Question

For a reaction, activation energy E$_a$=0 and the rate constant at 200K is 1.6×10$^6$s$^{-1}$. The rate constant at 400K will be-
[Given that gas constant]
R= 8.314 J K$^{-1}$ mol$^{-1}$

• A. 3.2 ×10$^4$ s$^{-1}$
• B. 1.6 × 10$^6$ s$^{-1}$
• C. 1.6 × 10$^3$ s$^{-1}$
• D. 3.2 × 10$^6$ s$^{-1}$

Answer 1

Answer: (B)

$log\left(\frac{K_{2}}{K_{1}}\right) = \frac{E}{2.303 R}\left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right)$
$ E_{a} = 0 $
$ log\left(\frac{K_{2}}{K_{1}}\right) = 0 $
$\frac{K_{2}}{K_{1}} = 10^{\circ} = 1$
$\Rightarrow K_{2} = K_{1} $
$ K_{2 } = 1.6 \times 10^{6} s^{-1}$ at $400K$