Question

Rate constant $k$ of a reaction varies with temperature according to the equation $\log k=$ constant $-\frac{E_{a}}{2.303 R} \times \frac{1}{T}$; where, $E_{a}$ is the energy of activation for the reaction. When a graph is plotted for $\log \,k \,v\, s \frac{1}{T}$ a straight line with a slope $-6670\, k$ is obtained. The activation energy for this reaction will be $\left(R=8.314 \,JK ^{-1}\, mol ^{-1}\right)$.

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

Answer 1

Answer: (B)

Slope of the line $=-\frac{E_{a}}{2303 \,R}=-6670 \,K$
$ E_{a} =2.303 \times 8.314\left( JK ^{-1}\, mol ^{-1}\right) \times 6670\, K $
$=127711.4 \,J \,mol ^{-1}=127.71 \,kJ \,mol ^{-1}$