Problem 38 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A first order reaction is \(50 \%\) complete in 30 minutes at \(27^{\circ} \mathrm{C}\) and in 10 minutes at \(47^{\circ} \mathrm{C}\). Calculate the reaction rate constant at \(27^{\circ} \mathrm{C}\) and the energy of activation of the reaction in \(\mathrm{kJ} / \mathrm{mole}\).

Step-by-Step Solution

Answer

The rate constant at 27°C is 0.0231 min⁻¹, and the activation energy is 43.83 kJ/mol.

Step 1: Use First Order Reaction Formula

For a first-order reaction, the formula for the rate constant is given by \( k = \frac{0.693}{t_{1/2}} \) where \( t_{1/2} \) is the half-life of the reaction. Here the reaction is \(50\%\) complete in 30 minutes at \(27^{\circ} \mathrm{C}\), meaning that \( t_{1/2} = 30 \) minutes.

Step 2: Calculate Reaction Rate Constant at 27°C

Substitute \( t_{1/2} = 30 \) minutes into the first order reaction rate formula: \( k_{27^{\circ} \mathrm{C}} = \frac{0.693}{30} = 0.0231 \text{ min}^{-1}.\)

Step 3: Use Arrhenius Equation for Activation Energy

The Arrhenius equation is given by \( k = A e^{-E_{a}/RT} \), which can be rearranged to \( \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \). By applying the equation for two temperatures, the equation becomes \( \ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \).

Step 4: Plug in Known Values to Arrhenius Equation

Let \( T_1 = 300 \text{ K} \) and \( T_2 = 320 \text{ K} \). We have \( k_{27^{\circ} \mathrm{C}} = 0.0231 \text{ min}^{-1} \) and need \( k_{47^{\circ} \mathrm{C}} \) which corresponds to 10 minutes. Similarly, calculate \( k_{47^{\circ} \mathrm{C}} = \frac{0.693}{10} = 0.0693 \text{ min}^{-1} \).

Step 5: Solve for Activation Energy \(E_a\)

With \( R = 8.314 \text{ J/mol} \cdot \text{K} \), plug in \( \ln \frac{0.0693}{0.0231} = \frac{E_a}{8.314} \left( \frac{1}{300} - \frac{1}{320} \right) \). Solve for \( E_a \).

Step 6: Final Calculation for \(E_a\)

Calculate \( \ln 3 = 1.0986 \), and \( \frac{1}{300} - \frac{1}{320} = \frac{20}{96000} = 0.0002083 \). Thus, \( E_a = 1.0986 \times \frac{8.314}{0.0002083} \). Hence, \( E_a = 43832.17 \text{ J/mol} = 43.83 \text{ kJ/mol}.\)

Key Concepts

First Order Reaction

In chemical kinetics, a first order reaction is one where the rate of the reaction is proportional to the concentration of a single reactant. This means that if the concentration of this reactant changes, the rate of reaction will change in direct proportion. A characteristic feature of first order reactions is their predictable half-life, which is the time it takes for half of the reactant to be consumed. To calculate the rate constant ( k ) for a first order reaction, we use the formula\( k = \frac{0.693}{t_{1/2}} \), where\( t_{1/2} \) is the half-life time of the reaction. In this case, since the reaction is 50% complete in 30 minutes at \(27^{\circ} \text{C}\), this duration represents the half-life for this reaction.

Reaction Rate Constant

The reaction rate constant ( k ) is a vital factor in determining how fast a reaction proceeds. For first order reactions, it is expressed in reciprocal time units, such as \(\text{min}^{-1}\) or \(\text{s}^{-1}\). The rate constant is unique for every reaction and can be influenced by conditions such as temperature. Calculating the rate constant at a specific temperature can help predict how quickly the reactants will turn into products under those conditions. In our example, when substituting \( t_{1/2} = 30 \) minutes into the first order reaction rate formula, we find that \( k_{27^{\circ} \text{C}} = 0.0231 \text{ min}^{-1} \). This means that at \(27^{\circ} \text{C}\), the reaction progresses at this quantified pace.

Activation Energy

Activation energy ( E_a ) is the minimum amount of energy required for a chemical reaction to occur. Think of it as a hill that the reactants need to climb over to proceed to products. Reactions with lower activation energies occur more readily, while those with higher energy barriers tend to be slower unless driven by external energy, like heat. Knowing the activation energy is crucial for understanding and controlling chemical reactions. In the given exercise, the activation energy was determined using the Arrhenius equation, resulting in \(43.83 \, \text{kJ/mol}\). This tells us the energy required to initiate the reaction, under the conditions provided.

Arrhenius Equation

The Arrhenius Equation relates the rate constant ( k ) of a reaction to the temperature ( T ) and activation energy ( E_a ). The formula \( k = A e^{-E_{a}/RT} \) illustrates how temperature influences the reaction rate: as temperature increases, the rate constant also increases, suggesting a faster reaction. By rearranging the Arrhenius equation into the form \( \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \), we introduce the natural logarithm, which facilitates easier interpretation and calculation. For practical applications, using two known temperatures and rate constants, we can derive the activation energy. In our scenario, this relationship was used to calculate \(E_a\) at different temperatures, exposing the sensitivity of this reaction to temperature variations.