Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
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Chapter 5, Problem 5G.21E

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant at 25oC has to be determined for the given reaction.

  2H2(g)+O2(g)2H2O(g)

Concept Introduction:

The equilibrium constant of a reaction can be calculated using the given expression,

  lnK=-ΔGroRTwhere,R=Gas constant (8.314JK-1mol-1)ΔGr°=StandardGibb'sfreeenergyT=TemperatureK= Equilibrium constant

(a)

Expert Solution
Check Mark

Answer to Problem 5G.21E

The equilibrium constant of the given reaction is 2.0×1080.

Explanation of Solution

Given reaction is

The combustion of hydrogen:

  2H2(g)+O2(g)2H2O(g)

Temperature of the reaction is 298K,

The ΔGf° value of H2O is 228.57kJmol-1

The ΔGf° value of H2 is 0

The ΔGf° value of O2 is 0

R=8.3145 JK-1mol-1

T=298K

Here, ΔGr° is calculated using given formula,

  ΔGr°=2ΔGf°(H2O(g))-[2ΔGf°(H2(g))+ΔGf°(O2(g))]=2×(228.57×103)0=457.14×103

Now, the equilibrium constant of the reaction is calculated as

  lnK=-ΔGroRT=457.14×103Jmol-1(8.3145JK-1mol-1)×(298K)=184.9

To calculate K value, inverse logarithm is taken for the above reaction.

  K=e184.9=2.0×1080

Therefore, the K value of given reaction is 2.0×1080.

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant at 25oC has to be determined for the given reaction.

  2CO(g)+O2(g)2CO2(g)

Concept introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 5G.21E

The equilibrium constant of the given reaction is 1×1090.

Explanation of Solution

Given reaction is

The oxidation of carbon monoxide:

  2CO(g)+O2(g)2CO2(g)

Temperature of the reaction is 298K,

The ΔGf° value of CO is 137.17kJmol-1

The ΔGf° value of CO2 is 394.36kJmol-1

The ΔGf° value of O2 is 0

R=8.3145 JK-1mol-1

T=298K

Here, ΔGr° is calculated using given formula,

  ΔGr°=2ΔGf°(CO2(g))-[2ΔGf°(CO(g))+ΔGf°(O2(g))]=2×(394.36×103)[2×(137.17×103)0]=5.14×105

Now, the equilibrium constant of the reaction is calculated as

  lnK=-ΔGroRT=5.14×105Jmol-1(8.3145JK-1mol-1)×(298K)=207.44

To calculate K value, inverse logarithm is taken for the above reaction.

  K=e207.44=1×1090

Therefore, the K value of given reaction is 1×1090.

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant at 25oC has to be determined for the given reaction.

  CaCO3(s)CaO(s)+CO2(g)

Concept introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 5G.21E

The equilibrium constant of the given reaction is 6.1×1022.

Explanation of Solution

Given reaction is

The decomposition of limestone:

  CaCO3(s)CaO(s)+CO2(g)

Temperature of the reaction is 298K,

The ΔGf° value of CO2 is 394.36kJmol-1

The ΔGf° value of CaO is 604.03kJmol-1

The ΔGf° value of CaCO3 is 1127.8kJmol-1

R=8.3145 JK-1mol-1

T=298K

Here, ΔGr° is calculated using given formula,

  ΔGr°=[ΔGf°CaO(s)+ΔGf°CO2(g)]-ΔGf°CaCO3(s)=(604.03×103394.36×103)(1127.8×103)=1.30×105

Now, the equilibrium constant of the reaction is calculated as

  lnK=-ΔGroRT=1.30×105Jmol-1(8.3145JK-1mol-1)×(298K)=52.47

To calculate K value, inverse logarithm is taken for the above reaction.

  K=e52.47=6.1×1022

Therefore, the K value of given reaction is 6.1×1022.

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Chapter 5 Solutions

Chemical Principles: The Quest for Insight

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