General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
bartleby

Videos

Question
Book Icon
Chapter 14, Problem 14.23QP

(a)

Interpretation Introduction

Interpretation:

The concentration of NaOBr after 22s from the given information has to be calculated.

Concept introduction:

Rate equation for the general reaction A+BProduct is,

Rate=krateconstat[A][B]

The rate of the reaction is proportinal to the concentration of A to the power of x, is [A]x

The rate of the reaction is proportional to the concentration of B to the power of y is [B]y

Then the rate equation becomes,

Rate=k[A]x[B]y

Order of this reaction is the sum of the powers to which all reactant concentrations appearing in the rate law are  raised.

Orderx+y

(a)

Expert Solution
Check Mark

Answer to Problem 14.23QP

The concentration of NaOBr after 22s from the given information is 0.034 M

Explanation of Solution

The given reaction is

2NaOBr(g)2NO(g)+Br2(g)

The reaction follows second order kinetics.

Rate constant of the given reaction is k=0.80M1s1

The concentration of NaOBr after 22s can be determined as follows,

For a second order reaction the relationship between concentrations of reactant and time is,

1[NaOBr]t=kt+1[NaOBr]01[NaOBr]t=(0.80/Ms)(22s)+10.086M1[NaOBr]t=29M1[NOBr] = 0.034 M

Therefore, the concentration of NaOBr after 22s is 0.034 M

(b)

Interpretation Introduction

Interpretation:

The half-life when [NaOBr]0=0.072Mand [NaOBr]0=0.054M has to be calculated.

Concept introduction:

Rate equation for the general reaction A+BProduct is,

Rate=krateconstat[A][B]

The rate of the reaction is proportinal to the concentration of A to the power of x, is [A]x

The rate of the reaction is proportional to the concentration of B to the power of y is [B]y

Then the rate equation becomes,

Rate=k[A]x[B]y

Order of this reaction is the sum of the powers to which all reactant concentrations appearing in the rate law are  raised.

Orderx+y

Half-life is the time required for one half of a reactant to react.

Half-life for a second order reaction is

t12=1k[A]0where[A]0istheinitialconcentrationofreactantA

(b)

Expert Solution
Check Mark

Answer to Problem 14.23QP

The half-life when [NaOBr]0=0.072Mand [NaOBr]0=0.054M is 23s

Explanation of Solution

The given reaction is

2NaOBr(g)2NO(g)+Br2(g)

The reaction follows second order kinetics.

Rate constant of the given reaction is k=0.80M1s1

The concentration of NaOBr after 22s can be determined as follows,

We know that for a second order half-life reaction and half-life of a second order reaction is dependent on initial concentration

t12=1k[A]0t12=1(0.80/Ms)(0.072M)t12=17 s

The half-life when [NaOBr]0=0.054M can be determined as follows,

t12=1k[A]0t12=1(0.80/Ms)(0.054M)t12=23 s

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Bookmarks Profiles Tab Window Help Chemical Formula - Aktiv Che X + → C 11 a app.aktiv.com Google Chrome isn't your default browser Set as default Question 12 of 16 Q Fri Feb 2 Verify it's you New Chrome availabl- Write the balanced molecular chemical equation for the reaction in aqueous solution for mercury(I) nitrate and chromium(VI) sulfate. If no reaction occurs, simply write only NR. Be sure to include the proper phases for all species within the reaction. 3 Hg(NO3)2(aq) + Cг2(SO4)3(aq) → 3 Hg₂SO (s) + 2 Cr(NO3), (aq) ean Ui mate co ence an climate bility inc ulnerabili women, main critic CLIMATE-INI ernational + 10 O 2 W FEB 1 + 4- 3- 2- 2 2 ( 3 4 NS 28 2 ty 56 + 2+ 3+ 4+ 7 8 9 0 5 (s) (1) Ch O 8 9 (g) (aq) Hg NR CI Cr x H₂O A 80 Q A DII A F2 F3 FA F5 F6 F7 F8 F9 #3 EA $ do 50 % 6 CO & 7 E R T Y U 8 ( 9 0 F10 34 F11 川 F12 Subr + delete 0 { P }
Deducing the reactants of a Diels-Alder reaction n the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? ? Δ • If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. • If your answer is no, check the box under the drawing area instead. Explanation Check Click and drag to start drawing a structure. >
Predict the major products of the following organic reaction: + Some important notes: A ? • Draw the major product, or products, of the reaction in the drawing area below. • If there aren't any products, because no reaction will take place, check the box below the drawing area instead. • Be sure to use wedge and dash bonds when necessary, for example to distinguish between major products that are enantiomers. Explanation Check Click and drag to start drawing a structure.

Chapter 14 Solutions

General Chemistry

Ch. 14.4 - Practice Exercise The first-order rate constant...Ch. 14.4 - Review of Concepts (a) What can you deduce about...Ch. 14.5 - Practice Exercise The reaction between NO2 and CO...Ch. 14.5 - Prob. 1RCCh. 14.6 - Prob. 1RCCh. 14 - Prob. 14.1QPCh. 14 - 15.2 Explain the difference between physical...Ch. 14 - Prob. 14.3QPCh. 14 - Prob. 14.4QPCh. 14 - Prob. 14.5QPCh. 14 - 14.6 Consider the reaction Suppose that at a...Ch. 14 - Prob. 14.7QPCh. 14 - 14.8 What are the units for the rate constants of...Ch. 14 - Prob. 14.9QPCh. 14 - Prob. 14.10QPCh. 14 - Prob. 14.11QPCh. 14 - Prob. 14.13QPCh. 14 - Prob. 14.14QPCh. 14 - Prob. 14.15QPCh. 14 - Prob. 14.16QPCh. 14 - Prob. 14.17QPCh. 14 - Prob. 14.18QPCh. 14 - Prob. 14.19QPCh. 14 - Prob. 14.20QPCh. 14 - 14.21 What is the half-life of a compound if 75...Ch. 14 - 14.22 The thermal decomposition of phosphine (PH3)...Ch. 14 - Prob. 14.23QPCh. 14 - Prob. 14.24QPCh. 14 - 14.25 Consider the first-order reaction A → B...Ch. 14 - Prob. 14.26QPCh. 14 - 14.27 Define activation energy. What role does...Ch. 14 - Prob. 14.28QPCh. 14 - Prob. 14.29QPCh. 14 - 14.30 As we know, methane burns readily in oxygen...Ch. 14 - Prob. 14.31QPCh. 14 - Prob. 14.32QPCh. 14 - Prob. 14.33QPCh. 14 - Prob. 14.34QPCh. 14 - Prob. 14.35QPCh. 14 - Prob. 14.36QPCh. 14 - Prob. 14.37QPCh. 14 - 14.38 The rate at which tree crickets chirp is 2.0...Ch. 14 - 14.39 The diagram here describes the initial state...Ch. 14 - Prob. 14.40QPCh. 14 - Prob. 14.41QPCh. 14 - Prob. 14.42QPCh. 14 - 14.43 Explain why termolecular reactions are...Ch. 14 - 14.44 What is the rate-determining step of a...Ch. 14 - Prob. 14.45QPCh. 14 - Prob. 14.46QPCh. 14 - Prob. 14.47QPCh. 14 - Prob. 14.48QPCh. 14 - Prob. 14.49QPCh. 14 - Prob. 14.50QPCh. 14 - Prob. 14.51QPCh. 14 - Prob. 14.52QPCh. 14 - Prob. 14.53QPCh. 14 - Prob. 14.54QPCh. 14 - Prob. 14.55QPCh. 14 - Prob. 14.56QPCh. 14 - Prob. 14.57QPCh. 14 - Prob. 14.58QPCh. 14 - Prob. 14.59QPCh. 14 - Prob. 14.60QPCh. 14 - Prob. 14.61QPCh. 14 - Prob. 14.62QPCh. 14 - Prob. 14.63QPCh. 14 - Prob. 14.64QPCh. 14 - Prob. 14.65QPCh. 14 - 14.66 The decomposition of N2O to N2 and O2 is a...Ch. 14 - Prob. 14.67QPCh. 14 - Prob. 14.68QPCh. 14 - 14.69 Consider the zero-order reaction a → B....Ch. 14 - Prob. 14.70QPCh. 14 - Prob. 14.72QPCh. 14 - Prob. 14.73QPCh. 14 - Prob. 14.74QPCh. 14 - Prob. 14.75QPCh. 14 - Prob. 14.76QPCh. 14 - Prob. 14.77QPCh. 14 - Prob. 14.78QPCh. 14 - Prob. 14.79QPCh. 14 - Prob. 14.80QPCh. 14 - Prob. 14.81QPCh. 14 - Prob. 14.82QPCh. 14 - 14.83 When a mixture of methane and bromine is...Ch. 14 - 14.84 Consider this elementary step: (a)...Ch. 14 - Prob. 14.85QPCh. 14 - Prob. 14.86QPCh. 14 - 14.87 In recent years ozone in the stratosphere...Ch. 14 - Prob. 14.88QPCh. 14 - Prob. 14.90QPCh. 14 - Prob. 14.91QPCh. 14 - Prob. 14.92QPCh. 14 - Prob. 14.93QPCh. 14 - Prob. 14.94QPCh. 14 - Prob. 14.95QPCh. 14 - Prob. 14.96QPCh. 14 - Prob. 14.97QPCh. 14 - Prob. 14.98QPCh. 14 - Prob. 14.100QPCh. 14 - Prob. 14.101QPCh. 14 - 14.102 Consider the potential energy profiles...Ch. 14 - Prob. 14.103QPCh. 14 - Prob. 14.104QPCh. 14 - 14.105 The activation energy for the...Ch. 14 - Prob. 14.106QPCh. 14 - Prob. 14.107SPCh. 14 - Prob. 14.108SPCh. 14 - Prob. 14.109SPCh. 14 - Prob. 14.110SPCh. 14 - Prob. 14.111SPCh. 14 - Prob. 14.112SPCh. 14 - Prob. 14.113SPCh. 14 - Prob. 14.114SPCh. 14 - Prob. 14.115SPCh. 14 - 14.116 To prevent brain damage, a drastic medical...Ch. 14 - Prob. 14.117SPCh. 14 - Prob. 14.118SP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Kinetics: Initial Rates and Integrated Rate Laws; Author: Professor Dave Explains;https://www.youtube.com/watch?v=wYqQCojggyM;License: Standard YouTube License, CC-BY