Problem 67 - A Complete Resource Book in Chemistry for JEE Main

Question

In a second-order reaction, if first-order is observed for both the reactants \(\mathrm{A}\) and \(\mathrm{B}\), then which one of the following reactant mixtures will provide the highest initial rate? (a) \(0.1 \mathrm{~mol}\) of \(\mathrm{A}\) and \(0.1 \mathrm{~mol}\) of in \(0.2\) litre solvent (b) \(1.0 \mathrm{~mol}\) of \(\mathrm{A}\) and \(1.0 \mathrm{~mol}\) of in one litre solvent (c) \(0.2 \mathrm{~mol}\) of \(\mathrm{A}\) and \(0.2 \mathrm{~mol}\) of in \(0.1\) litre solvent (d) \(0.1 \mathrm{~mol}\) of \(\mathrm{A}\) and \(0.1 \mathrm{~mol}\) of in \(0.1\) litre solvent

Step-by-Step Solution

Answer

Option (c) provides the highest initial rate with \(4.0k\).

Step 1: Identify the reaction order

In this exercise, it's given that the reaction is second-order, with first-order for each reactant A and B. This means the rate law for the reaction can be expressed as \( \text{Rate} = k[A][B] \), where \( k \) is the rate constant, and \([A]\) and \([B]\) are the concentrations of reactants A and B respectively.

Step 2: Calculate initial concentrations

Convert the given moles of reactants into concentration by using the formula \( [ ext{Concentration}] = \frac{ ext{moles}}{ ext{volume}} \). For each option:- (a) \([A] = \frac{0.1}{0.2} = 0.5 \text{ M}\) and \([B] = 0.5 \text{ M}\).- (b) \([A] = \frac{1.0}{1.0} = 1.0 \text{ M}\) and \([B] = 1.0 \text{ M}\).- (c) \([A] = \frac{0.2}{0.1} = 2.0 \text{ M}\) and \([B] = 2.0 \text{ M}\).- (d) \([A] = \frac{0.1}{0.1} = 1.0 \text{ M}\) and \([B] = 1.0 \text{ M}\).

Step 3: Calculate rate for each option

Substitute the concentrations into the rate law to find the initial rate for each option:- (a) \( \text{Rate} = k(0.5)(0.5) = 0.25k \).- (b) \( \text{Rate} = k(1.0)(1.0) = 1.0k \).- (c) \( \text{Rate} = k(2.0)(2.0) = 4.0k \).- (d) \( \text{Rate} = k(1.0)(1.0) = 1.0k \).

Step 4: Compare the rates

Compare the calculated rates for each scenario. Option (c) has the highest rate of \(4.0k\) compared to the other options, which have lower initial rates.

Key Concepts

Second-order reaction

A second-order reaction involves the concentration of one or more reactants raised to the second power. In this specific exercise, we are dealing with a reaction where each reactant A and B contributes to first-order kinetics. This means that the overall reaction is second-order because the rate of reaction is dependent on the product of the concentrations of these two reactants.

In simple terms, a second-order reaction's rate is governed by the formula \[ \text{Rate} = k[A][B] \]where \([A]\) and \([B]\) are the molar concentrations of reactants A and B respectively, and \(k\) is the rate constant. Understanding the concept of order is crucial for predicting how changes in concentrations impact the reaction speed.

The second-order kinetics imply that doubling the concentration of either reactant while keeping the other constant will double the rate of reaction.

Rate law

The rate law is a mathematical description that shows how the rate of a reaction depends on the concentration of its reactants. For a second-order reaction involving reactants A and B, the rate law is expressed as \( \text{Rate} = k[A][B] \). This formula indicates two important aspects:

• The reaction rate is directly proportional to the product of the concentrations of the reactants.
• The constant \(k\) is specific to the reaction and must be determined experimentally.

The exponents in the rate law correspond to the order of the reaction with respect to each reactant. For this particular reaction, both exponents are 1, indicating first-order behavior for each reactant but second-order overall.

By examining the rate law, chemists can make predictions about how different concentrations will affect the speed and progression of the reaction.

Concentration calculation

Calculating the concentration of reactants is a key step in understanding reaction rates. The concentration \([ ext{Concentration} ]\) is calculated using the formula:\[ [\text{Concentration}] = \frac{\text{moles}}{\text{volume}} \]where "moles" refers to the number of moles of the reactant, and "volume" is the amount of solvent, usually in liters.

Taking option (c) from the exercise as an example, you would convert 0.2 moles of reactant A in a 0.1-liter solution to a concentration by dividing 0.2 by 0.1, resulting in a concentration of \(2.0 \text{ M}\), or molarity. Similarly, you do the same for reactant B.

These calculated concentrations are crucial for plugging into the rate law to determine the rate of reaction.

Initial rate

The initial rate of a reaction is the speed at which reactants are consumed when the reaction begins, where conditions like concentration are known and controlled.

It's calculated by substituting the initial concentrations into the rate law equation. For example, if the concentrations of reactants A and B are both 2.0 M, as in option (c), then the initial rate is calculated as:\[ \text{Rate} = k(2.0)(2.0) = 4.0k \]This means the initial rate is 4.0 times the rate constant \(k\).

The initial rate is vital for understanding how quickly a reaction will proceed at the start, which can help in experimental design and understanding the kinetics of the reaction.

Rate constant

The rate constant \(k\) is a crucial component in the rate law equation, acting as a proportionality factor that relates the concentrations of reactants to the rate of reaction. It remains constant for a given reaction at a specific temperature but changes if the temperature is altered.

It's determined experimentally, often by measuring the initial rate of reaction at various concentrations and rearranging the rate law to solve for \(k\). For a second-order reaction like the one in the exercise, the units of \(k\) typically are \( ext{L mol}^{-1} ext{s}^{-1} \), reflecting the overall reaction order.

Understanding and calculating the rate constant helps chemists predict the reaction's behavior under different situations, driving decisions in both industrial and laboratory settings.