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: Dinitrogen pentoxide (N2O5) decomposes to NO2 and O2 at relatively low temperatures in the following reaction: 2N2O5(aq) → 4NO2(aq) + O2(g) This reaction is carried out in a CCl 4 solution at 45°C. The data are described in the table as well as the plots of the concentration versus t, ln concentration versus t, and 1/concentration versus t and results are below. Determine the rate law and calculate the rate constant. a. Zero order/ 5.3 x10-5 s -1     b. Second order/ 4.8 x 10−4 s −1 M-1      c. First order/5.3 x 10-5 s -1 M-1 d. Second order/ 2.9 x 10−4 s −1 M-1               e. First order/4.8 x 10−4 s −1

Solution: Dinitrogen pentoxide (N2O5) decomposes to NO2 and O2 at relatively low temperatures in the following reaction: 2N2O5(aq) → 4NO2(aq) + O2(g) This reaction is carried out in a CCl 4 solution at 45°C.

Problem

Dinitrogen pentoxide (N2O5) decomposes to NO2 and O2 at relatively low temperatures in the following reaction:

2N2O5(aq) → 4NO2(aq) + O2(g)

This reaction is carried out in a CCl 4 solution at 45°C. The data are described in the table as well as the plots of the concentration versus t, ln concentration versus t, and 1/concentration versus t and results are below. Determine the rate law and calculate the rate constant.

a. Zero order/ 5.3 x10-5 s -1     b. Second order/ 4.8 x 10−4 s −1 M-1      c. First order/5.3 x 10-5 s -1 M-1

d. Second order/ 2.9 x 10−4 s −1 M-1               e. First order/4.8 x 10−4 s −1