Problem 194 - A Complete Resource Book in Chemistry for JEE Main

Question

Two first order reactions \(\mathrm{A}\) and \(\mathrm{B}\) have the same frequency factor and activation energy of \(\mathrm{A}\) exceeds that of \(\mathrm{B}\) by \(10.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\). If \(\mathrm{A}\) is \(30 \%\) complete in \(60 \mathrm{~min}\). at \(100^{\circ} \mathrm{C}\), how long will the \(\mathrm{B}\) take for \(70 \%\) decomposition. [Given : \(\log 3=0.4771, \log 7=0.845\), antilog \(0.4633=2.9]\)

Step-by-Step Solution

Answer

B takes about 69.84 min for 70% decomposition.

Step 1: Understanding First Order Reaction Kinetics

For a first-order reaction, the rate constant \( k \) can be determined using the equation: \[ k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \]\[ \ln \left( \frac{[A]_0}{[A]} \right) = \ln(1 - \text{fraction decomposed}) \].

Step 2: Calculate Rate Constant for Reaction A

For reaction A, which is \(30\%\) complete, the fraction decomposed is \(0.3\). Therefore, \( \ln(1 - 0.3) = \ln(0.7) = -0.3567 \). Using the formula for the rate constant, \[ k_A = \frac{-0.3567}{60} \approx -0.005945 \].

Step 3: Relationship of Activation Energies

The activation energy of A exceeds B by \(10.46 \text{ kJ/mol} \). If \(E_a_A\) and \(E_a_B\) are the activation energies for A and B respectively, then \(E_a_A = E_a_B + 10.46\).

Step 4: Arrhenius Equation for the Rate Constants

The Arrhenius equation for the rate constants is: \[ k = Ae^{-E_a/RT} \] Given that the frequency factors are the same for both, let’s denote them by \(A\). Then, the equation for both A and B will be:- For A: \( k_A = A e^{-E_a_A / RT} \)- For B: \( k_B = A e^{-E_a_B / RT} \).

Step 5: Calculate Rate Constant for Reaction B

Using the difference in activation energies and the Arrhenius equation, \[ \frac{k_B}{k_A} = e^{(E_a_A - E_a_B) / RT} \].Let \(T = 100 + 273 = 373\) K. Then, \( \frac{k_B}{k_A} = e^{-10.46 \times 1000 / 8.314 \times 373} \approx e^{-3.368} \approx \frac{1}{2.9} \).Hence, \(k_B \approx k_A \cdot 2.9 \approx -0.005945 \cdot 2.9 \approx -0.017241\).

Step 6: Calculate Time for 70% Decomposition of B

For reaction B, which needs \(70\%\) completion, the fraction decomposed is \(0.7\). Calculate:\[ \ln(0.3) = -1.203972 \]Using \( k_B \) from Step 5, solve for time \(t_B\):\[ t_B = \frac{-1.203972}{-0.017241} \approx 69.84 \text{ min} \]

Key Concepts

Rate Constant

In first-order reactions, the rate constant is a crucial factor that helps us describe how fast a reaction proceeds. It is denoted by the symbol \( k \). This constant is used in various equations to relate concentration changes over time. Specifically, for a first-order reaction, the rate constant can be determined using:

• \( k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \)
• Where \( [A]_0 \) is the initial concentration, and \( [A] \) is the concentration after time \( t \).

The equation indicates that the reaction's rate does not depend on the concentration of the reactants. Even as the concentration decreases, the rate constant \( k \) remains the same.
This idea is important for calculating how much of a reactant will turn into products over time, which we explored with the case of reaction A starting 30% complete in 60 minutes.

Activation Energy

Activation energy is the minimum energy required for a chemical reaction to proceed. It's like the hurdle reactants need to overcome to transform into products. In our problem, the activation energy of reaction A is higher than that of reaction B by 10.46 kJ/mol. Activation energy plays a vital role in influencing the reaction rate:

• Higher activation energies mean slower reactions because less reactant energy will exceed the necessary threshold.
• Conversely, lower activation energies facilitate faster reactions.

This energy barrier can be visualized as a small hill that reactants must climb; the taller the hill, the harder it is to climb, impacting the rate accordingly.
Thus, comparing activation energies lets us predict which reaction will be faster, assuming other factors like temperature and frequency factor are constant.

Arrhenius Equation

The Arrhenius equation links the rate constant \( k \) with activation energy and temperature. It is expressed as:
\[ k = A e^{-E_a/RT} \]

• Where \( A \) is the frequency factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
• This formula shows us that as temperature increases, the exponential term increases, thereby increasing \( k \). This means reactions generally go faster at higher temperatures.

In our calculations, we used the Arrhenius equation to determine how the rate constants for A and B compare by considering their activation energies. Since both reactions have the same frequency factor, the major difference in their speeds comes from their activation energies.
Utilizing the Arrhenius equation gives insight into how sensitive a reaction's speed is to changes in temperature and activation energy, making it a central concept in chemical kinetics.