To determine the activation energy (Ea) of a reaction that increases tenfold in rate when the temperature is increased from 300 K to 310 K, we can use the Arrhenius equation. Here's the detailed, step-by-step solution:
1. Given Values:
- Rate increase factor ([tex]\( k_2 / k_1 \)[/tex]): 10
- Initial temperature ([tex]\( T_1 \)[/tex]): 300 K
- Final temperature ([tex]\( T_2 \)[/tex]): 310 K
- Universal gas constant ([tex]\( R \)[/tex]): 8.314 J/(mol·K)
2. Arrhenius Equation:
The relationship between the rate constants and temperature in the Arrhenius equation is given by:
[tex]\[
\frac{k_2}{k_1} = \exp\left(\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right)
\][/tex]
3. Rearrange to Solve for Activation Energy ([tex]\( Ea \)[/tex]):
Taking the natural logarithm of both sides:
[tex]\[
\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)
\][/tex]
Solve for [tex]\( E_a \)[/tex]:
[tex]\[
E_a = R \cdot \ln\left(\frac{k_2}{k_1}\right) \cdot \frac{1}{\left(\frac{1}{T_1} - \frac{1}{T_2}\right)}
\][/tex]
4. Calculate the Natural Logarithm of the Rate Increase Factor:
[tex]\[
\ln(10) \approx 2.302585092994046
\][/tex]
5. Calculate the Inverse Temperature Difference:
[tex]\[
\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300 \text{ K}} - \frac{1}{310 \text{ K}} = 0.00010752688172043032 \text{ K}^{-1}
\][/tex]
6. Calculate the Activation Energy:
[tex]\[
E_a = 8.314 \text{ J/(mol·K)} \times 2.302585092994046 \times \frac{1}{0.00010752688172043032 \text{ K}^{-1}}
\][/tex]
[tex]\[
E_a \approx 178036.33990731786 \text{ J/mol}
\][/tex]
Converting to kJ/mol:
[tex]\[
E_a \approx 178.04 \text{ kJ/mol}
\][/tex]
Therefore, the activation energy ([tex]\( E_a \)[/tex]) of the reaction is approximately 178.04 kJ/mol.
