Problem 88 - A Complete Resource Book in Chemistry for JEE Main

Question

The experimental rate law for a reaction \(2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow\) Product, is \(\mathrm{V} \alpha \mathrm{C}_{\mathrm{A}} \mathrm{C}_{\mathrm{B}}^{-2} .\) If the concentration of both \(\mathrm{A}\) and are doubled the rate of reaction increases by a factor of (a) \(\sqrt{2}\) (b) 2 (c) \(2 . \sqrt{2}\) (d) 4

Step-by-Step Solution

Answer

None of the options match due to error in increase-decrease balance. Initial miscalculation. Options do not fit.

Step 1: Understand the Rate Law

The given rate law is \( V \propto C_A C_B^{-2} \). This implies that the rate of reaction is directly proportional to the concentration of \( A \) and inversely proportional to the square of the concentration of \( B \).

Step 2: Express the Rate Law Mathematically

Initially, the rate can be expressed as \( V = k C_A C_B^{-2} \) for some proportionality constant \( k \). This equation represents how the concentrations of \( A \) and \( B \) influence the rate \( V \).

Step 3: Consider the Effect of Doubling Concentrations

If the concentrations of both \( A \) and \( B \) are doubled, the new concentrations are \( 2C_A \) and \( 2C_B \). Substitute these into the rate law: \( V' = k (2C_A) (2C_B)^{-2} \).

Step 4: Simplify the New Rate Expression

Simplify the expression to \( V' = k (2C_A) \left( \frac{1}{4C_B^2} \right) \). This gives \( V' = \frac{k C_A}{2 C_B^2} \).

Step 5: Compare the New Rate to the Original Rate

Originally, \( V = k \frac{C_A}{C_B^2} \). The new rate \( V' = \frac{k C_A}{2 C_B^2} \) shows that \( V' = \frac{V}{2} \).

Step 6: Find the Increase Factor of the Rate

Initially, the rate is \( V \). After doubling, the rate becomes \( V' = \frac{V}{2} \), indicating the rate decreases by a factor of \( \frac{1}{2} \), not increases. Verify options: \( \sqrt{2}, 2, 2\sqrt{2}, 4 \) do not match initial error indication leading to incorrect choices.

Key Concepts

Rate Law

When discussing chemical kinetics, rate laws play a crucial role. A rate law expresses the rate of a chemical reaction in terms of the concentration of its reactants. It's a mathematical equation that links the concentration of the reactants to the speed at which products are formed. In the given problem, the rate law is expressed as \( V \propto C_A C_B^{-2} \). This indicates that:

• The reaction rate is directly proportional to the concentration of \( A \).
• The reaction rate is inversely proportional to the square of the concentration of \( B \).

Thus, as the concentration of \( A \) increases, the reaction rate increases. Conversely, as the concentration of \( B \) increases, the rate decreases because it's in the denominator with a power of \(-2\). Understanding how the concentrations of reactants affect the rate of reaction is key to mastering chemical kinetics.

Reaction Rate

The reaction rate is a measure of how quickly a chemical reaction proceeds. It is typically expressed in terms of concentration change per unit time. In the context of our exercise, we're considering the original rate \( V \) and the new rate \( V' \) after changing reactant concentrations.Initially, the rate \( V \) is determined by \( V = k C_A C_B^{-2} \), where \( k \) is a constant. When we double the concentrations of both \( A \) and \( B \), we adjust these values in the equation to find the new rate.This new expression, \( V' = k (2C_A) (2C_B)^{-2} \), exemplifies how sensitive the rate is to changes in concentration. After simplification, it results in:- \( V' = \frac{k C_A}{2 C_B^2} \)From this, we can compare with the original to see how the rate changes. The importance lies in understanding that the rate is affected not just by the quantity of each reactant but also by the configuration of the rate law itself.

Proportional Relationships

Proportional relationships in rate laws define how changes in reactant concentrations can affect the reaction rate. In our rate law example, the relationships are:- Directly proportional to \( C_A \): The reaction rate increases linearly with an increase in \( A \)'s concentration.- Inversely proportional to \( C_B^2 \): The rate decreases quadratically as \( B \)'s concentration increases. When you double the concentration of \( A \) to \( 2C_A \) and \( B \) to \( 2C_B \), these relationships guide how each change contributes to the new rate:

• \( 2C_A \) doubles the influence from \( A \).
• \((2C_B)^{-2}\) reduces \( B \)'s contribution by a factor of four, due to squaring and inversion.

Hence, manipulating concentrations while understanding their proportional effects allows us to predict whether the rate will increase or decrease, a fundamental skill when dealing with chemical kinetics.