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# 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.

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

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 $y=mx+c$
$\Rightarrow m = -\large\frac{E_a}{R}$
$\Rightarrow E_a = -mR = - (-1.01 \times 10^{4}\;K) \times (8.314\; J/K·mol) = 83971\;Jmol^{-1} = 84\;kJmol^{-1}$