World of Chemistry
World of Chemistry
7th Edition
ISBN: 9780618562763
Author: Steven S. Zumdahl
Publisher: Houghton Mifflin College Div
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Chapter 17, Problem 67A
Interpretation Introduction

Interpretation:

The equilibrium constant of the given reaction have to be determined.

Concept Introduction:

Chemical equilibrium: Chemical equilibrium is a state of a chemical reaction when the rate of forward reaction and backward reaction is equal. The chemical reaction follows the law of mass action. According to the law of mass action, the rate of a chemical reaction is directly proportional to the product of active masses of the reactants.

   aA + bB cC + dD

   Rateoftheforwardreaction r f = k f [A] a [B] b

   Rateofthebackwardreaction r b = k b [C] c [D] d

   k f and k b aretherateconstantsofforwardandbackwardreactionrespectively

   Atequilibriumtherateoftheforwardreaction=therateofthebackwardreaction

   k f [A] a [B] b = k b [C] c [D] d

   k f k b = K eq = [C] c [D] d [A] a [B] b

   K eq istheequilibriumconstantofthereaction.

Expert Solution & Answer
Check Mark

Answer to Problem 67A

The equilibrium constant of the given reaction is 1.0.

Explanation of Solution

Data given: At initially [N2] = 5.0 M and [H2] = 4.0 M and at equilibrium [H2] = 1.0 M

The given chemical equation is-

  N2(g) + 3H2(g) 2NH3(g)

   Whent=0,[ N 2 ]=5.0Mand[ H 2 ]=4.0M

   Atequilibrium,[ N 2 ]=(5.0-x)M,[ H 2 ]=(4.0-3x)Mand[N H 3 ]=2xM

   Giventhat,[ H 2 ]=(4.0-3x)M=1.0M

   So,thevalueofx=1

   Atequilibrium,[ N 2 ]=4.0M,[ H 2 ]=1.0Mand[N H 3 ]=2.0M

   K eq = [N H 3 ] 2 [ N 2 ] [ H 2 ] 3

   = (2.0) 2 (4.0)× (1.0) 3 =1.0

Chapter 17 Solutions

World of Chemistry

Ch. 17.2 - Prob. 4RQCh. 17.2 - Prob. 5RQCh. 17.3 - Prob. 1RQCh. 17.3 - Prob. 2RQCh. 17.3 - Prob. 3RQCh. 17.3 - Prob. 4RQCh. 17.3 - Prob. 5RQCh. 17.3 - Prob. 6RQCh. 17.3 - Prob. 7RQCh. 17 - Prob. 1ACh. 17 - Prob. 2ACh. 17 - Prob. 3ACh. 17 - Prob. 4ACh. 17 - Prob. 5ACh. 17 - Prob. 6ACh. 17 - Prob. 7ACh. 17 - Prob. 8ACh. 17 - Prob. 9ACh. 17 - Prob. 10ACh. 17 - Prob. 11ACh. 17 - Prob. 12ACh. 17 - Prob. 13ACh. 17 - Prob. 14ACh. 17 - Prob. 15ACh. 17 - Prob. 16ACh. 17 - Prob. 17ACh. 17 - Prob. 18ACh. 17 - Prob. 19ACh. 17 - Prob. 20ACh. 17 - Prob. 21ACh. 17 - Prob. 22ACh. 17 - Prob. 23ACh. 17 - Prob. 24ACh. 17 - Prob. 25ACh. 17 - Prob. 26ACh. 17 - Prob. 27ACh. 17 - Prob. 28ACh. 17 - Prob. 29ACh. 17 - Prob. 30ACh. 17 - Prob. 31ACh. 17 - Prob. 32ACh. 17 - Prob. 33ACh. 17 - Prob. 34ACh. 17 - Prob. 35ACh. 17 - Prob. 36ACh. 17 - Prob. 37ACh. 17 - Prob. 38ACh. 17 - Prob. 39ACh. 17 - Prob. 40ACh. 17 - Prob. 41ACh. 17 - Prob. 42ACh. 17 - Prob. 43ACh. 17 - Prob. 44ACh. 17 - Prob. 45ACh. 17 - Prob. 46ACh. 17 - Prob. 47ACh. 17 - Prob. 48ACh. 17 - Prob. 49ACh. 17 - Prob. 50ACh. 17 - Prob. 51ACh. 17 - Prob. 52ACh. 17 - Prob. 53ACh. 17 - Prob. 54ACh. 17 - Prob. 55ACh. 17 - Prob. 56ACh. 17 - Prob. 57ACh. 17 - Prob. 58ACh. 17 - Prob. 59ACh. 17 - Prob. 60ACh. 17 - Prob. 61ACh. 17 - Prob. 62ACh. 17 - Prob. 63ACh. 17 - Prob. 64ACh. 17 - Prob. 65ACh. 17 - Prob. 66ACh. 17 - Prob. 67ACh. 17 - Prob. 68ACh. 17 - Prob. 69ACh. 17 - Prob. 1STPCh. 17 - Prob. 2STPCh. 17 - Prob. 3STPCh. 17 - Prob. 4STPCh. 17 - Prob. 5STPCh. 17 - Prob. 6STPCh. 17 - Prob. 7STP
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