Q12 / 180 · KCET 2016 · Chemistry
The activation energy of a chemical reaction can be determined by,
- Correct answerAevaluating rate constants at two different temperatures.
- BO changing the concentration of reactants.
- Cevaluating the concentration of reactants at two different temperatures.
- Devaluating rate constant at standard temperature,
Solution
Activation energy of a chemical reaction can be determined by measuring the values of rate constant at two
differenttemperatures
Let \( k_{1} \) and \( k_{2} \) are the rate constant for the reaction at two different temperature, that is, \( T_{1} \) and \( T_{2} \) respectively.
\( \log k_{1}=\log A-\frac{E_{a}}{2.303 R T_{1}} \rightarrow(1) \) \( \log k_{2}=\log A-\frac{E_{a}}{2.303 R T_{2}} \rightarrow(2) \)
Subtract Eq. (1) from Eq. (2), we get
\[ \begin{array}{l} \log k_{2}-\log k_{1}=\frac{E_{a}}{2.303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right] \\ \log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right] \\ \log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right] \\ E_{a}=2.303 R \log \frac{k_{2}}{k_{1}} \times\left[\frac{T_{1} T_{2}}{T_{2}-T_{1}}\right] \end{array} \]
differenttemperatures
Let \( k_{1} \) and \( k_{2} \) are the rate constant for the reaction at two different temperature, that is, \( T_{1} \) and \( T_{2} \) respectively.
\( \log k_{1}=\log A-\frac{E_{a}}{2.303 R T_{1}} \rightarrow(1) \) \( \log k_{2}=\log A-\frac{E_{a}}{2.303 R T_{2}} \rightarrow(2) \)
Subtract Eq. (1) from Eq. (2), we get
\[ \begin{array}{l} \log k_{2}-\log k_{1}=\frac{E_{a}}{2.303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right] \\ \log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right] \\ \log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right] \\ E_{a}=2.303 R \log \frac{k_{2}}{k_{1}} \times\left[\frac{T_{1} T_{2}}{T_{2}-T_{1}}\right] \end{array} \]
Hence, the correct answer is option: 'A'.