Question

What is the activation energy for a reaction if its rate doubles when the temperature is raised from \(20^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\) ? \(\left(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 2=0.3010\right)\) (1) \(269 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (2) \(34.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (3) \(15.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (4) \(342 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Short answer

The activation energy is approximately 34.7 kJ mol⁻¹.

Step-by-step solution

- Understanding the given data

Identify the given data: Temperature change from 20°C to 35°C, R = 8.314 J mol⁻¹ K⁻¹, and log 2 = 0.3010. Recognize that temperature must be converted to Kelvin by adding 273.15.

- Convert temperatures to Kelvin

Convert 20°C to Kelvin: T₁ = 20 + 273.15 = 293.15 K. Convert 35°C to Kelvin: T₂ = 35 + 273.15 = 308.15 K.

- Set up the Arrhenius equation

Use the Arrhenius equation in logarithmic form: \( \frac{k_2}{k_1} = 2 \). Since k₂ = 2k₁, we use the relation: \( \frac{\text{ln}(k_2/k_1)}{\text{ln}(2)} = \frac{E_a}{R} \times \bigg( \frac{1}{T_1} - \frac{1}{T_2} \bigg) \).

- Substitute the known values

Substitute the known values into the Arrhenius equation: \( \text{ln}(2) = \frac{E_a}{8.314} \times \bigg( \frac{1}{293.15} - \frac{1}{308.15} \bigg) \). Simplify: \( 0.3010 = \frac{E_a}{8.314} \times \bigg( \frac{15.00}{293.15 \times 308.15} \bigg) \).

- Calculate the activation energy

Calculate the activation energy: \( E_a = 0.3010 \times 8.314 \times \frac{293.15 \times 308.15}{15.00} \). This simplifies to: \( E_a ≈ 34,849 \text{ ... }\rightarrow ≈ 34.7 \text{kJ} \text{mol⁻¹} \).

Key concepts

Arrhenius equation

The Arrhenius equation is a formula that helps us understand how the rate of a chemical reaction depends on temperature. It is expressed as:
\[k = A e^{-E_a / (RT)}\text{, where:}\]

• \(k\): Reaction rate constant
• \(A\): Pre-exponential factor (frequency of collisions)
• \(E_a\): Activation energy
• \(R\): Universal gas constant (8.314 J/mol·K)
• \(T\): Temperature in Kelvin

When using the equation, notice that an increase in temperature leads to an increase in the reaction rate. This is because higher temperatures provide more energy, helping reactant molecules overcome the activation energy barrier faster.
The logarithmic form used to solve the exercise is helpful for comparing reaction rates at different temperatures. We use the natural logarithm instead of the exponential form:
\[ \text{ln}(\frac{k_2}{k_1}) = - \frac{E_a}{R} \bigg( \frac{1}{T_2} - \frac{1}{T_1} \bigg) \]
It’s crucial to be comfortable with both forms of the equation to solve problems efficiently.

temperature conversion to Kelvin

Temperature in chemistry is often required to be in Kelvin (K) rather than Celsius (°C). The conversion is straightforward: just add 273.15 to the Celsius value.

• From Celsius to Kelvin: \(T(K) = T(°C) + 273.15 \)
• For example: \(20^{\text{°C}} + 273.15 = 293.15 \text{K} \)
• Similarly, for \(35^{\text{°C}}:\)
  \(35 + 273.15 = 308.15 \text{K}\)

Using Kelvin is important because the Kelvin scale starts at absolute zero, where molecular motion theoretically stops. It provides a true measure of thermal energy, making calculations in thermodynamics and kinetics more accurate and universal.

reaction rate

The reaction rate tells us how fast a reaction occurs. In general, it's influenced by several factors like temperature, pressure, concentration of reactants, and presence of catalysts.
When you raise the temperature, you speed up the molecules, making them collide more often and with more energy. This often results in a higher reaction rate.
In the given problem, we know the rate doubles when the temperature increases from \(20^{\text{°C}}\) to \(35^{\text{°C}}\). This doubling effect is critical to calculating the activation energy using the Arrhenius equation.
Knowing how to connect changes in reaction rates with temperature changes helps in understanding the kinetics of the reaction. It tells us how molecules' movements and energy distributions affect how quickly products are formed from reactants.