Question

Graph between $\log\, k$ and $1 / T\left[k\right.$ is rate constant $\left( s ^{-1}\right)$ and $T$ and temperature $\left.( K )\right]$ is a straight line with $OX =5, \theta=\tan ^{-1}(1 / 2.303)$. Hence $-E_{ a }$ will be

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

Answer 1

Answer: (C)

$\log k=\log A-\frac{E_{ a }}{2.303 R T}(y=c+m x)$
Slope $=\frac{-E_{ a }}{2.303 R}=\frac{1}{2.303}$ (given) $\left(\tan \theta=\frac{1}{2.303}\right)$
$-E_{ a }=2.303 R \times$ Slope
$=2.303 \times \frac{R}{2.303}=R=2\, cal$