ExamBro
ExamBro
GUJCET · Chemistry · Chemical Kinetics

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 \(T_2 k\) ?

  1. A \(\log _{10} \frac{k_2}{k_1}=\frac{ E _a}{2.303 R }\left[\frac{1}{T_2}-\frac{1}{T_1}\right]\)
  2. B \(\log _{10} 2=\frac{ E _a}{2.303 R }\left[\frac{1}{T_1}-\frac{1}{T_2}\right]\)
  3. C \(\log _{10} \frac{k_1}{k_2}=\frac{ E _a}{2.303 R }\left[\frac{1}{T_1}-\frac{1}{T_2}\right]\)
  4. D \(\log _{10} \frac{1}{2}=\frac{ E _a}{2.303}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\log _{10} 2=\frac{ E _a}{2.303 R }\left[\frac{1}{T_1}-\frac{1}{T_2}\right]\)

Step-by-step Solution

Detailed explanation

B