Problem 152 - A Complete Resource Book in Chemistry for JEE Main

Question

The energy of activation and specific rate constant for a first order reaction at \(25^{\circ} \mathrm{C}\) are \(100 \mathrm{~kJ} / \mathrm{mol}\) and \(3.46\) \(\times 10^{-5} \mathrm{sec}^{-1}\) respectively. Determine the temperature at which half life of reaction is 2 hour. \(2 \mathrm{~N}_{2} \mathrm{O}_{3} \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}\) \(\begin{array}{ll} \left.\text { (in } \mathrm{CCl}_{4}\right) & \left.\text { (in } \mathrm{CCl}_{4}\right)\end{array}\) (a) \(300 \mathrm{~K}\) (b) \(302 \mathrm{~K}\) (c) \(304 \mathrm{~K}\) (d) \(306 \mathrm{~K}\)

Step-by-Step Solution

Answer

The temperature is 306 K, option (d).

Step 1: Understanding First-Order Reaction and Half-Life

For a first-order reaction, the half-life formula is given by: \( t_{1/2} = \frac{0.693}{k} \). Where \( t_{1/2} \) is the half-life, and \( k \) is the rate constant. We need to find the temperature at which the half-life is 2 hours or 7200 seconds.

Step 2: Rearrange the Half-Life Formula

Rearrange the half-life formula to solve for \( k \): \( k = \frac{0.693}{t_{1/2}} \). Plug in the given half-life, \( t_{1/2} = 7200 \, \text{seconds} \).

Step 3: Calculate the Rate Constant Using Half-Life

Using the rearranged formula, calculate \( k \): \[ k = \frac{0.693}{7200} \approx 9.625 \times 10^{-5} \text{ sec}^{-1} \].This is the new rate constant at the unknown temperature.

Step 4: Arrhenius Equation Setup

The Arrhenius equation relates the rate constants at two different temperatures. It is given by: \[ \frac{k_2}{k_1} = e^{\frac{-E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)} \]Where \( k_1 \) is the initial rate constant at 25°C (or 298 K), \( k_2 \) is the rate constant calculated in Step 3, \( E_a = 100,000 \text{ J/mol} \), and \( R = 8.314 \text{ J/mol K} \). \( T_1 = 298 \text{ K} \) and \( T_2 = \text{unknown} \).

Step 5: Solve for Unknown Temperature \(T_2\)

Substitute all known values and solve for \( T_2 \):\[ \log \left( \frac{9.625 \times 10^{-5}}{3.46 \times 10^{-5}} \right) = \frac{-100000}{8.314} \left( \frac{1}{T_2} - \frac{1}{298} \right) \].Calculate the logarithms and rearrange to find \( T_2 \).

Step 6: Final Calculation of \(T_2\)

After rearranging and solving the equation, obtain the value for \( T_2 \). This involves detailed calculation, often assisted by a calculator:\[ T_2 \approx 306 \text{ K} \].

Step 7: Choose the Correct Option

Compare the calculated temperature of 306 K with the given options: (a) 300 K, (b) 302 K, (c) 304 K, (d) 306 K. The correct choice is (d) 306 K.

Key Concepts

First Order Reaction

A first-order reaction is a type of chemical reaction where the rate is directly proportional to the concentration of one reactant. This means that the reaction's speed decreases as the reactant gets used up. A key characteristic of first-order reactions is their half-life, which is the time required for half of the reactant to be consumed.
The half-life of a first-order reaction is constant and independent of the initial concentration of the reactant. It's calculated using the formula:

• \( t_{1/2} = \frac{0.693}{k} \),

where \( k \) is the rate constant. This formula reflects that even as the concentration changes, the time it takes for the concentration to halve remains constant.
In the context of the problem, we needed to determine the temperature at which the half-life of the reaction would be 2 hours. Thus, our task involved rearranging this formula to find the new rate constant at this specific half-life duration.

Arrhenius Equation

The Arrhenius Equation is a crucial formula in chemical kinetics that describes how the rate constant \( k \) changes with temperature. It is expressed as:

• \( k = A e^{-\frac{E_a}{RT}} \)

In this equation:

• \( A \) is the pre-exponential factor, which accounts for the frequency of collisions and the proper orientation for reactions.
• \( E_a \) is the activation energy, or the minimum amount of energy needed for a reaction to occur.
• \( R \) is the gas constant, 8.314 J/mol·K.
• \( T \) is the temperature in Kelvin.

The temperature dependence of reaction rates is captured by the Arrhenius Equation. Higher temperatures typically increase reaction rates as they provide the energy needed to surpass the activation energy barrier.
In our problem, we used the Arrhenius Equation in a comparative form, which allowed us to connect the rate constants at two different temperatures. This comparison enabled the calculation of the unknown temperature where the reaction's half-life is 2 hours.

Activation Energy

Activation energy (\( E_a \)) is the minimum energy that reacting molecules must possess for a reaction to occur. It represents an energy barrier that reactants must overcome to transform into products.
Understanding activation energy is crucial because it determines the sensitivity of the reaction rate to temperature changes. The higher the activation energy, the more significant the effect of temperature on the rate.
In our problem, the given activation energy was 100 kJ/mol. This value was integral to using the Arrhenius equation, as it appeared in the exponential factor that modified how the rate constant changed with temperature. With this energy barrier known, we could determine the temperature that allows the reaction to proceed at a different rate corresponding to a specific half-life.