   Chapter 18, Problem 32PS

Chapter
Section
Textbook Problem

Using values of ΔfH° and S°, calculate ΔrG° for each of the following reactions at 25 °C. (a) 2 Na(s) + 2 H2O(ℓ) → 2 NaOH(aq) + H2(g) (b) 6 C(graphite) + 3 H2(g) → C6H6(ℓ) Which of these reactions is (are) predicted to be product-favored at equilibrium? Are the reactions enthalpy- or entropy-driven?

(a)

Interpretation Introduction

Interpretation:

The standard free energy change ΔG for the given reaction should be calcualted using SoandΔfHo values.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔG. It is related to entropy and entropy by the following expression,

ΔG=ΔHTΔS

Here, ΔH is the change in enthalpy and ΔS is the change in entropy.

Explanation

The standard free energy change for the reaction of sodium and water at 298 K is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

2Na(s)+2H2O(l)2NaOH(aq)+H2(g)ΔfH°(kJ/mol)0285.83469.150S(J/Kmol)51.2169.9548.1130.7

ΔrH°=nΔfH°(products)nΔfH°(reactants)=[[(2 mol NaOH(aq)/mol-rxn)ΔfH°[NaOH(aq)]+(1 mol H2(g)/mol-rxn)ΔfH°[H2(g)] ][(2 mol Na(s)/mol-rxn)ΔfH°[Na(s)]+(2 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]] ]

Substituting the enthalpy values

ΔrH°=[[(2 mol NaOH(aq)/mol-rxn)(469.15 kJ/mol)+(1 mol H2(g)/mol-rxn)(0 kJ/mol) ][(2 mol Na(s)/mol-rxn)(0 kJ/mol)+(2 mol H2O(l)/mol-rxn)(285.83 kJ/mol)] ] =366

(b)

Interpretation Introduction

Interpretation:

The standard free energy change ΔG for the given reaction should be calcualted using SoandΔfHo values.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔG. It is related to entropy and entropy by the following expression,

ΔG=ΔHTΔS

Here, ΔH is the change in enthalpy and ΔS is the change in entropy.

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