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: Consider this initial-rate data at a certain temperature for the reaction described by2 NOBr (g) => 2 NO (g) + Br2 (g)Determine the value and units of the rate constant k =?

Solution: Consider this initial-rate data at a certain temperature for the reaction described by2 NOBr (g) => 2 NO (g) + Br2 (g)Determine the value and units of the rate constant k =?

Problem

Consider this initial-rate data at a certain temperature for the reaction described by

2 NOBr (g) => 2 NO (g) + Br(g)

Determine the value and units of the rate constant k =?

Solution

The rate law can be expressed by:

$r=k[reactant]^n$

From the reaction given:

$2\ NOBr_{(g)} \rightarrow2NO_{(g)} +Br_{2(g)}$

The rate law then becomes:

$r = k [2\ NOBr]^n$

k will be the rate constant. n is the order of the reaction. Before solving for k, we must solve for n.

Substitute the values $1.01\times10^2$ for r and 0.600 for reactant in the first rate expression:

View the complete written solution...