Chemists commonly use a rule of thumb that an increase of \(10 \space\mathrm{K}\) in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to \(35^{\circ} \mathrm{C} ?\)

First, we need to convert the given temperatures from Celsius to Kelvin. To do this, we will add 273.15 to each temperature: \( T_1 = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \space\mathrm{K} \) \( T_2 = 35^{\circ} \mathrm{C} + 273.15 = 308.15 \space\mathrm{K} \)

The rule of thumb states that an increase of 10 K in temperature doubles the rate of the reaction. We can write this as: \( \frac{rate_2}{rate_1} = 2 \)

The Arrhenius equation states that the rate constant k of a reaction depends upon the activation energy E and temperature T as follows: \( k = A \times e^{\frac{-E}{RT}} \) Here, A is the pre-exponential or frequency factor, E is the activation energy, R is the gas constant (\(8.314 \space\mathrm{J\;mol^{-1}K^{-1}}\)), and T is the temperature.

Now, we can write the equation for the ratio of rate constants at the two different temperatures: \( \frac{k_2}{k_1} = \frac{A \times e^{\frac{-E}{R \times 308.15}}}{A \times e^{\frac{-E}{R \times 298.15}}} \) We know that \( \frac{rate_2}{rate_1} = 2 \), and since the rate constant is directly proportional to the rate of reaction, we can substitute it in: \( 2 = \frac{A \times e^{\frac{-E}{R \times 308.15}}}{A \times e^{\frac{-E}{R \times 298.15}}} \)

Now, we need to solve the equation for E. First, we can simplify the equation by canceling the pre-exponential factor A: \( 2 = \frac{e^{\frac{-E}{R \times 308.15}}}{e^{\frac{-E}{R \times 298.15}}} \) Next, we take the natural logarithm of both sides: \( \ln{2} = \ln{\left(\frac{e^{\frac{-E}{R \times 308.15}}}{e^{\frac{-E}{R \times 298.15}}}\right)} \) \( \ln{2} = \frac{-E}{R \times 308.15} + \frac{E}{R \times 298.15} \) Multiplying the whole equation by \(R \times 308.15 \times 298.15\) to get rid of the fractions: \( \ln{2} \times R \times 308.15 \times 298.15 = -E \times 298.15 + E \times 308.15 \) Now, combining the terms with E: \( E(308.15 - 298.15) = \ln{2} \times R \times 308.15 \times 298.15 \) Finally, we can solve for the activation energy E: \( E = \frac{\ln{2} \times R \times 308.15 \times 298.15}{10} \) Inserting the value of R (\(8.314 \space\mathrm{J\;mol^{-1}K^{-1}}\)) and calculating the result: \( E \approx 5284.4 \space\mathrm{J\;mol^{-1}} \) So, the activation energy must be approximately \(5284.4 \space\mathrm{J\;mol^{-1}}\) for the rule of thumb to be true for a temperature increase from 25 to \(35^{\circ} \mathrm{C}\).