Problem 135 - A Complete Resource Book in Chemistry for JEE Main

Question

When concentrations of the reactants is increased sixteen times, the rate becomes two times. The reaction is of (a) \(1 / 4\) order (b) fourth-order (c) third-order (d) \(1 / 8\) order

Step-by-Step Solution

Answer

The reaction is of \( 1/4 \) order.

Step 1: Understand the question

The question asks us to determine the order of a reaction when the concentration of the reactants is increased sixteen times (i.e., multiplied by 16) and the rate becomes two times (i.e., multiplied by 2).

Step 2: Define the rate law expression

The rate law expression for a reaction is generally given by \( \text{Rate} = k[A]^n \), where \( k \) is the rate constant, \([A]\) is the concentration of the reactants, and \( n \) is the order of the reaction we need to find.

Step 3: Set up equations based on the rate change

Initially, let the rate be \( ext{Rate}_1 = k[A]^n \). When the concentration increases 16 times, the rate is \( ext{Rate}_2 = k(16[A])^n = 2 \times ext{Rate}_1 = 2k[A]^n \).

Step 4: Relate the equations

Relate the modified rate with the initial rate: \( k(16[A])^n = 2k[A]^n \). This implies \( 16^n \times k[A]^n = 2k[A]^n \).

Step 5: Solve for reaction order n

Cancel \( k[A]^n \) from both sides: \( 16^n = 2 \). Take the logarithm of both sides: \( n \cdot \log(16) = \log(2) \). Simplify using \( \log(16) = 4 \log(2) \), which gives \( 4n = 1 \). Thus, \( n = \frac{1}{4} \).

Step 6: Conclude the order of reaction

The calculated order of reaction \( n \) is \( \frac{1}{4} \), which means the reaction order is \( 1/4 \).

Key Concepts

Rate Law

To comprehend reaction mechanisms in chemistry, the rate law is fundamental. It expresses the relationship between the rate of a chemical reaction and the concentration of reactants. The general form of the rate law is given by the expression \( \text{Rate} = k[A]^n \), where \( k \) is the rate constant, \([A]\) represents the concentration of the reactants, and \( n \) denotes the order of the reaction. The order of the reaction is an experimental quantity and tells how the concentration of reactants affects the rate.

Some critical points to consider about rate laws include:

• The value of \( n \) does not need to be an integer; it can also be a fraction.
• The rate constant \( k \) is specific to reactions at a particular temperature.
• Determining the rate law requires conducting experiments to observe changes in reaction rates as reagent concentrations change.

Understanding these basics helps in determining the order of the reaction when given changes in concentrations and reaction rates.

Reaction Kinetics

Reaction kinetics is the branch of chemistry that deals with understanding how and why reactions occur at certain rates. It focuses on the speed of chemical reactions and the influence of various factors on this speed. Reaction order is a crucial concept in kinetics, indicating how the rate is dependent on the concentration.

Factors affecting reaction kinetics include:

• Concentration of reactants - as seen in the rate law, changes in concentration can directly impact reaction speed.
• Temperature - often, increasing the temperature increases the reaction rate.
• Presence of a catalyst - catalysts act to lower the activation energy required for the reaction, usually speeding it up.
• The physical state of the reactants - it influences how easily molecules interact.

Understanding reaction kinetics involves a detailed study of these factors to accurately predict reaction behavior and adjust conditions to control reaction speeds effectively.

Logarithms in Chemistry

Logarithms are a powerful tool in chemistry, especially when dealing with reaction orders and rates. Taking logarithms of both sides of an equation simplifies a problem by converting multiplicative relationships into additive ones. This is particularly useful in the context of rate laws when determining the order of a reaction.

For instance, if you have an expression like \( 16^n = 2 \) and need to solve for \( n \), taking the logarithm of both sides simplifies the process:

• Using the logarithmic identity \( \log(a^b) = b\log(a)\), this converts \( 16^n = 2 \) into \( n \times \log(16) = \log(2)\).
• Substituting \( \log(16) = 4 \log(2) \) simplifies the expression to \( 4n = 1 \), leading to \( n = \frac{1}{4} \).
• This use of logarithms provides a straightforward method to solve exponential equations in chemistry.

Understanding how to apply logarithms in these contexts can greatly aid in analyzing complex reactions and uncovering their characteristics.