In other words, more molecules will have enough energy to exceed the activation energy required for them to react.

So, use the two sets of data you were given to find the value of ${E}_{a}$

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

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

$\textcolor{b l u e}{\left(1\right)} - \textcolor{b l u e}{\left(2\right)} \implies \ln \left({k}_{1}\right) - \ln \left({k}_{2}\right) = \cancel{\ln \left(A\right)} - {E}_{a} / \left(R {T}_{1}\right) - \cancel{\ln \left(A\right)} + {E}_{a} / \left(R {T}_{2}\right)$

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

ln( (55cancel("s"^(-1)))/(100cancel("s"^(-1)))) = (1/310 - 1/300)cancel("K"^(-1)) * E_a/(8.314"J"/("mol" * cancel("K"))

E_a = (8.314"J"/"mol" * ln(0.55))/(0.000107527) = "46225 J/mol"#

Now use this value to determine the rate constant at 350 K

$\ln \left(\frac{55}{k} _ 3\right) = \left(\frac{1}{350} - \frac{1}{300}\right) \cdot \frac{46225}{8.314}$

$\ln \left(\frac{55}{k} _ 3\right) = - 2.64756$

$\frac{55}{k} _ 3 = {e}^{- 2.64756} = 0.070824 \implies {k}_{3} = \frac{55}{0.070824} = {\text{776.6 s}}^{- 1}$

Rounded to one sig fig, the number of sig figs given for ${T}_{1}$ and ${k}_{2}$, the answers will be

${E}_{a} = \textcolor{g r e e n}{\text{50 kJ/mol}}$

${k}_{3} = \textcolor{g r e e n}{{\text{800 s}}^{- 1}}$