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. \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}\) (in \(\left.\mathrm{CCl}_{4}\right) \quad\) (in \(\left.\mathrm{CCl}_{4}\right)\) (a) \(300 \mathrm{~K}\) (b) \(302 \mathrm{~K}\) (c) \(304 \mathrm{~K}\) (d) \(306 \mathrm{~K}\)

Step-by-Step Solution

Answer

The temperature is 304 K (option c).

Step 1: Find the Half-life Equation for a First-order Reaction

The half-life (\( t_{1/2} \)) for a first-order reaction is given by the formula:\[t_{1/2} = \frac{0.693}{k}\]where \( k \) is the specific rate constant of the reaction. We are given the half-life and need to find the rate constant for this condition.

Step 2: Convert Half-life from Hours to Seconds

The given half-life is 2 hours. Convert this into seconds because the rate constant we have is in seconds:\[2 \text{ hours} = 2 \times 60 \times 60 = 7200 \text{ seconds}\]

Step 3: Calculate Rate Constant Using Given Half-life

Using the half-life equation:\[t_{1/2} = \frac{0.693}{k}\]Substitute the half-life of 7200 seconds:\[7200 = \frac{0.693}{k}\]Solve for \( k \):\[k = \frac{0.693}{7200} \approx 9.63 \times 10^{-5} \text{ sec}^{-1}\]

Step 4: Use Arrhenius Equation to Find the New Temperature

The Arrhenius equation relates the rate constant to temperature:\[k = A \cdot e^{-E_a/(RT)}\]Rewriting for two temperatures and using given and calculated rate constants:\[\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]Using \(k_1 = 3.46 \times 10^{-5}\) and \(k_2 = 9.63 \times 10^{-5}\), activation energy \(E_a = 100 \times 10^3 \text{ J/mol}\), and gas constant \(R = 8.314 \text{ J/mol K}\). Substitute into the equation.

Step 5: Solve for Temperature T2

Compute\[\ln\left(\frac{9.63 \times 10^{-5}}{3.46 \times 10^{-5}}\right) = \frac{100000}{8.314}\left(\frac{1}{298} - \frac{1}{T_2}\right)\]Calculate:\[\ln\left(2.783\right) = 12.02 \left(\frac{1}{298} - \frac{1}{T_2}\right)\]\(\ln(2.783) \approx 1.023\), then solve for \(T_2\):\[T_2 = \left(\frac{1}{\frac{1}{298} - \frac{1.023}{12.02}}\right)\]Solving gives \(T_2 \approx 304 \text{ K}\).

Key Concepts

Activation Energy

Activation energy is a critical concept in the study of chemical reactions. It represents the minimum amount of energy required for a reaction to occur. This concept helps explain why some reactions occur rapidly, while others happen slowly or not at all.
In a first-order reaction, such as the one in the exercise, the activation energy influences how quickly the reaction proceeds. A higher activation energy typically means a slower reaction, as more energy is needed for the reactants to transition into products.
In the exercise, the activation energy is provided as 100 kJ/mol. This value helps determine the influence of temperature on the rate of reaction. It is a key value when applying the Arrhenius Equation, which relates the rate constant of a reaction to activation energy, temperature, and other factors.

Rate Constant

The rate constant, often denoted by the letter \( k \), is a numerical value that represents the speed at which a chemical reaction proceeds. It is influenced by factors such as temperature, pressure, and the presence of catalysts.
For first-order reactions, the rate constant has the units of \( ext{s}^{-1} \). In the given problem, the rate constant at \( 25^{ ext{o}} ext{C} \) is \( 3.46 \times 10^{-5} ext{ sec}^{-1} \). This represents how often reactant molecules successfully collide and transform into products under specified conditions.
The value of the rate constant can change with temperature, a relationship described by the Arrhenius Equation. Knowing the rate constant allows us to predict how long it will take for a certain proportion of a substance to react.

Arrhenius Equation

The Arrhenius Equation is a formula that expresses the relationship between the rate constant \( k \), activation energy \( E_a \), temperature \( T \), and the pre-exponential factor \( A \). It is typically written as:
\[ k = A \, e^{-E_a/(RT)} \]
Where:

• \( k \) is the rate constant.
• \( A \) is the pre-exponential factor, representing collision frequency.
• \( e \) is the base of the natural logarithm.
• \( E_a \) is the activation energy in joules per mole.
• \( R \) is the universal gas constant: \( 8.314 \, ext{J/mol K} \).
• \( T \) is the temperature in Kelvin.

The equation is crucial for understanding how a change in temperature affects the rate of a reaction. In the exercise, the Arrhenius Equation is rearranged to find the temperature \( T_2 \) when the rate constant changes due to different half-life conditions.

Half-life of Reaction

Half-life is the time required for half of the reactant in a reaction to be consumed. For first-order reactions, the half-life is a constant, irrespective of the initial concentration of reactants.
The formula for the half-life of a first-order reaction is:
\[ t_{1/2} = \frac{0.693}{k} \]
This formula shows that the half-life is inversely proportional to the rate constant \( k \). Hence, when \( k \) increases, the half-life decreases and vice versa.
In the exercise, the half-life is given as 2 hours, which is converted to 7200 seconds to use in calculations. By applying the half-life formula, the rate constant for the new condition is found, which is then utilized to solve for the new temperature using the Arrhenius Equation.