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