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

Step-by-Step Solution

Answer

The temperature at which the reaction has a half-life of 2 hours is 302 K (option b).

Step 1: Understand the Reaction Type

The problem specifies a first-order reaction, which means that the rate of reaction depends only on the concentration of one reactant. The general form of the rate law for a first-order reaction can be expressed as \( r = k[A] \), but for half-life calculations, we use the integrated rate law specific to first-order reactions.

Step 2: Chart the Given Data and Required Formula

Given data includes activation energy \( E_a = 100 \text{ kJ/mol} = 100,000 \text{ J/mol} \), rate constant \( k = 3.46 \times 10^{-5} \text{ sec}^{-1} \), and the desired half-life \( t_{1/2} = 2 \text{ hours} = 7200 \text{ seconds} \). The formula for the half-life of a first-order reaction is \( t_{1/2} = \frac{0.693}{k} \). To find the new rate constant corresponding to the 2-hour half-life, rearrange the formula to get \( k_{new} = \frac{0.693}{7200} \).

Step 3: Calculate the New Rate Constant

Use the half-life formula to find the new rate constant: \( k_{new} = \frac{0.693}{7200} = 9.625 \times 10^{-5} \text{ sec}^{-1} \).

Step 4: Use the Arrhenius Equation

The Arrhenius equation relates the rate constant \( k \) to temperature \( T \): \( k = A e^{-E_a/RT} \), where \( R = 8.314 \text{ J/mol}\cdot\text{K} \) and \( A \) is the frequency factor. Since the frequency factor \( A \) remains constant, you can use the relation \( \frac{k_1}{k_2} = e^{\frac{-E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)} \), where \( k_1 = 3.46 \times 10^{-5} \text{ sec}^{-1} \) at \( T_1 = 298 \text{ K} \) (25°C) and \( k_2 = 9.625 \times 10^{-5} \text{ sec}^{-1} \).

Step 5: Solve for the New Temperature

Substitute the known values into the Arrhenius relation to solve for \( T_2 \): \[ \ln\left( \frac{3.46 \times 10^{-5}}{9.625 \times 10^{-5}} \right) = \frac{-100,000}{8.314} \left( \frac{1}{298} - \frac{1}{T_2} \right) \]Calculate the left side: \( \ln(0.3594) = -1.0238 \). Solve for \( T_2 \):\[ -1.0238 = \frac{-100,000}{8.314} \left( \frac{1}{298} - \frac{1}{T_2} \right) \]\[ 1.0238 \times 8.314 = 100,000 \left( \frac{1}{298} - \frac{1}{T_2} \right) \]\[ \frac{1}{T_2} = \frac{1}{298} - \frac{1.0238 \times 8.314}{100,000} \]Calculate \( T_2 \) to find \( T_2 \approx 302 \text{ K} \).

Step 6: Match With Given Options

The calculated temperature \( T_2 \approx 302 \text{ K} \) matches option (b).

Key Concepts

Activation Energy

Activation energy, often denoted as \(E_a\), is the minimum energy that reactant molecules need to reach the transition state during a chemical reaction. This energy barrier must be overcome for a reaction to proceed.

In the context of the given exercise, the activation energy for the reaction is provided as \(100 \text{kJ/mol}\). This means that each mole of reactant needs this amount of energy to transition towards forming the products.

• A higher activation energy means that fewer molecules have the necessary energy to react at a given temperature.
• Conversely, a lower activation energy implies that more molecules can successfully collide with enough energy to overcome the barrier.

This concept is crucial for understanding why reactions occur more rapidly at higher temperatures: increased thermal energy allows a greater fraction of molecules to surpass the activation barrier.

Arrhenius Equation

The Arrhenius equation is a formula that describes how the rate constant \(k\) of a reaction depends on the temperature \(T\) and the activation energy \(E_a\). It is given by:\[k = A e^{-E_a/RT}\]where:

• \(k\) is the rate constant.
• \(A\) is the frequency factor, representing the number of times reactants approach each other with the correct orientation.
• \(e\) is the base of natural logarithms.
• \(R\) is the universal gas constant, approximately \(8.314 \text{J/mol}\cdot\text{K}\).
• \(T\) is the absolute temperature in Kelvin.

This relationship shows that as temperature increases, \(k\) also increases, suggesting faster reaction rates.

For the exercise, the equation is rearranged to find the new temperature that achieves a specified half-life. By comparing two rate constants at different temperatures, the equation helps us understand how much temperature change is needed to achieve a certain reaction rate.

Rate Constant

The rate constant, represented as \(k\), is a proportionality constant that links the rate of a chemical reaction to the concentration of reactants. For first-order reactions, the rate is directly proportional to the concentration of one reactant.

In the exercise, the rate constant for the reaction at \(25^{\circ} \mathrm{C}\) is given as \(3.46 \times 10^{-5} \mathrm{sec}^{-1}\). This low value suggests a slow reaction speed at that temperature.

When the half-life of the reaction changes (such as when it becomes 2 hours as desired in the problem), it indicates that \(k\) must change as well. Solving for a new rate constant \(k_{new}\) helps us adjust for the desired half-life, enabling the exercise to solve for the temperature required to shift the reaction's pace.

Half-Life Calculation

The half-life \(t_{1/2}\) of a reaction is the time required for half of the reactant to be consumed. For first-order reactions, it's expressed using the formula:\[t_{1/2} = \frac{0.693}{k}\]This denotes that the half-life is inversely proportional to the rate constant \(k\).

The exercise involves calculating a new half-life by adjusting \(t_{1/2}\) to match a specific duration, 2 hours in this case. Converting 2 hours into seconds gives \(7200 \text{ seconds}\). By rearranging the half-life formula, we find \(k_{new}\) to meet this timeline.

• This rearrangement yields \(k_{new} = \frac{0.693}{7200}\).
• Finding \(k_{new}\) helps us further manipulate the Arrhenius equation to find out at what temperature this half-life is possible.

Understanding these calculations reinforces how reaction conditions, such as temperature, directly influence reaction timelines and speeds.