   # A reaction has the following experimental rate equation: Rate = k [A] 2 [B]. If the concentration of A is doubled and the concentration of B is halved, what happens to the reaction rate? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 14, Problem 49GQ
Textbook Problem
64 views

## A reaction has the following experimental rate equation: Rate = k[A]2[B]. If the concentration of A is doubled and the concentration of B is halved, what happens to the reaction rate?

Interpretation Introduction

Interpretation:

The response of the reaction rate with respect to change in concentration of reactant present in the reaction should be determined using the given rate law.

Concept introduction:

Rate law: It is generally the rate equation that consists of the reaction rate with the concentration or the pressures of the reactants and constant parameters.

Rate constant: The rate constant for a chemical reaction is the proportionality term in the chemical reaction rate law which gives the relationship between the rate and the concentration of the reactant present in the chemical reaction.

Rate order: It is represented by the exponential term of the respective reactant present in the rate law and the overall order of the reaction is the sum of all the exponents of all reactants present in the chemical reaction. The order of the reaction is directly proportional to the concentration of the reactants.

### Explanation of Solution

Given:

Rate=k[A]2[B]

Rate=k[A]2[B]If concentration of A is doubled and cocentration of B is reduced by 12theresulting rate is as follows, Rate = 22×1

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