Problem 5 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the reaction of \(\mathrm{H}_{2}\) with \(\mathrm{I}_{2}\), the rate constant is \(2.5 \times 10^{-4} \mathrm{dm}^{3} \mathrm{~mol}^{-1} \mathrm{~s}^{-}\) \({ }^{1}\) at \(327{ }^{\circ} \mathrm{C}\) and \(1.0 \mathrm{dm}^{3} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\) at \(527^{\circ} \mathrm{C}\). The activation energy for the reaction, in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is : \(\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\) (a) 166 (b) 150 (c) 72 (d) 59

Step-by-Step Solution

Answer

The activation energy is 166 kJ/mol, option (a).

Step 1: Identify the Given Values

We are given two rate constants: \( k_1 = 2.5 \times 10^{-4} \, \mathrm{dm}^3 \, \mathrm{mol}^{-1} \, \mathrm{s}^{-1} \) at \( T_1 = 327^{\circ} \mathrm{C} \) and \( k_2 = 1.0 \, \mathrm{dm}^3 \, \mathrm{mol}^{-1} \, \mathrm{s}^{-1} \) at \( T_2 = 527^{\circ} \mathrm{C} \). We need to convert these temperatures to Kelvin: \( T_1 = 327 + 273.15 = 600.15 \mathrm{K} \) and \( T_2 = 527 + 273.15 = 800.15 \mathrm{K} \). The gas constant \( R \) is provided as \( 8.314 \, \mathrm{J} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1} \).

Step 2: Select the Arrhenius Equation

The Arrhenius equation relates the rate constants at two different temperatures to the activation energy \( E_a \). It is given by: \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] We will use this equation to solve for \( E_a \).

Step 3: Calculate the Natural Logarithm of the Rate Constant Ratio

First, calculate the ratio \( \frac{k_2}{k_1} = \frac{1.0}{2.5 \times 10^{-4}} = 4000 \). Then, compute the natural logarithm of this ratio: \[ \ln(4000) \approx 8.294 \]

Step 4: Calculate the Temperature Difference Term

Compute the inverse temperature difference: \[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{600.15} - \frac{1}{800.15} \] Calculate each term: \[ \frac{1}{600.15} \approx 0.001666 \] \[ \frac{1}{800.15} \approx 0.00125 \] Therefore, \[ \frac{1}{600.15} - \frac{1}{800.15} \approx 0.000416 \]

Step 5: Solve for Activation Energy

Substitute the values into the Arrhenius equation: \[ 8.294 = \frac{E_a}{8.314} \times 0.000416 \] Rearrange to solve for \( E_a \): \[ E_a = \frac{8.294}{0.000416} \times 8.314 \] Calculate \( E_a \): \[ E_a \approx 166060.04 \, \mathrm{J} \, \mathrm{mol}^{-1} \] Convert this to \( \mathrm{kJ} \): \[ E_a \approx 166 \mathrm{kJ} \, \mathrm{mol}^{-1} \]

Step 6: Finalize the Solution

The activation energy for the reaction is calculated as \( 166 \mathrm{kJ} \, \mathrm{mol}^{-1} \). Compare this result with the multiple-choice options: (a) 166, (b) 150, (c) 72, (d) 59. The correct answer is (a) 166.

Key Concepts

Arrhenius Equation

The Arrhenius Equation is a crucial tool in chemical kinetics, helping us understand how reaction rates vary with temperature. It is represented by: \[ k = A e^{-\frac{E_a}{RT}} \] where:

• \( k \) is the rate constant.
• \( A \) is the frequency factor, a constant for each chemical reaction.
• \( E_a \) represents the activation energy.
• \( R \) is the gas constant (8.314 \( \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} \)).
• \( T \) is the temperature in Kelvin.

The equation shows that the rate constant \( k \) increases exponentially with a rise in temperature if activation energy \( E_a \) is positive. A higher rate constant means that the reaction occurs faster. This relationship is especially useful in understanding why chemical reactions occur faster at higher temperatures.
The Arrhenius equation also allows us to calculate the activation energy when given two rate constants at different temperatures by rearranging it to find the natural log of the ratio of the rate constants:\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] This form helps to determine the activation energy, which is the minimum energy required for reactants to transform into products.

Rate Constant

The rate constant \( k \) is a critical parameter in kinetics, describing the speed of a chemical reaction. Each chemical reaction has its unique rate constant that serves as a proportional constant in the rate equation. It connects the reactant concentrations to the rate of reaction. Mathematically, for a general reaction:\[ \text{Rate} = k [A]^m [B]^n \] where:

• \([A]\) and \([B]\) are the concentrations of the reactants.
• \(m\) and \(n\) are the orders of the reactions with respect to \([A]\) and \([B]\).
• \(k\) varies with temperature and is generally expressed in units that depend on the overall order of the reaction.

The units of \(k\) can help define the reaction order. For example, in a reaction involving H₂ and I₂, the rate constant might be expressed in units of \( \mathrm{dm}^3 \mathrm{mol}^{-1} \mathrm{s}^{-1} \), indicating a second order reaction. Rate constants increase with temperature, which is why reactions take place faster at higher temperatures. Understanding rate constants helps not only in predicting how quickly a reaction proceeds but also in designing chemical processes.

Temperature Conversion

In chemical calculations, particularly those involving the Arrhenius equation, temperatures must be converted to Kelvin. This is crucial because Kelvin is the standard scientific unit for temperature, aligning proportionally with the kinetic energy of particles. The Kelvin scale begins at absolute zero, the point where theoretically all kinetic motion ceases, thus aligning with calculations involving energy.
To convert Celsius to Kelvin, you simply add 273.15:\[ T(\mathrm{K}) = T(^{\circ}\mathrm{C}) + 273.15 \] For instance, in our example, converting 327°C to Kelvin becomes:\[ 327 + 273.15 = 600.15 \, \mathrm{K} \] and 527°C becomes:\[ 527 + 273.15 = 800.15 \, \mathrm{K} \] Using Kelvin in the Arrhenius equation allows for accurate and consistent calculations of rate constants and activation energies.

Gas Constant

The gas constant \( R \) is a foundational constant in chemistry, connecting various fundamental equations such as the ideal gas law and the Arrhenius equation. It is denoted in units of \( \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} \) when dealing with energy-based reactions, making it crucial for calculating activation energy. The value of the gas constant used here is 8.314 \( \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} \), which ensures the units in calculations align correctly.
When using the Arrhenius equation, the gas constant helps relate the temperature (in Kelvin) with energy terms, such as activation energy and temperature effects on reaction rates.

• This constant allows scientists to predict how changes in temperature and energy affect reaction rates.
• Its consistency with energy units makes it invaluable for calculations across various chemistry fields.

Thus, the gas constant is a crucial bridge between theoretical concepts and practical chemistry applications, helping to translate physical conditions into action rates.