General Chemistry: Atoms First
General Chemistry: Atoms First
2nd Edition
ISBN: 9780321809261
Author: John E. McMurry, Robert C. Fay
Publisher: Prentice Hall
bartleby

Videos

Question
Book Icon
Chapter 13, Problem 13.38SP

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant expression (Kc) for given reaction has to bewritten.

Concept introduction:

Equilibrium constant(Kc):

Equilibrium constant (Kc) is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

AB

Rate of forward reaction = Rate of reverse reactionkf[A]a=kr[B]b

On rearranging,

[A]a[B]b=kfkr=Kc

Where,

kf is the rate constant of the forward reaction.

kr is the rate constant of the reverse reaction.

Kc is the equilibrium constant.

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant expression (Kc) for given reaction has to be written.

Concept introduction:

Equilibrium constant(Kc):

Equilibrium constant (Kc) is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

AB

Rate of forward reaction = Rate of reverse reactionkf[A]a=kr[B]b

On rearranging,

[A]a[B]b=kfkr=Kc

Where,

kf is the rate constant of the forward reaction.

kr is the rate constant of the reverse reaction.

Kc is the equilibrium constant.

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant expression (Kc) for given reaction has to be written.

Concept introduction:

Equilibrium constant(Kc):

Equilibrium constant (Kc) is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

AB

Rate of forward reaction = Rate of reverse reactionkf[A]a=kr[B]b

On rearranging,

[A]a[B]b=kfkr=Kc

Where,

kf is the rate constant of the forward reaction.

kr is the rate constant of the reverse reaction.

Kc is the equilibrium constant.

Blurred answer
Students have asked these similar questions
Show work with explanation needed. Don't give Ai generated solution
7. Calculate the following for a 1.50 M Ca(OH)2 solution. a. The concentration of hydroxide, [OH-] b. The concentration of hydronium, [H3O+] c. The pOH d. The pH
A first order reaction is 46.0% complete at the end of 59.0 minutes. What is the value of k? What is the half-life for this reaction? HOW DO WE GET THERE? The integrated rate law will be used to determine the value of k. In [A] [A]。 = = -kt What is the value of [A] [A]。 when the reaction is 46.0% complete?

Chapter 13 Solutions

General Chemistry: Atoms First

Ch. 13.6 - Prob. 13.11CPCh. 13.6 - Prob. 13.12PCh. 13.6 - Prob. 13.13PCh. 13.6 - Prob. 13.14PCh. 13.6 - Prob. 13.15PCh. 13.6 - Prob. 13.16PCh. 13.8 - Prob. 13.17PCh. 13.9 - Prob. 13.18PCh. 13.9 - Prob. 13.19CPCh. 13.10 - Prob. 13.20PCh. 13.10 - Prob. 13.21PCh. 13.10 - Prob. 13.22CPCh. 13.11 - Prob. 13.23PCh. 13.11 - Prob. 13.24PCh. 13.11 - Prob. 13.25PCh. 13 - Consider the interconversion of A molecules (red...Ch. 13 - Prob. 13.27CPCh. 13 - Prob. 13.28CPCh. 13 - Prob. 13.29CPCh. 13 - Prob. 13.30CPCh. 13 - Prob. 13.31CPCh. 13 - Prob. 13.32CPCh. 13 - Prob. 13.33CPCh. 13 - Prob. 13.34CPCh. 13 - Prob. 13.35CPCh. 13 - Prob. 13.36CPCh. 13 - The following pictures represent the initial and...Ch. 13 - Prob. 13.38SPCh. 13 - Prob. 13.39SPCh. 13 - Prob. 13.40SPCh. 13 - Prob. 13.41SPCh. 13 - Prob. 13.42SPCh. 13 - Prob. 13.43SPCh. 13 - Prob. 13.44SPCh. 13 - Prob. 13.45SPCh. 13 - Prob. 13.46SPCh. 13 - Prob. 13.47SPCh. 13 - Prob. 13.48SPCh. 13 - Prob. 13.49SPCh. 13 - Prob. 13.50SPCh. 13 - Prob. 13.51SPCh. 13 - Prob. 13.52SPCh. 13 - Prob. 13.53SPCh. 13 - Prob. 13.54SPCh. 13 - Prob. 13.55SPCh. 13 - Prob. 13.56SPCh. 13 - Prob. 13.57SPCh. 13 - Prob. 13.58SPCh. 13 - Prob. 13.59SPCh. 13 - Prob. 13.60SPCh. 13 - Prob. 13.61SPCh. 13 - Prob. 13.62SPCh. 13 - Prob. 13.63SPCh. 13 - Prob. 13.64SPCh. 13 - Prob. 13.65SPCh. 13 - Prob. 13.66SPCh. 13 - Prob. 13.67SPCh. 13 - Prob. 13.68SPCh. 13 - Prob. 13.69SPCh. 13 - Prob. 13.70SPCh. 13 - Prob. 13.71SPCh. 13 - Prob. 13.72SPCh. 13 - Prob. 13.73SPCh. 13 - Gaseous indium dihydride is formed from the...Ch. 13 - Prob. 13.75SPCh. 13 - Prob. 13.76SPCh. 13 - Prob. 13.77SPCh. 13 - Prob. 13.78SPCh. 13 - Prob. 13.79SPCh. 13 - Prob. 13.80SPCh. 13 - Prob. 13.81SPCh. 13 - The value of Kc for the reaction of acetic acid...Ch. 13 - In a basic aqueous solution, chloromethane...Ch. 13 - Prob. 13.84SPCh. 13 - Prob. 13.85SPCh. 13 - Prob. 13.86SPCh. 13 - Prob. 13.87SPCh. 13 - Prob. 13.88SPCh. 13 - Prob. 13.89SPCh. 13 - Prob. 13.90SPCh. 13 - Prob. 13.91SPCh. 13 - Prob. 13.92SPCh. 13 - Consider the endothermic reaction Fe3+ (aq) + Cl...Ch. 13 - Prob. 13.94SPCh. 13 - Prob. 13.95SPCh. 13 - Prob. 13.96SPCh. 13 - Prob. 13.97SPCh. 13 - Prob. 13.98CHPCh. 13 - Prob. 13.99CHPCh. 13 - Prob. 13.100CHPCh. 13 - Prob. 13.101CHPCh. 13 - Prob. 13.102CHPCh. 13 - Prob. 13.103CHPCh. 13 - Prob. 13.104CHPCh. 13 - Prob. 13.105CHPCh. 13 - Refining petroleum involves cracking large...Ch. 13 - Prob. 13.107CHPCh. 13 - Prob. 13.108CHPCh. 13 - Prob. 13.109CHPCh. 13 - Prob. 13.110CHPCh. 13 - At 1000 K, Kp = 2.1 106 and H = 107.7 kJ for the...Ch. 13 - Consider the gas-phase decomposition of NOBr: 2...Ch. 13 - At 100C, Kc = 4.72 for the reaction 2 NO2(g) ...Ch. 13 - Prob. 13.114CHPCh. 13 - Prob. 13.115CHPCh. 13 - Prob. 13.116CHPCh. 13 - Prob. 13.117CHPCh. 13 - Prob. 13.118CHPCh. 13 - Prob. 13.119CHPCh. 13 - Prob. 13.120CHPCh. 13 - Prob. 13.121CHPCh. 13 - Prob. 13.122CHPCh. 13 - Prob. 13.123CHPCh. 13 - Prob. 13.124CHPCh. 13 - Prob. 13.125MPCh. 13 - Prob. 13.126MPCh. 13 - The equilibrium constant Kc for the gas-phase...Ch. 13 - Prob. 13.128MPCh. 13 - Prob. 13.129MPCh. 13 - Prob. 13.130MPCh. 13 - Prob. 13.131MPCh. 13 - Prob. 13.132MPCh. 13 - Consider the sublimation of mothballs at 27C in a...Ch. 13 - Prob. 13.134MPCh. 13 - Prob. 13.135MPCh. 13 - For the decomposition reaction PCl5(g) PCl3(g) +...Ch. 13 - Prob. 13.137MP
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: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:OpenStax
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemical Equilibria and Reaction Quotients; Author: Professor Dave Explains;https://www.youtube.com/watch?v=1GiZzCzmO5Q;License: Standard YouTube License, CC-BY