Question

The rate constant of a reaction is $1.5\times {10}^{-3}at{25}^{\circ }Cand2.1\times {10}^{-2}at{60}^{\circ }C$. The activation energy is:

• A. $\frac{35}{333}R\times lo{g}_e\frac{2.1\times {10}^{-2}}{1.5\times {10}^{-2}}$
• B. $\frac{298\times 333}{35}R\times lo{g}_e\frac{21}{1.5}$
• C. $\frac{298\times 333}{35}R\times lo{g}_{e}2.1$
• D. None of these

Answer 1

Answer: (B)

$\frac{298\times 333}{35}R\times lo{g}_e\frac{21}{1.5}$

$T_1$ $=273+25=298$ K
$T_2$ $=273+60=333$ K
$log\frac{k_2}{k_1}=\frac{E_a}{2.3R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
or $lo{g}_e\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
$lo{g}_e\frac{2.1\times {10}^{-2}}{1.5\times {10}^{-3}}=\frac{E_a}{R}\times \left(\frac{35}{333\times 298}\right)$
$\therefore E_a=\frac{298\times 333}{35}\times R\times lo{g}_e\frac{21}{1.5}$