Problem 60 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The reaction: \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{NO}_{2}+\mathrm{OH}^{-}\) \(\rightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}^{-}-\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}\) obeys the rate law for pseudo first-order kinetics in the presence of a large excess of hydroxide ion. If \(1 \%\) of nitro ethane undergoes reaction in half a minute when the reactant concentration is \(0.002 \mathrm{M}\), what is the pseudo first-order rate constant? (a) \(2 \times 10^{-2} \mathrm{~min}^{-1}\) (b) \(6 \times 10^{-3} \mathrm{~min}^{-1}\) (c) \(4 \times 10^{-2} \mathrm{~min}^{-1}\) (d) \(1 \times 10^{-2} \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

\(k \approx 1 \times 10^{-2} \text{ min}^{-1}\)

Step 1: Identify the Given Information

Recognize that the reaction follows pseudo first-order kinetics and that 1% of nitroethane reacts in 30 seconds. Note that the initial concentration of nitroethane is 0.002 M, and a large excess of hydroxide ion ensures it remains constant.

Step 2: Calculate the Remaining Concentration

Calculate the concentration of nitroethane left after 30 seconds. Since 1% reacts, 99% remains. Hence, remaining concentration = (99/100) * 0.002 M = 0.00198 M.

Step 3: Apply the Pseudo First-Order Rate Law

Use the formula for first-order kinetics: \(ln(\frac{[A]_0}{[A]}) = kt\), where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant. We will use this formula to find \(k\) by plugging in the respective values.

Step 4: Calculate the Rate Constant (k)

Input the values into the first-order kinetics equation: \(ln(\frac{0.002}{0.00198}) = k(0.5)\), and solve for \(k\).

Step 5: Solve for k

Divide both sides of the equation by 0.5 to isolate \(k\): \(k = \frac{ln(0.002/0.00198)}{0.5}\), and calculate \(k\).

Key Concepts

Rate Constant Calculation

Understanding how to calculate the rate constant is essential when studying the dynamics behind chemical reactions. The rate constant, often designated by the symbol 'k', provides vital information about the speed at which a reaction proceeds. In pseudo first-order kinetics, the calculation revolves around a situation where one reactant is in large excess, often a solvent, and its concentration effectively remains constant during the reaction.

To obtain the rate constant 'k', we use the formula for first-order kinetics: \(ln(\frac{[A]_0}{[A]}) = kt\), where \([A]_0\) is the initial concentration, \([A]\) is the concentration at a specific time 't', and 'k' is the rate constant we aim to find. By rearranging the equation and solving for 'k', we get \(k = \frac{ln([A]_0/[A])}{t}\). The use of natural logarithm (ln) allows us to compare the exponential decay of a reactant's concentration over time.

In our example problem, the initial concentration is 0.002 M, and after 30 seconds (0.5 minutes), 1% of nitroethane has reacted, leaving a concentration of 0.00198 M. Plugging these values into the equation enables us to solve for the pseudo first-order rate constant 'k'. This constant is fundamental as it can predict the behavior of the reaction under various conditions and durations.

Chemical Kinetics

Chemical kinetics is the area of chemistry concerned with the speed or rate of a chemical reaction, as well as the mechanisms by which reactions occur. It provides insights into the steps that take place from the reactants turning into products and which factors affect the reaction rate—like temperature, concentration, and presence of catalysts.

Kinetics can answer questions such as 'How fast does a reaction proceed?' and 'What processes are controlling this speed?' The concept of reaction order, such as first-order, second-order or zero-order, plays a crucial role in chemical kinetics, as it describes the dependence of the reaction rate on the concentration of the reactants.

Pseudo First-Order Kinetics

In pseudo first-order reactions, while the overall order might not be one, the reaction behaves as first-order with respect to one reactant due to the large excess of another reactant. This simplification is useful as it allows for easier analysis and interpretation of the experimental data.

First-Order Reaction

A first-order reaction is one where the rate is directly proportional to the concentration of one reactant. As time progresses, the concentration of this reactant decreases exponentially. A classic characteristic of first-order kinetics is that the half-life of the reaction, the time it takes for half of the reactant to be consumed, is constant and independent of its initial concentration.

The mathematical expression of a first-order reaction is \(rate = k[A]\), where 'k' is the rate constant, and '[A]' is the concentration of the reactant. By integrating this rate law over time, we derive the equation \(ln(\frac{[A]_0}{[A]}) = kt\), which we frequently apply in rate constant calculations.

Understanding this concept allows predicting how long it will take for a certain percentage of a reactant to be consumed. In the given problem, knowing that a pseudo first-order reaction was taking place enabled precise calculation of the rate constant for the reaction – an essential step to comprehending the chemical kinetics behind the process.