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 13.4, Problem 13.7P
Interpretation Introduction

Interpretation:

The equilibrium constant Kp and Kc of given reaction at given temperature has to beidentified.

Concept introduction:

Equilibrium constant(Kp):

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant (Kp) is defined as ratio of partial pressure of products to partial pressure of reactants. Each partial pressure 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.

aAbB

Rate of forward reaction = Rate of reverse reactionkfPAa=krPBa

On rearranging,

PAaPBb=kfkr=Kp

Where,

kf is the rate constant of the forward reaction.

kr is the rate constant of the reverse reaction.

Kp is the equilibrium constant.

The pressure of each species of ideal gas and its molar concentration are directly proportional to each other.

PAV = nARTPA = nAVRT=[A]RT

In same way PB =[B]RT. Thus, equilibrium constant Kp is given as

Kp=[B]b[A]a×(RT)b-aKp = Kc(RT)Δn

Where,

R is gas constant

T is absolute temperature

Δn is difference in moles of products to reactants in gas phase.

Blurred answer
Students have asked these similar questions
The following reaction can be used to make H2 for the synthesis of ammonia from the greenhouse gases carbon dioxide and methane: CH4 (g) + CO2 (g) ⇌ 2 CO (g) + 2 H2 (g) At 1200. K, Kp = 3.548 × 106. What is the partial pressure of CH4 at equilibrium when 10.00 atm of CH4 reacts with 15.00 atm of CO2 at 1200. K?
The atmosphere contains 78% N2(g), 21% O2(g), and traces of other gases. These gases react at high temperature to form nitrogen monoxide, an important pollutant. N2(g) + O2(g) = 2NO (g) For this reaction at a certain temperature, Kc = 3.8 x 10^-5. If the initial concentration of N2(g) is 0.78 mol/L, and O2(g) is 0.21 mol/L, what will be the equilibrium concentration of NO(g) at this temperature?
Ozone is formed from oxygen.         3 O 2(g) 2 O 3(g) Calculate the value of K p at a temperature of 1.0 K, given that K c = 10.0  at that same temperature

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: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
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
Chemical Equilibria and Reaction Quotients; Author: Professor Dave Explains;https://www.youtube.com/watch?v=1GiZzCzmO5Q;License: Standard YouTube License, CC-BY