EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119227946
Author: Willard
Publisher: VST
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Chapter 16, Problem 53AE

(a)

Interpretation Introduction

Interpretation:

Moles of HI from reaction mixture of H2 and I2 has to be determined.

Concept Introduction:

The reactant in a reaction that tells about amount of product formed in a reaction is termed as limiting reagent. It helps to determine the yield of reaction.

The quantity of species that gives relation between reactants and products is determined by the stoichiometry of a reaction. Suppose the general reaction,

  2A+B3C

In the above reaction, two moles of A reacts with one mole of B to form three moles of C. Therefore, stoichiometric ratio of A and B is 2:2, of A and C is 2:3 and of B and C is 1:3.

(b)

Interpretation Introduction

Interpretation:

Moles of HI, H2 and I2 in equilibrium reaction mixture has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

Equilibrium constant for below reaction has to be calculated.

  H2(g)+I2(g)2HI(aq)

Concept Introduction:

For equilibrium reaction rate of forward reaction is equal to rate of reverse direction. Consider a general reaction that is as follows:

  aA+bBcC+dD

The expression for equilibrium constant is written as follows:

  Keq=[C]c[D]d[A]a[B]b

Here,

Keq is equilibrium constant

[A], [B], [C] and [D] are concentration of substances in reaction

a, b, c and d are coefficients of substances

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Chapter 16 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

Ch. 16.6 - Prob. 16.11PCh. 16.6 - Prob. 16.12PCh. 16.7 - Prob. 16.13PCh. 16.7 - Prob. 16.14PCh. 16.7 - Prob. 16.15PCh. 16.8 - Prob. 16.16PCh. 16 - Prob. 1RQCh. 16 - Prob. 2RQCh. 16 - Prob. 3RQCh. 16 - Prob. 4RQCh. 16 - Prob. 5RQCh. 16 - Prob. 6RQCh. 16 - Prob. 7RQCh. 16 - Prob. 8RQCh. 16 - Prob. 9RQCh. 16 - Prob. 10RQCh. 16 - Prob. 11RQCh. 16 - Prob. 12RQCh. 16 - Prob. 13RQCh. 16 - Prob. 14RQCh. 16 - Prob. 15RQCh. 16 - Prob. 16RQCh. 16 - Prob. 17RQCh. 16 - Prob. 18RQCh. 16 - Prob. 19RQCh. 16 - Prob. 20RQCh. 16 - Prob. 21RQCh. 16 - Prob. 22RQCh. 16 - Prob. 23RQCh. 16 - Prob. 24RQCh. 16 - Prob. 25RQCh. 16 - Prob. 26RQCh. 16 - Prob. 27RQCh. 16 - Prob. 1PECh. 16 - Prob. 2PECh. 16 - Prob. 3PECh. 16 - Prob. 4PECh. 16 - Prob. 5PECh. 16 - Prob. 6PECh. 16 - Prob. 7PECh. 16 - Prob. 8PECh. 16 - Prob. 9PECh. 16 - Prob. 10PECh. 16 - Prob. 11PECh. 16 - Prob. 12PECh. 16 - Prob. 13PECh. 16 - Prob. 14PECh. 16 - Prob. 15PECh. 16 - Prob. 16PECh. 16 - Prob. 17PECh. 16 - Prob. 18PECh. 16 - Prob. 19PECh. 16 - Prob. 20PECh. 16 - Prob. 21PECh. 16 - Prob. 22PECh. 16 - Prob. 23PECh. 16 - Prob. 24PECh. 16 - Prob. 25PECh. 16 - Prob. 26PECh. 16 - Prob. 27PECh. 16 - Prob. 28PECh. 16 - Prob. 29PECh. 16 - Prob. 30PECh. 16 - Prob. 31PECh. 16 - Prob. 32PECh. 16 - Prob. 33PECh. 16 - Prob. 34PECh. 16 - Prob. 35PECh. 16 - Prob. 36PECh. 16 - Prob. 37PECh. 16 - Prob. 38PECh. 16 - Prob. 39PECh. 16 - Prob. 40PECh. 16 - Prob. 41PECh. 16 - Prob. 42PECh. 16 - Prob. 43PECh. 16 - Prob. 44PECh. 16 - Prob. 45PECh. 16 - Prob. 46PECh. 16 - Prob. 47PECh. 16 - Prob. 48PECh. 16 - Prob. 49AECh. 16 - Prob. 50AECh. 16 - Prob. 51AECh. 16 - Prob. 52AECh. 16 - Prob. 53AECh. 16 - Prob. 54AECh. 16 - Prob. 55AECh. 16 - Prob. 56AECh. 16 - Prob. 57AECh. 16 - Prob. 58AECh. 16 - Prob. 59AECh. 16 - Prob. 60AECh. 16 - Prob. 61AECh. 16 - Prob. 62AECh. 16 - Prob. 63AECh. 16 - Prob. 64AECh. 16 - Prob. 65AECh. 16 - Prob. 66AECh. 16 - Prob. 67AECh. 16 - Prob. 68AECh. 16 - Prob. 69AECh. 16 - Prob. 70AECh. 16 - Prob. 71AECh. 16 - Prob. 72AECh. 16 - Prob. 73AECh. 16 - Prob. 74AECh. 16 - Prob. 75AECh. 16 - Prob. 76AECh. 16 - Prob. 77AECh. 16 - Prob. 78AECh. 16 - Prob. 79AECh. 16 - Prob. 80AECh. 16 - Prob. 81AECh. 16 - Prob. 83AECh. 16 - Prob. 84AECh. 16 - Prob. 85AECh. 16 - Prob. 86CECh. 16 - Prob. 87CECh. 16 - Prob. 88CECh. 16 - Prob. 89CE
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY