Chemistry: The Central Science, Books a la Carte Edition & Solutions to Red Exercises for Chemistry & Mastering Chemistry with Pearson eText -- Access Card  Package
Chemistry: The Central Science, Books a la Carte Edition & Solutions to Red Exercises for Chemistry & Mastering Chemistry with Pearson eText -- Access Card Package
1st Edition
ISBN: 9780134024516
Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus
Publisher: PEARSON
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Chapter 5, Problem 34E

(a)

Interpretation Introduction

To determine: The value of equilibrium constant for the given reaction at 298K .

Solution: The value of equilibrium constant for the given reaction is 7.856×1015 .

Explanation:

Given

The balanced chemical equation for the given reaction is,

    SO2(g)+2H2S(g)3S(s)+2H2O(g)

The temperature of the reaction is 298K .

The standard free energy change of the given reaction is calculated by the formula,

    ΔG0=nΔG0( products)mΔG0( reactants)=(2ΔG H 2 O( g )0+3ΔG S( s )0)(ΔG SO 2 ( g )0+2ΔG H 2 S( g )0)

Substitute the values of ΔGH2O(g)0 , ΔGS(s)0 , ΔGSO2(g)0 and ΔGH2S(g)0 from appendix C in equation.

    ΔH0=(2×228.57kJ/mol+3×0kJ/mol)(1×300.4kJ/mol+2×33.01kJ/mol)=90.72kJ/mol

The conversion of free energy from kJ/mol into J/mol is done as,

    1kJ/mol=103J/mol

Hence, the conversion of free energy from 90.72kJ/mol into J/mol is,

    90.72kJ/mol=90.72×103J/mol

The equilibrium constant for the given reaction is calculated by the formula,

    ΔG0=RTlnKlnK=ΔG0RT

Where,

  • ΔG0 is the standard free energy of the reaction.
  • R is the gas constant (8.314J/molK) .
  • T is the temperature of the reaction.
  • K is the equilibrium constant of the reaction.
Substitute the values of ΔG0 , T and R in above equation.
    lnK=( 90.72× 10 3 J/mol)8.314J/molK×298K=36.6

The equilibrium constant of the reaction is calculated by taking exponential of the above equation.

    K=e36.6=7.856×1015

Conclusion:

The value of equilibrium constant for the given reaction is 7.856×1015 .

(b)

Interpretation Introduction

To determine: If the given reaction is feasible to remove SO2 or not.

(c)

Interpretation Introduction

To determine: The equilibrium pressure of SO2 gas at 298K .

(d)

Interpretation Introduction

To determine: If the given reaction is more or less effective at higher temperature.

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Chapter 5 Solutions

Chemistry: The Central Science, Books a la Carte Edition & Solutions to Red Exercises for Chemistry & Mastering Chemistry with Pearson eText -- Access Card Package

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