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

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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
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Kinetics: Initial Rates and Integrated Rate Laws; Author: Professor Dave Explains;https://www.youtube.com/watch?v=wYqQCojggyM;License: Standard YouTube License, CC-BY