Question

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C ? (R = 8.314 J mol^-1 K^-1)

• A. 342 kJ mol^-1
• B. 269 kJ mol^-1
• C. 34.7 kJ mol^-1
• D. 15.1 kJ mol^-1

Answer 1

The Correct Option is C

To determine the activation energy (Ea) of a reaction where the rate doubles when the temperature is raised from 20°C to 35°C, we use the Arrhenius equation. The Arrhenius equation for the rate constant k is given by:

k = A \exp\left(-\frac{E_a}{RT}\right)

Where:

• A is the pre-exponential factor
• E_a is the activation energy
• R is the universal gas constant (8.314 \, \text{J mol}^{-1} \text{K}^{-1})
• T is the temperature in Kelvin

Given that the rate doubles, the relation between the two rate constants at two different temperatures is:

\frac{k_2}{k_1} = 2

The formula for the ratio of rate constants is derived from the Arrhenius equation:

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

Taking natural logarithm on both sides:

\ln(2) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

Converting the given temperatures to Kelvin:

• T_1 = 20\,^{\circ}\text{C} = 293\,\text{K}
• T_2 = 35\,^{\circ}\text{C} = 308\,\text{K}

Substituting the known values into the equation:

\ln(2) = -\frac{E_a}{8.314}\left(\frac{1}{308} - \frac{1}{293}\right)

0.693 = -\frac{E_a}{8.314} \times (-0.000176)

Solving for E_a:

E_a = \frac{0.693 \times 8.314}{0.000176} = 34705 \, \text{J mol}^{-1}

Convert E_a to kJ/mol:

E_a = 34.7 \, \text{kJ mol}^{-1}

Thus, the activation energy for the reaction is 34.7 kJ mol^-1.

Therefore, the correct answer is: 34.7 kJ mol^-1.