A reaction is first order in \(A\) and second order in \(B\) : (i) Write differential rate equation. (ii) How is the rate affected when the concentration of \(B\) is tripled? (iii) How is the rate affected when the concentration of both \(A\) and \(B\) is doubled?

Short Answer

Expert verified
The rate law is Rate = k[A][B]^2. Tripling [B] increases the rate by a factor of 9. Doubling both [A] and [B] increases the rate by a factor of 8.

Step by step solution

01

Writing the Differential Rate Equation

For a reaction that is first order in reactant A and second order in reactant B, the rate of reaction can be expressed by the rate law equation: Rate = k[A]^1[B]^2. Here, k is the rate constant, [A] is the concentration of reactant A, and [B] is the concentration of reactant B.
02

Effect of Tripling the Concentration of B

If the concentration of B is tripled, the new rate can be found by plugging into the rate equation (Rate' = k[A]^1[3B]^2). The term [3B]^2 means (3^2)[B]^2 which is equal to 9[B]^2. Therefore, tripling the concentration of B will increase the rate of the reaction by a factor of 9.
03

Effect of Doubling the Concentration of A and B

Doubling the concentration of both A and B means we need to calculate the new rate (Rate'' = k[2A]^1[2B]^2). The term [2A]^1 remains 2[A] and [2B]^2 becomes 4[B]^2. Multiplying these changes gives us a new rate that is 8 times the original rate (2 * 4 = 8). Therefore, doubling the concentration of A and B increases the rate of reaction by a factor of 8.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Rate Equation
Understanding the differential rate equation is crucial for analyzing the kinetics of a chemical reaction. Essentially, it relates the rate of change in concentration of reactants over time. The equation is tailored to the reaction order with respect to each reactant.

In our exercise, given a reaction first order in reactant A and second order in reactant B, the differential rate equation is formulated as: \[ \text{Rate} = k[A]^1[B]^2 \].

This equation reveals that the rate is directly proportional to the concentration of A and the square of the concentration of B. The powers of the concentrations ([A] and [B]) are indicative of the reaction order, which in turn defines the relationship between the concentration of reactants and the rate of the reaction. The presence of the rate constant 'k' in the equation denotes that this proportionality is unique for every reaction under specific conditions.
Concentration Effect on Reaction Rate
The rate of a chemical reaction is heavily influenced by the concentrations of its reactants. According to the rate law, modifying the concentration of one or more reactants will change the reaction rate.

For a reaction like the one in our exercise, where the order with respect to B is second, tripling the concentration of B would exponentially affect the rate. To visualize, if [B] is tripled: \[ \text{Rate}' = k[A][3B]^2 \], this simplifies to \[ \text{Rate}' = k[A](9[B]^2) \]. Consequently, the rate becomes nine times faster, demonstrating a squared dependency on the concentration of B, due to its second-order nature.

This concept is an indispensable part of chemical kinetics, as it allows chemists to control and optimize reaction rates for industrial and laboratory processes.
Rate Constant
The rate constant 'k' in the rate equation is a crucial factor that determines the speed of a chemical reaction at a given temperature. While reactant concentrations can vary, the rate constant remains fixed under constant temperature and pressure, characteristic of the specific reaction.

In our given problem: \[ \text{Rate} = k[A]^1[B]^2 \], 'k' is unaffected by changes in concentration levels but is sensitive to temperature changes. The magnitude of 'k' reflects the rate at which reactants transform into products, and different reactions have different rate constants. Knowing 'k' and the reaction order allows chemists to predict how changes in concentration or conditions will affect the reaction rate.

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Most popular questions from this chapter

In a reaction; \(2 A \longrightarrow\) Products, the concentration of \(A\) decreases from \(0.5\) mol litre \(^{-1}\) to \(0.4\) mol litre \(^{-1}\) in 10 minutes. Calculate rate during this interval.

A chemical reaction \(2 A \longrightarrow 4 B+C ;\) in gaseous phase shows an increase in concentration of \(B\) by \(5 \times 10^{-3} M\) in 10 second. Calculate: (a) rate of appearance of \(B\), (b) rate of the reaction, (c) rate of disappearance of \(A\).

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) takes place according to I order as $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} $$ Calculate : (a) The rate constant, if instantaneous rate is \(1.4 \times 10^{-6}\) mol litre \(^{-1}\) \(\mathrm{sec}^{-1}\) when concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.04 \mathrm{M}\) (b) The rate of reaction when concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(1.20 \mathrm{M}\) (c) The concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) when the rate of reaction will be \(2.45 \times 10^{-5} \mathrm{~mol}\) litre \(^{-1} \mathrm{sec}^{-\mathrm{i}}\).

The reaction; \(2 \mathrm{NO}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{NOBr}\), is supposed to follow the following mechanism, (i) \(\mathrm{NO}+\mathrm{Br}_{2} \stackrel{\text { fast }}{\longrightarrow} \mathrm{NOBr}_{2}\) (ii) \(\mathrm{NOBr}_{2}+\mathrm{NO} \stackrel{\text { slow }}{\longrightarrow} 2 \mathrm{NOBr}\) Suggest the rate law expression.

The reaction; \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\), shows an increase in concentration of \(\mathrm{NO}_{2}\) by \(20 \times 10^{-3}\) mol litre \(^{-1}\) in 5 second. Calculate (a) rate of appearance of \(\mathrm{NO}_{2}\), (b) rate of reaction and (c) rate of disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}\).

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