Ask Questions, Get Answers

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

The rate constant (k) for a reaction was determined as a function of temperature. A plot of $ln\; k$ versus $ \large\frac{1}{T}$ (in $K$) is linear and has a slope of $-1.01 \times 10^{4}\;K$. Calculate the activation energy for the reaction.

$\begin{array}{1 1}84\;kJmol^{-1} \\ 48\;kJmol^{-1} \\ 24 \;kJmol^{-1} \\ 96\;kJmol^{-1} \end{array}$

Can you answer this question?

1 Answer

0 votes
Answer:$ 84\;kJmol^{-1}$
Taking natural logarithms of the Arrhenius equation, $k=Ae^{\large\frac{-E_a}{RT}}$ we get:
$ln\; k=ln A-\large\frac{E_a}{RT}$ which we can compare to $y=mx+c$
$\Rightarrow m = -\large\frac{E_a}{R}$
$\Rightarrow E_a = -mR = - (-1.01 \times 10^{4}\;K) \times (8.314\; J/K·mol) = 83971\;Jmol^{-1} = 84\;kJmol^{-1}$
answered Jul 25, 2014 by balaji.thirumalai

Related questions

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