Problem 32 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The rate law for a reaction between the substances \(\mathrm{A}\) and \(\mathrm{B}\) is given by rate \(=\) \(K[\mathrm{~A}]^{n}[\mathrm{~B}]^{m} .\) On doubling the concentration of \(\mathrm{A}\) and halving the concentration of \(\mathrm{B}\), the ratio of the new rate to the earlier rate of the reaction will be as (a) \(1 / 2^{m+n}\) (b) \((m+n)\) (c) \((n-m)\) (d) \(2^{(n-m)}\)

Step-by-Step Solution

Answer

The ratio of the new rate to the earlier rate of the reaction is \(2^{(n-m)}\).

Step 1: Understanding the rate law

The rate law for a reaction is an equation that relates the rate of the reaction to the concentrations of the reactants. For this reaction, the rate law is given as rate = k[A]^n[B]^m, where k is the rate constant, [A] and [B] are the concentrations of substances A and B, and n and m are the reaction orders with respect to A and B.

Step 2: Adjusting concentrations of A and B

We are told that the concentration of A is doubled, so [A] becomes 2[A]. The concentration of B is halved, so [B] becomes [B]/2.

Step 3: Writing the new rate equation

The new rate of the reaction after doubling [A] and halving [B] can be expressed as new rate = k(2[A])^n([B]/2)^m.

Step 4: Simplifying the new rate equation

Simplify the new rate equation by distributing the exponents: new rate = k(2^n[A]^n)(2^{-m}[B]^m) = k2^{n-m}[A]^n[B]^m.

Step 5: Comparing the new rate with the original rate

To find the ratio of the new rate to the original rate, we need to divide the new rate by the original rate: \(\frac{\text{new rate}}{\text{original rate}} = \frac{k2^{n-m}[A]^n[B]^m}{k[A]^n[B]^m} = 2^{n-m}\).

Key Concepts

Chemical Kinetics

Chemical kinetics is the study of the speed or rate at which chemical reactions occur and the factors that affect these rates. Understanding kinetics is crucial as it helps chemists control reactions, whether to speed them up or slow them down for various industrial or laboratory processes.

Factors that influence the rate of a chemical reaction include the concentration of reactants, temperature, presence of a catalyst, and surface area of reactants. By analyzing how these factors alter reaction rates, chemists can deduce the mechanism of the reaction—how reactants actually transform into products at the molecular level. This insight can be vital for developing new chemical processes, ensuring safety in industrial settings, and in many other applications of chemistry.

Reaction Order

The reaction order is a key concept in kinetics that indicates the power law dependence of the rate on the concentration of each reactant. In the rate law expression rate = k[A]ⁿ[B]^m, the exponents 'n' and 'm' represent the order of the reaction with respect to reactants A and B.

For instance, if 'n' is 1, the reaction is said to be first-order with respect to A; if it's 2, the reaction is second-order in A, and so on. It's worth noting that reaction order is not necessarily related to the stoichiometry of the reaction; it must be determined experimentally. The overall order of the reaction is the sum of the orders with respect to each reactant, in this case, n+m. Reaction order plays a pivotal role in predicting how changes in concentration affect the reaction rate.

Concentration Effect on Reaction Rate

The concentration of reactants is a vital factor in chemical kinetics, as it can significantly impact the rate at which a chemical reaction proceeds. A basic principle is that an increase in the concentration of a reactant will typically lead to an increase in the reaction rate, which is intuitive as more reactant molecules increase the likelihood of successful collisions leading to product formation.

For the rate law given as rate = k[A]ⁿ[B]^m, changing the concentration of A or B will alter the rate according to their respective orders in the reaction (referred to as 'n' for A and 'm' for B). Therefore, doubling the concentration of A, if the reaction is first-order in A, will double the rate. However, if it's second-order in A, the rate will quadruple. Conversely, halving the concentration of B will decrease the rate, but by how much depends on the order 'm'. This effect is crucial for controlling chemical reactions in various practical applications, like industrial synthesis and pharmacokinetics.