Question

Which equation is true to calculate the energy of activation, if the rate of reaction is doubled by increasing temperature from $T_1 \, K \, to \, T_2$ K?

• A. $log_{ 10} \frac{ k_1 }{ k_2 } = \frac{ E_a }{ 2.303 \, R} \bigg[ \frac{1}{ T_1} - \frac{ 1}{ T_2} \bigg ]$
• B. $log_{ 10} \frac{ k_2 }{ k_1} = \frac{ E_a }{ 2.303 \, R} \bigg[ \frac{1}{ T_2} - \frac{ 1}{ T_1} \bigg ]$
• C. $log_{ 10} \frac{ 1 }{ 2 } = \frac{ E_a }{ 2.303 } \bigg[ \frac{1}{ T_2} - \frac{ 1}{ T_1} \bigg ]$
• D. $log_{ 10 } 2 = \frac{ E_a }{ 2.303 \, R} \bigg[ \frac{1}{ T_1} - \frac{ 1}{ T_2} \bigg ]$

Answer 1

Answer: (D)

$log_{ 10 } 2 = \frac{ E_a }{ 2.303 \, R} \bigg[ \frac{1}{ T_1} - \frac{ 1}{ T_2} \bigg ]$

Answer 2

If rate of reaction is doubled by increasing
temperature from $T_1 \, K \, to \, T_2$ K.
At \, T_1, \hspace15mm \, k_1 = k \, (say)
and T_2, \, \hspace18mm \, k_2 = 2 \, k
On applying,
$log_{ 10} \frac{ k_2 }{ k_1} = \frac{ E_a }{ 2.303 \, R} \bigg[ \frac{1}{ T_1} - \frac{ 1}{ T_2} \bigg ]$
$log_{ 10} \frac{ 2k }{ k} = \frac{ E_a }{ 2.303 \, R} \bigg[ \frac{1}{ T_1} - \frac{ 1}{ T_2} \bigg ]$
$log_{ 10} 2 = \frac{ E_a }{ 2.303 \, R} \bigg[ \frac{1}{ T_1} - \frac{ 1}{ T_2} \bigg ]$