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## Sagot :

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.