Chemists commonly use a rule of thumb that an increase in temperature of 10 K doubles the reaction rate. What must the activation energy be for this statement to be true if the change in temperature is from 25°C to 35°C?

• A. Approximately 26 kJ/mol
• B. Approximately 78 kJ/mol
• C. Approximately 104 kJ/mol
• D. Approximately 52 kJ/mol

Answer: (D)

Verified step by step guidance

1. Convert the given temperatures from Celsius to Kelvin. Since the change is from 25°C to 35°C, the temperatures in Kelvin are 298 K and 308 K respectively.
2. Use the Arrhenius equation to relate the rate constants at two different temperatures: \( k_2 = k_1 \cdot e^{\frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)} \), where \( E_a \) is the activation energy, \( R \) is the gas constant (8.314 J/mol·K), and \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.
3. According to the problem, the reaction rate doubles, so \( \frac{k_2}{k_1} = 2 \). Substitute this into the Arrhenius equation: \( 2 = e^{\frac{-E_a}{R} \left( \frac{1}{308} - \frac{1}{298} \right)} \).
4. Take the natural logarithm of both sides to solve for \( E_a \): \( \ln(2) = \frac{-E_a}{R} \left( \frac{1}{308} - \frac{1}{298} \right) \).
5. Rearrange the equation to solve for \( E_a \): \( E_a = -R \cdot \ln(2) \cdot \left( \frac{1}{308} - \frac{1}{298} \right)^{-1} \). Calculate this expression to find the activation energy in kJ/mol.