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 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 will take approximately 86.6 minutes for 70% decomposition.

Step 1: Write the Basic First Order Reaction Formula

For a first order reaction, the rate constant \(k\) is expressed using the formula \(k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right)\). Here, \([A_0]\) is the initial concentration and \([A]\) is the concentration after time \(t\).

Step 2: Calculate the Rate Constant for Reaction A

Given that reaction A is 30% complete in 60 minutes, 30% completion means that 30% of \([A_0]\) has reacted, leaving 70% of \([A_0]\) unreacted. Thus, \( \frac{[A]}{[A_0]} = 0.7\). By substituting these values into the formula for \(k_A\):\[k_A = \frac{2.303}{60} \log \left( \frac{1}{0.7} \right) = \frac{2.303}{60} \cdot 0.1549 = 0.00594 \ min^{-1}\]

Step 3: Relate the Rate Constants Using Activation Energy Difference

The activation energy difference for the reactions of A and B is \(10.46 \ \text{kJ/mol}\). The Arrhenius equation \(k = A e^{-E_a/(RT)}\) shows that the rate constant and activation energy are related. Since \(A\) is the same for both reactions, \[\frac{k_A}{k_B} = e^{-(E_{a_A} - E_{a_B})/(RT)}\] Substituting the given activation energy difference: \[\frac{k_A}{k_B} = e^{-10460/(8.314 \times 373)}=0.467\]

Step 4: Solve for Rate Constant of Reaction B

Use the relation from Step 3 to solve for \(k_B\): \[k_B = \frac{k_A}{0.467} = \frac{0.00594}{0.467} = 0.0127 \ min^{-1}\]

Step 5: Calculate the Time for 70% Completion of B

For 70% decomposition of B, 30% remains: \( \frac{[B_0]}{[B]} = \frac{1}{0.3} \). Substitute these values into the first-order rate equation for B:\[t_{B} = \frac{2.303}{k_B} \log \left( \frac{1}{0.3} \right) = \frac{2.303}{0.0127} \cdot 0.4771 = 86.6 \ min\]

Key Concepts

Activation Energy

Activation energy is the minimum energy required to initiate a chemical reaction. It's like the energy boost needed to start a roller coaster on its track.

In the context of first-order reactions, activation energy dictates how quickly or slowly a reaction can proceed. The higher the activation energy, the slower the reaction generally is, because molecules need more energy to overcome the initial barrier.

For our exercise, we learned that the activation energy for reaction A is 10.46 kJ/mol greater than for reaction B. This difference plays a crucial role in the differing rates of the two reactions.

• Reaction A is slower because more energy is needed, acting as a larger hurdle to overcome.
• Reaction B, with lower activation energy, progresses faster. This is evident as we calculated that B takes less time to reach a higher percentage of completion compared to A.

Understanding this concept is essential, as it highlights the relationship between activation energy and the speed of a reaction.

Rate Constant Calculation

The rate constant, symbolized as \(k\), is a fundamental factor that illustrates the speed of a reaction. It is a specific value that quantifies how fast reactants turn into products.

For first-order reactions, the rate constant k can be calculated using the formula: \[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \] where

• \([A_0]\) is the initial concentration,
• \([A]\) is the concentration after time \(t\).

In the provided solution, we used this formula to determine the rate constant for reactions A and B.

For reaction A, knowing that it is 30% complete in 60 minutes, we found that 70% remains unreacted. Plugging in these numbers, we determined \(k_A = 0.00594 \, \text{min}^{-1}\).

As a next step, by understanding that the concentration of B changes similarly, we calculated \(k_B = 0.0127\, \text{min}^{-1}\).

These calculations are vital as they set the baseline for comparing two reactions and understanding their completion times.

Arrhenius Equation

The Arrhenius equation is a cornerstone in the world of chemical kinetics. It links the rate constant \(k\), temperature \(T\), and activation energy \(E_a\) in a compact exponential form:
\[ k = A e^{-E_a/(RT)} \]

• \(A\) is the frequency factor (pre-exponential factor) representing the number of times reactants approach the activation state per unit time.
• \(E_a\) is the activation energy.
• \(R\) is the universal gas constant, and
• \(T\) is the temperature in Kelvin.

This equation showcases how temperature and activation energy influence the rate constant and, consequently, the reaction rate. A lower \(E_a\) or a higher \(T\) generally results in a larger rate constant, speeding up the reaction.

For the reactions in the exercise, we examined how the constant frequency factor and differing activation energies of A and B affect \(k\). By employing the equation, we derived the relation between \(k_A\) and \(k_B\), showing that despite having the same pre-exponential factor, reaction B with a lower \(E_a\) had a higher \(k\), thus decomposing faster.

The Arrhenius equation thus becomes a powerful tool to predict and compare reaction behaviors across different conditions.