Question

The rate of the chemical reaction doubles when the temperature is raised from \(298\,K\) to \(308\,K\). Calculate activation energy \((E_a)\) for this reaction assuming that it does not change with temperature. (Given : \(R = 8.314\,J\,mol^{-1}\,K^{-1}\), \(\log 2 = 0.30\))

Hint:

For temperature dependence of reaction rates, remember the Arrhenius equation: \[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] If the rate doubles, then \[ \frac{k_2}{k_1}=2 \] and use \[ \log 2 = 0.3010 \approx 0.30 \] to calculate the activation energy.

Answer 1

Concept: The effect of temperature on the rate of a chemical reaction is explained by the Arrhenius equation. \[ k=Ae^{-E_a/RT} \] where \[ k=\text{rate constant} \] \[ A=\text{Arrhenius constant (frequency factor)} \] \[ E_a=\text{activation energy} \] \[ R=\text{gas constant} \] \[ T=\text{absolute temperature} \] For two different temperatures \(T_1\) and \(T_2\), the Arrhenius equation can be written in logarithmic form as \[ \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Since the rate of a reaction is directly proportional to the rate constant, the ratio of rates can be substituted in place of the ratio of rate constants.

Step 1: Writing the given data. Initial temperature: \[ T_1=298\,K \] Final temperature: \[ T_2=308\,K \] The rate doubles when temperature increases from \(298\,K\) to \(308\,K\). Therefore, \[ \frac{k_2}{k_1}=2 \] Gas constant: \[ R=8.314\,J\,mol^{-1}\,K^{-1} \] Also, \[ \log 2=0.30 \]

Step 2: Applying the Arrhenius equation. Using \[ \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Substituting the known values, \[ 0.30 = \frac{E_a}{2.303\times8.314} \left( \frac{1}{298} - \frac{1}{308} \right) \]

Step 3: Calculating the temperature term. \[ \frac{1}{298}-\frac{1}{308} = \frac{308-298}{298\times308} \] \[ = \frac{10}{91784} \] \[ = 1.0895\times10^{-4} \] Thus, \[ 0.30 = \frac{E_a}{2.303\times8.314} \times1.0895\times10^{-4} \]

Step 4: Calculating \(2.303R\). \[ 2.303\times8.314 = 19.147 \] Therefore, \[ 0.30 = \frac{E_a}{19.147} \times1.0895\times10^{-4} \] Multiplying both sides by \(19.147\), \[ 5.7441 = E_a(1.0895\times10^{-4}) \]

Step 5: Calculating the activation energy. \[ E_a = \frac{5.7441}{1.0895\times10^{-4}} \] \[ E_a = 5.272\times10^4\,J\,mol^{-1} \] \[ E_a = 52720\,J\,mol^{-1} \] Converting into kilojoules, \[ E_a = 52.72\,kJ\,mol^{-1} \] Hence, \[ \boxed{E_a\approx52.8\,kJ\,mol^{-1}} \]

Step 6: Verification of the result. The reaction rate doubles for a temperature increase of \(10\,K\). For such reactions, activation energies are commonly found in the range of \(50-60\,kJ\,mol^{-1}\). The obtained value \[ 52.8\,kJ\,mol^{-1} \] is therefore chemically reasonable and consistent with the Arrhenius theory.

Final Answer: \[ \boxed{E_a\approx52.8\,kJ\,mol^{-1}} \]