Question

Rate constant $k$ of a reaction varies with temperature according to the equation $\log k=constant-\frac{E_a}{2.303R}\times \frac{1}{T}$ where $E_a$ is the energy of activation for the reaction. When a graph is plotted for $\log k$ vs $\frac{1}{T}$ a straight line with a slope $-6670k$ is obtained. The activation energy for this reaction will be:

($R=8.314J{K}^{-1}{mol}^{-1})$

• A. $122.65kJ{mol}^{-1}$
• B. $127.71kJ{mol}^{-1}$
• C. $142.34kJ{mol}^{-1}$
• D. $150.00kJ{mol}^{-1}$

Answer 1

The correct option is B $127.71kJ{mol}^{-1}$

Slope of the line $=-\frac{E_a}{2.303R}=-6670K$

$E_a=2.303\times 8.314\left(J{K}^{-1}{mol}^{-1}\right)\times 6670K$ $=127711.4J{mol}^{-1}=127.71kJ{mol}^{-1}$