   Chapter 21, Problem 24PS

Chapter
Section
Textbook Problem

Using data in Appendix L, calculate ΔrH°, ΔrG°, and ΔrS° for the reaction of carbon and water to give CO and H2 at 298 K.

Interpretation Introduction

Interpretation:

To determine the ΔrH°,ΔrG° and ΔrS° for the production of hydrogen by reaction of carbon and water.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrG. The expression for the free energy change is:

ΔrG°=nΔfG°(products)nΔfG°(reactants)

The expression for the enthalpy change is:

ΔrH°=nΔfH°(products)nΔfH°(reactants)

The expression for the entropy change is:

ΔrS°=nS°(products)nS°(reactants)

Explanation

The ΔrG°,ΔrH° and ΔrS° for the given reaction is calculated below.

Given:

Refer to Appendix L for the values of standard free energy values.

The given reaction is,

C(s) + H2O(g)CO(g)+H2(g)

The ΔrG° for C(s) is 0 kJ/mol.

The ΔrG° for H2O(g) is 228.59 kJ/mol.

The ΔrG° for CO(g) is 137.168 kJ/mol.

The ΔrG° for H2(g) is 0 kJ/mol.

The expression for the free energy change is:

ΔrG°=nΔfG°(products)nΔfG°(reactants)=[[(1 mol CO(g)/mol-rxn)ΔfG°[CO(g)]+(1 mol H2(g)/mol-rxn)ΔfG°[H2(g)]][(1 mol C(s)/mol-rxn)ΔfG°[C(s)]+(1 mol H2O(g)/mol-rxn)ΔfG°[H2O(g)]] ]

Substitute the values,

ΔrG°=[[(1 mol CO(g)/mol-rxn)(137.168 kJ/mol)+(1 mol H2(g)/mol-rxn)(0 kJ/mol)][(1 mol C(s)/mol-rxn)(0 kJ/mol)+(1 mol H2O(g)/mol-rxn)(228.59 kJ/mol)] ]=+91.42 kJ/mol-rxn

The ΔrH° for C(s) is 0 kJ/mol.

The ΔrH° for H2O(g) is 241.83 kJ/mol.

The ΔrH° for CO(g) is 110.525 kJ/mol.

The ΔrH° for H2(g) is 0 kJ/mol

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