This question involves calculating the activation energy of a reaction using the Arrhenius equation, given that the rate doubles when the temperature increases from 300 K to 310 K.

Key Concept: The Arrhenius equation relates the rate constant (k) of a reaction to the temperature (T) and activation energy (E_a):

$k=A{e}^{-\frac{E_a}{RT}}$

where A is the pre-exponential factor, R is the gas constant, and T is the temperature in Kelvin.

For two different temperatures T₁ and T₂, the ratio of rate constants is given by:

$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}(\frac{1}{T_1}-\frac{1}{T_2})$

Given:

• T₁ = 300 K, T₂ = 310 K
• k₂ = 2k₁ (since the rate doubles)
• R = 8.314 J K^-1 mol^-1
• log 2 = 0.301

Step 1: Substitute the known values into the equation:

$\log \left(\frac{2k_1}{k_1}\right)=\frac{E_a}{2.303\times 8.314}(\frac{1}{300}-\frac{1}{310})$

Step 2: Simplify the left side (log(2) = 0.301) and calculate the temperature difference term:

$0.301=\frac{E_a}{19.147}\left(\frac{310-300}{300\times 310}\right)=\frac{E_a}{19.147}\left(\frac{10}{93000}\right)$

Step 3: Further simplify the fraction:

$0.301=\frac{E_a}{19.147}\times \frac{1}{9300}$

Step 4: Rearrange to solve for E_a:

$E_a=0.301\times 19.147\times 9300$

Step 5: Calculate the value:

First, 0.301 × 19.147 ≈ 5.763

Then, 5.763 × 9300 ≈ 53605.9 J/mol

Convert to kJ/mol: E_a ≈ 53.6 kJ/mol

Final Answer: The activation energy is 53.6 kJ mol^-1.

Related Topics & Formulae

Arrhenius Equation: Fundamental equation describing the temperature dependence of reaction rates. The two-point form is crucial for solving problems where rate constants at two temperatures are compared.

Activation Energy (E_a): The minimum energy required for a reaction to occur. A higher E_a means the reaction is more sensitive to temperature changes.

Key Formula:

$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}(\frac{1}{T_1}-\frac{1}{T_2})$

This formula is derived by taking the logarithm of the ratio of the Arrhenius equations at two different temperatures.