Question

Figure shows a graph in $log_{10} K\, vs\,\frac{1}{T}$ where $K$ is rate constant and $T$ is temperature. The straight line $BC$ has slope, $\left|tan\theta\right|=\frac{1}{2.303}$ and an intercept of $5$ on y-axis. Thus $E_A$, the energy of activation is:

• A. $2.303 \times 2\, cal$
• B. $2/2.303 \,cal$
• C. $2 \,cal$
• D. None of these

Answer 1

Answer: (C)

$log_{10}K vs.\frac{1}{T}$ is linear;
slope$=\frac{-E_{A}}{2.303R}$ intercept $= log_{10} A$
Since, $\left|tan \theta\right|=\frac{1}{2.303}$
$\therefore \frac{E_{A}}{2.303R}=\frac{1 }{2.303} \Rightarrow E_{a}=R=2 cal$