Problem 222 - A Complete Resource Book in Chemistry for JEE Main

Question

The rate of a reaction doubles when its temperature changes from \(300 \mathrm{~K}\) to \(310 \mathrm{~K}\). Activation energy of such a reaction will be: \(\left(\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right.\) and \(\log 2=0.301\) ) (a) \(58.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(60.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(53.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(48.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The activation energy is approximately \(53.6 \text{ kJ/mol}\).

Step 1: Understanding the Problem

We need to determine the activation energy of a reaction given that its rate doubles when the temperature increases from 300 K to 310 K. We are provided with the universal gas constant \( R = 8.314 \mathrm{~JK^{-1}~mol^{-1}} \) and \( \log 2 = 0.301 \).

Step 2: Utilizing the Arrhenius Equation

According to the Arrhenius equation, the rate constant \( k \) is related to temperature \( T \) and activation energy \( E_a \): \[ k = A e^{-E_a / (RT)} \]where \( A \) is the pre-exponential factor.

Step 3: Relating Rate Constants to Temperature Change

When the rate doubles, the ratio of the rate constants (\( k_2/k_1 \)) can be expressed using the Arrhenius equation for two different temperatures \( T_1 = 300 \mathrm{~K} \) and \( T_2 = 310 \mathrm{~K} \): \[\frac{k_2}{k_1} = \frac{A e^{-E_a / (R \times 310)}}{A e^{-E_a / (R \times 300)}} = e^{E_a (\frac{1}{300} - \frac{1}{310})/R}\]Given that \( \frac{k_2}{k_1} = 2 \), we have:\[2 = e^{(E_a / R) \left(\frac{1}{300} - \frac{1}{310}\right)}\]

Step 4: Taking Logarithms

Take the natural logarithm of both sides to solve for \( E_a \):\[\ln(2) = \frac{E_a}{R} \left(\frac{1}{300} - \frac{1}{310}\right) \]Convert \( \ln(2) \) using the given \( \log 2 = 0.301 \) into base \( e \):\[\ln(2) = 0.301 \times \ln(10) = 0.301 \times 2.303 = 0.694\]

Step 5: Solving for Activation Energy

Now substitute back and solve for \( E_a \):\[0.694 = \frac{E_a}{8.314} \left(\frac{1}{300} - \frac{1}{310}\right)\]Calculate \( \left(\frac{1}{300} - \frac{1}{310}\right) \):\[\frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300}\]

Step 6: Calculating Final Expression for Activation Energy

Substitute and solve for \( E_a \):\[0.694 = \frac{E_a}{8.314} \times \frac{1}{9300}\]Rearranging gives:\[E_a = 0.694 \times 8.314 \times 9300\]Calculate \( E_a \):\[E_a = 53.57 \text{ kJ/mol}\]Rounded to one decimal place, \( E_a \approx 53.6 \text{ kJ/mol} \).

Key Concepts

Arrhenius Equation

The Arrhenius equation is a fundamental relationship in chemical kinetics, linking the rate constant of a reaction to the temperature and activation energy. It is given by:

• \( k = A e^{-E_a / (RT)} \)

Here, \( k \) is the rate constant, \( A \) is the pre-exponential factor or frequency factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the absolute temperature in Kelvin. This equation shows how the rate of a reaction increases with temperature, assuming the activation energy remains constant. A smaller activation energy or higher temperature will result in a faster reaction rate due to a larger value of \( k \). The Arrhenius equation is essential for understanding the dependence of reaction rates on temperature.

Temperature Effect on Reaction Rate

Temperature plays a critical role in determining the rate of chemical reactions. In general, increasing the temperature increases the kinetic energy of the reactant molecules, leading to more frequent and more energetic collisions. This increase in collision frequency and energy results in a higher rate of reaction.

• Higher temperatures mean molecules move faster, increasing collision chances.
• The activation energy barrier is more easily overcome at elevated temperatures.

For many reactions, it is observed that the reaction rate approximately doubles for every 10-degree Celsius (or Kelvin) rise in temperature. This is a rule of thumb and can be explained through the Arrhenius equation, where a small change in temperature causes an exponential increase in the rate constant \( k \), thus speeding up the reaction.

Rate Constant Calculation

The rate constant, \( k \), is a crucial parameter in the study of reaction kinetics, representing the proportionality factor in the rate law. Calculating \( k \) involves understanding its relationship with temperature and activation energy, as dictated by the Arrhenius equation.

• The formula \( k = A e^{-E_a / (RT)} \) is used, where \( A \) and \( E_a \) are determined experimentally.
• Thermodynamic data like activation energy and temperature are key inputs.

When working with experimental data, plotting ln(\( k \)) against 1/\( T \) provides a straight line with a slope of \(-E_a/R\) and intercept of ln(\( A \)). This linear relationship allows for the determination of \( E_a \) and the pre-exponential factor \( A \), which in turn helps in calculating \( k \) at any given temperature.

Kinetics

Kinetics is the study of the rate at which chemical reactions occur and the factors affecting these rates. It involves examining how different conditions like temperature, pressure, concentration, and the presence of catalysts influence reaction speed.

• Kinetics helps in determining how fast a reaction will proceed under given conditions.
• Understanding kinetics is essential for optimizing industrial chemical processes, where controlling reaction rates is crucial for efficiency.

The foundational tool for analyzing reaction kinetics is the rate law, which expresses the rate of a reaction as a function of the concentration of reactants. Additionally, understanding activation energy through the Arrhenius equation aids in elucidating the energy barrier that must be overcome for a reaction to proceed, providing insights into how conditions can be altered to accelerate reactions.