Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 11, Problem 59QRT

(a)

Interpretation Introduction

Interpretation:

The rate law has to be written for the given elementary reaction.

Concept Introduction:

For elementary reactions, rate law can be written using stoichiometry.  The order of unimolecular reaction is one and bimolecular reaction is two.  In a bimolecular reaction if the reactants are different then order with respect to each reactant will be one in order to make the overall order two.

(b)

Interpretation Introduction

Interpretation:

The rate law has to be written for the given elementary reaction.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The rate law has to be written for the given elementary reaction.

Concept Introduction:

Refer to part (a).

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Students have asked these similar questions
Consider the following reaction: 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g) (a) The rate law for this reaction is second order in NO(g) and first-order in H2(g). What is the rate law for this reaction? (b) If the rate constant for this reaction at a certain temperature is 9.70e+04, what is the reaction rate when [NO(g)] = 0.0560 M and [H2(g)] = 0.119 M?Rate =  M/s.(c) What is the reaction rate when the concentration of NO(g) is doubled, to 0.112 M while the concentration of H2(g) is 0.119 M?Rate =  M/s
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Consider the following reaction: 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g) (a) The rate law for this reaction is second order in NO(g) and first order in H2(g). What is the rate law for this reaction?(b) If the rate constant for this reaction at a certain temperature is 79200, what is the reaction rate when [NO(g)] = 0.0852 M and [H2(g)] = 0.137 M?Rate =____  M/s.(c) What is the reaction rate when the concentration of NO(g) is doubled, to 0.170 M while the concentration of H2(g) is 0.137 M?Rate = ____ M/s

Chapter 11 Solutions

Chemistry: The Molecular Science

Ch. 11.3 - Prob. 11.6PSPCh. 11.3 - Prob. 11.7PSPCh. 11.4 - Prob. 11.6ECh. 11.4 - Prob. 11.7CECh. 11.4 - Prob. 11.8PSPCh. 11.4 - Prob. 11.8CECh. 11.5 - Prob. 11.9PSPCh. 11.5 - The frequency factor A is 6.31 108 L mol1 s1 and...Ch. 11.6 - Prob. 11.10CECh. 11.7 - Prob. 11.11ECh. 11.7 - The Raschig reaction produces the industrially...Ch. 11.7 - Prob. 11.12ECh. 11.8 - The oxidation of thallium(I) ion by cerium(IV) ion...Ch. 11.9 - Prob. 11.11PSPCh. 11.9 - Prob. 11.14CECh. 11 - An excellent way to make highly pure nickel metal...Ch. 11 - Prob. 1QRTCh. 11 - Prob. 2QRTCh. 11 - Prob. 3QRTCh. 11 - Prob. 4QRTCh. 11 - Prob. 5QRTCh. 11 - Prob. 6QRTCh. 11 - Prob. 7QRTCh. 11 - Prob. 8QRTCh. 11 - Prob. 9QRTCh. 11 - Prob. 10QRTCh. 11 - Prob. 11QRTCh. 11 - Cyclobutane can decompose to form ethylene: The...Ch. 11 - Prob. 13QRTCh. 11 - Prob. 14QRTCh. 11 - For the reaction 2NO2(g)2NO(g)+O2(g) make...Ch. 11 - Prob. 16QRTCh. 11 - Prob. 17QRTCh. 11 - Ammonia is produced by the reaction between...Ch. 11 - Prob. 19QRTCh. 11 - Prob. 20QRTCh. 11 - The reaction of CO(g) + NO2(g) is second-order in...Ch. 11 - Nitrosyl bromide, NOBr, is formed from NO and Br2....Ch. 11 - Prob. 23QRTCh. 11 - Prob. 24QRTCh. 11 - Prob. 25QRTCh. 11 - For the reaction these data were obtained at 1100...Ch. 11 - Prob. 27QRTCh. 11 - Prob. 28QRTCh. 11 - Prob. 29QRTCh. 11 - Prob. 30QRTCh. 11 - Prob. 31QRTCh. 11 - Prob. 32QRTCh. 11 - For the reaction of phenyl acetate with water the...Ch. 11 - When phenacyl bromide and pyridine are both...Ch. 11 - The compound p-methoxybenzonitrile N-oxide, which...Ch. 11 - Prob. 36QRTCh. 11 - Radioactive gold-198 is used in the diagnosis of...Ch. 11 - Prob. 38QRTCh. 11 - Prob. 39QRTCh. 11 - Prob. 40QRTCh. 11 - Prob. 41QRTCh. 11 - Prob. 42QRTCh. 11 - Prob. 43QRTCh. 11 - Prob. 44QRTCh. 11 - Prob. 45QRTCh. 11 - Prob. 46QRTCh. 11 - Prob. 47QRTCh. 11 - Prob. 48QRTCh. 11 - Prob. 49QRTCh. 11 - Prob. 50QRTCh. 11 - Prob. 51QRTCh. 11 - Prob. 52QRTCh. 11 - For the reaction of iodine atoms with hydrogen...Ch. 11 - Prob. 54QRTCh. 11 - The activation energy Ea is 139.7 kJ mol1 for the...Ch. 11 - Prob. 56QRTCh. 11 - Prob. 57QRTCh. 11 - Prob. 58QRTCh. 11 - Prob. 59QRTCh. 11 - Prob. 60QRTCh. 11 - Prob. 61QRTCh. 11 - Prob. 62QRTCh. 11 - Prob. 63QRTCh. 11 - Which of the reactions in Question 62 would (a)...Ch. 11 - Prob. 65QRTCh. 11 - Prob. 66QRTCh. 11 - Prob. 67QRTCh. 11 - Prob. 68QRTCh. 11 - Prob. 69QRTCh. 11 - Prob. 70QRTCh. 11 - Prob. 71QRTCh. 11 - For the reaction the rate law is Rate=k[(CH3)3CBr]...Ch. 11 - Prob. 73QRTCh. 11 - Prob. 74QRTCh. 11 - Prob. 75QRTCh. 11 - For this reaction mechanism, write the chemical...Ch. 11 - Prob. 77QRTCh. 11 - Prob. 78QRTCh. 11 - Prob. 79QRTCh. 11 - When enzymes are present at very low...Ch. 11 - Prob. 81QRTCh. 11 - The reaction is catalyzed by the enzyme succinate...Ch. 11 - Prob. 83QRTCh. 11 - Many biochemical reactions are catalyzed by acids....Ch. 11 - Prob. 85QRTCh. 11 - Prob. 86QRTCh. 11 - Prob. 87QRTCh. 11 - Prob. 88QRTCh. 11 - Prob. 89QRTCh. 11 - Prob. 90QRTCh. 11 - Prob. 91QRTCh. 11 - Prob. 92QRTCh. 11 - Prob. 93QRTCh. 11 - Prob. 94QRTCh. 11 - Nitryl fluoride is an explosive compound that can...Ch. 11 - Prob. 96QRTCh. 11 - Prob. 97QRTCh. 11 - For a reaction involving the decomposition of a...Ch. 11 - Prob. 99QRTCh. 11 - Prob. 100QRTCh. 11 - Prob. 101QRTCh. 11 - This graph shows the change in concentration as a...Ch. 11 - Prob. 103QRTCh. 11 - Prob. 104QRTCh. 11 - Prob. 105QRTCh. 11 - Prob. 106QRTCh. 11 - Prob. 107QRTCh. 11 - Prob. 108QRTCh. 11 - Prob. 109QRTCh. 11 - Prob. 110QRTCh. 11 - Prob. 111QRTCh. 11 - Prob. 112QRTCh. 11 - Prob. 113QRTCh. 11 - Prob. 114QRTCh. 11 - Prob. 115QRTCh. 11 - Prob. 116QRTCh. 11 - Prob. 118QRTCh. 11 - Prob. 119QRTCh. 11 - In a time-resolved picosecond spectroscopy...Ch. 11 - If you know some calculus, derive the integrated...Ch. 11 - If you know some calculus, derive the integrated...Ch. 11 - (Section 11-5) A rule of thumb is that for a...Ch. 11 - Prob. 11.BCPCh. 11 - Prob. 11.CCPCh. 11 - Prob. 11.DCP
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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