Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

If a temperature increase from 20$^\circ$C to 35$^\circ$C triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

$\begin{array}{1 1} 55\;kJmol^{-1} \\ 83\;kJmol^{-1} \\ 72\;kJmol^{-1} \\ 68\;kJmol^{-1} \end{array}$

Can you answer this question?

1 Answer

0 votes
Answer: $ 55\;kJmol^{-1}$
We have to find the activation barrier, i.e, the activation energy, give two temperatures $T_1$ and $T_2$.
Its also mentioned that the rate constant triples, so $k_2 = 3k_1$
From Arrhenius equation, $k=Ae^{\large\frac{-E_a}{RT}}$, we get:
$\Rightarrow \large\frac{k_2}{k_1} $$ = \large\frac{Ae^{\large\frac{-E_a}{RT_1}}}{Ae^{\large\frac{-E_a}{RT_2}}}$
Using the fact that $\large\frac{e^x}{e^y}$$ = e^{x-y}$ and taking the natural log on both sides and simplifying, we arrive at:
$ ln\large\frac{k_2}{k_1}$$ = \large\frac{E_a}{R}$$ (\large\frac{1}{T_1}$$ - \large\frac{1}{T_2}$$)$
Substituting the values for $k_2 = 3k_1$ and $T_1, T_2$ (in $^\circ$K), we get:
$\Rightarrow ln 3 = 1.0986 = \large\frac{E_a}{8.314\;J\;K.mol^{-1}}$$ (\large\frac{1}{293.15}$$ - \large\frac{1}{308.15}$$)$
Solving for $E_a$, we get $E_a = 55006\;Jmol^{-1} = 55\;kJmol^{-1}$
answered Jul 25, 2014 by balaji.thirumalai
edited Jul 25, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App