Problem 41 - Chemistry and Chemical Reactivity

Question

Calculate the activation energy, \(E_{\mathrm{a}},\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ from the observed rate constants: \(k\) at \(25^{\circ} \mathrm{C}=3.46 \times\) \(10^{-5} s^{-1}\) and \(k\) at \(55^{\circ} \mathrm{C}=1.5 \times 10^{-3} \mathrm{s}^{-1}.\)

Step-by-Step Solution

Answer

The activation energy, \(E_{a},\) is approximately 103.6 kJ/mol.

Step 1: Understand the Arrhenius Equation

The Arrhenius equation is given by \( k = A e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant (8.314 J/(mol·K)), and \( T \) is the temperature in Kelvin.

Step 2: Convert Temperatures to Kelvin

Convert the temperatures from Celsius to Kelvin. \[ T_1 = 25 + 273.15 = 298.15 \text{ K} \]\[ T_2 = 55 + 273.15 = 328.15 \text{ K} \]

Step 3: Rearrange the Arrhenius Equation

Use the natural logarithm form of the Arrhenius equation: \[ \ln k_2 - \ln k_1 = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

Step 4: Substitute Values

Substitute the given rate constants and temperatures into the equation:\[ \ln(1.5 \times 10^{-3}) - \ln(3.46 \times 10^{-5}) = -\frac{E_a}{8.314} \left(\frac{1}{328.15} - \frac{1}{298.15}\right) \]

Step 5: Calculate the Left Side of the Equation

Calculate the difference in natural logarithms:\[ \ln(1.5 \times 10^{-3}) \approx -6.502 \]\[ \ln(3.46 \times 10^{-5}) \approx -10.267 \]\[ -6.502 + 10.267 = 3.765 \]

Step 6: Calculate the Right Side of the Equation

Calculate the temperature reciprocal difference:\[ \frac{1}{328.15} - \frac{1}{298.15} \approx -0.000303 \]

Step 7: Solve for Activation Energy

Set the equations equal:\[ 3.765 = \frac{E_a}{8.314} \times 0.000303 \]Rearrange to solve for \( E_a \):\[ E_a = \frac{3.765 \times 8.314}{0.000303} \]\[ E_a \approx 103607 \text{ J/mol} = 103.6 \text{ kJ/mol} \]

Key Concepts

Arrhenius Equation

The Arrhenius Equation plays a crucial role in understanding the relationship between the rate of a chemical reaction and the temperature at which it occurs. In chemistry, we often want to predict how fast a reaction will proceed, and the Arrhenius equation provides a way to do that. The equation is given by:\[ k = A e^{-\frac{E_a}{RT}} \]

• \(k\) is the rate constant, which gives us an idea of the reaction speed.
• \(A\) is the pre-exponential factor, related to the frequency of collisions resulting in a reaction.
• \(E_a\) represents the activation energy needed for the reaction to occur.
• \(R\) is the universal gas constant, 8.314 J/(mol·K).
• \(T\) is the temperature in Kelvin.

Using this equation, we can see that as temperature increases, the rate constant \(k\) increases too, because the negative exponent decreases, meaning that more molecules have enough energy to overcome the activation barrier.

Rate Constants

Rate constants \(k\) are vital in understanding how quickly a chemical reaction proceeds. The value of \(k\) indicates the speed of a reaction at a certain temperature. A higher \(k\) implies a faster reaction. Rate constants are specific to particular reactions and can change with varying conditions such as temperature.In the exercise given, two rate constants are provided at different temperatures:

• At 25°C, \(k = 3.46 \times 10^{-5} \, s^{-1}\).
• At 55°C, \(k = 1.5 \times 10^{-3} \, s^{-1}\).

These values indicate that the reaction becomes much faster as the temperature increases from 25°C to 55°C. This change is predictable by the Arrhenius equation, which ties rate constants to temperature and activation energy.

Temperature Conversion

In many scientific calculations, temperature must be expressed in Kelvin, especially when using the Arrhenius equation. This is because Kelvin is the SI unit for temperature and offers a direct relationship between temperature changes and energy.Here's how you convert Celsius to Kelvin:\[ T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \]For the given temperatures:

• 25°C is converted to \(298.15\, \text{K}\).
• 55°C becomes \(328.15\, \text{K}\).

By converting to Kelvin, we ensure consistent and accurate use of the Arrhenius equation and other calculations involving temperature and energy.

Natural Logarithm Calculation

The natural logarithm (ln) is commonly used in chemistry, especially in relation to the Arrhenius equation because it helps linearize exponential functions. This property is particularly helpful when comparing two rate constants to determine the activation energy of a reaction.When you take the logarithm of the Arrhenius equation, it becomes useful for finding the difference between rate constants at two temperatures:\[ \ln k_2 - \ln k_1 = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]In the exercise example, the calculation was:

• \( \ln(1.5 \times 10^{-3}) \approx -6.502 \)
• \( \ln(3.46 \times 10^{-5}) \approx -10.267 \)
• Difference: \(3.765 \)

This numerical representation allows us to solve for activation energy by substituting known values into the transformed equation.