Question

The slope of the line for the graph of \(\log k\) vs \(1 / T\) for the reaction, \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+1 / 2 \mathrm{O}_{2}\) is \(-5000\). Calculate the energy of activation of the reaction. (a) \(95.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(9.57 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(957 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(0.957 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)

Short answer

41.57 \( \mathrm{kJ/mol} \) (option is not listed).

Step-by-step solution

Understanding the Problem

We are dealing with a reaction where we need to find the activation energy (Ea). The slope of the graph of \( \log k \) versus \( 1/T \) is given as \(-5000\). This graph represents an Arrhenius plot.

Applying the Arrhenius Equation

The Arrhenius equation is \( k = Ae^{-\frac{E_a}{RT}} \). Taking the natural logarithm on both sides gives \( \ln k = \ln A - \frac{E_a}{R} \left(\frac{1}{T}\right) \). Comparing this with the line equation \( y = mx + c \), the slope \( m = -\frac{E_a}{R} \).

Relating the Slope to Activation Energy

In the slope \( m = -5000 \), it corresponds to \( -\frac{E_a}{R} = -5000 \), hence \( \frac{E_a}{R} = 5000 \).

Calculating the Energy of Activation

To find the activation energy, rearrange the formula: \( E_a = 5000 \times R \). Use the gas constant \( R = 8.314 \) J/mol·K to calculate \( E_a \).

Performing the Final Calculation

Substitute \( R = 8.314 \) J/mol·K into the equation and compute: \[ E_a = 5000 \times 8.314 \] \[ E_a = 41570 \text{ J/mol} \]. Convert Joules to kilojoules by dividing by 1000: \[ 41.57 \text{ kJ/mol} \].

Key concepts

Arrhenius equation

The Arrhenius equation is a fundamental formula used to understand how the rate of a chemical reaction depends on temperature. It is expressed as: \( k = Ae^{-\frac{E_a}{RT}} \), where:

• \( k \) is the rate constant of the reaction.
• \( A \) is the pre-exponential factor, also known as the frequency factor.
• \( E_a \) is the activation energy of the reaction.
• \( R \) is the universal gas constant.
• \( T \) is the temperature in Kelvin.

Understanding this equation helps us see that the rate constant \( k \) increases with temperature and decreases with higher activation energies because of the exponential nature of \( e^{-\frac{E_a}{RT}} \). Taking the natural logarithm of both sides transforms it into a linear relationship: \( \ln k = \ln A - \frac{E_a}{R} \left(\frac{1}{T}\right) \). In this form, the equation resembles a straight line equation of the form \( y = mx + c \), which helps us analyze reaction rates graphically.

Arrhenius plot

An Arrhenius plot is a graph that visually represents the linear relationship derived from the Arrhenius equation. It is a plot of \( \log k \) or \( \ln k \) versus \( \frac{1}{T} \). The slope of the line in this plot is a key factor in determining the activation energy \( E_a \) of a reaction. This slope is equivalent to \(-\frac{E_a}{R}\), where \( R \) is the universal gas constant. Hence, changes in the slope indicate changes in activation energy. The y-intercept of an Arrhenius plot, given by \( \ln A \), provides insight into the frequency factor of the reaction. The linearity of the plot makes it straightforward to calculate activation energy by simply determining the slope. Researchers often use Arrhenius plots to quickly compare different reactions or to assess how a reaction's rate constant is affected by temperature.

Slope calculation

Calculating the slope of an Arrhenius plot allows us to find the activation energy of a reaction. In this specific example, given a slope of \(-5000\), we start with the relationship \( m = -\frac{E_a}{R} \). To find \( E_a \), rearrange this equation: \( E_a = -m \times R \). Inserting the known values, the calculation proceeds as:- The given slope \( m = -5000 \)- The universal gas constant \( R = 8.314 \) J/mol·K.So, \( E_a = 5000 \times 8.314 \) J/mol, leading to:\( E_a = 41570 \) J/mol To express this in kilojoules per mole, divide by 1000, resulting in:\( 41.57 \) kJ/mol. This result allows us to identify the closest choice among possible answers for activation energy, concluding the calculation process effectively.