Ch.13 - Chemical KineticsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: A reaction has a rate constant of 0.0185 s -1 and the only reactant has an initial concentration of 0.135 M. What is the concentration in molarity of this reactant after 125 seconds?A. 4.84 x 10-5 B.

Problem

A reaction has a rate constant of 0.0185 s -1 and the only reactant has an initial concentration of 0.135 M. What is the concentration in molarity of this reactant after 125 seconds?

A. 4.84 x 10-5 

B. 0.0134

C. 0.103

D. 0.310

E. 4.84

 

Solution

We’re being asked to calculate the concentration of reactant remaining from 0.135 M after 125 seconds. The rate constant of the reaction is 0.0185 s–1.


We don’t know the order of the reaction but we can quickly figure it out from the rate constant. 


Recall that the unit for the rate constant is given by:


k=Mn-1·s-1


where n = order of the reaction

The given rate constant only has –1 as its units, which means n = 1 and the reaction follows first-order kinetics.


The integrated rate law for a first-order reaction is as follows:


ln[A]t=-kt+ln[A]0


where: 

[A]t = concentration at time t

k = rate constant

t = time

[A]0 = initial concentration


Solution BlurView Complete Written Solution