# Sodium reacts violently with water according to the equation Na(s) + H 2 O( ℓ ) → NaOH(aq) + ½ H 2 (g) Without doing calculations, predict the signs of Δ r H ° and Δ r S ° for the reaction. Verify your prediction with a calculation.

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

Chapter
Section

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18, Problem 47GQ
Textbook Problem
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## Sodium reacts violently with water according to the equationNa(s) + H2O(ℓ) → NaOH(aq) + ½ H2(g)Without doing calculations, predict the signs of ΔrH° and ΔrS° for the reaction. Verify your prediction with a calculation.

Interpretation Introduction

Interpretation:

The signs for ΔrS° and ΔrH° with respect to the reaction of sodium with water should be predicted without using calculations.

Concept introduction:

The standard entropy change for any reaction is the sum of standard molar entropies of products, subtracted from the sum of standard molar entropies of reactants. The standard molar entropies are multiplied by the stoichiometric coefficient which is as per the balanced equation.

ΔrS°=nS°(products)nS°(reactants)

The enthalpy change for any reaction can be expressed similarly,

ΔrH°=nH°(products)nH°(reactants)

The sign of ΔrH° tells if the reaction is exothermic or endothermic.

### Explanation of Solution

The reaction of sodium with water is an exothermic reaction. Sodium reacts violently with water. Thus, ΔrH° should be negative.

There is an increase in number of moles of gases after the reaction of sodium and water. Since a gas is formed, thus ΔrS° should be positive.

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

Na(s)+H2O(l)NaOH(aq)+12H2(g)ΔfH°(kJ/mol)0285.83469.150S(J/Kmol)51.2169.9548.1130.7

ΔrH°=nΔfH°(products)nΔfH°(reactants)=[[(1 mol NaOH(aq)/mol-rxn)ΔfH°[NaOH(aq)]+(0.5 mol H2(g)/mol-rxn)ΔfH°[H2(g)] ][(1 mol Na(s)/mol-rxn)ΔfH°[Na(s)]+(1 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]] ]

Substituting the enthalpy values,

ΔrH°=[[(1 mol NaOH(aq)/mol-rxn)(469.15 kJ/mol)+(0.5 mol H2(g)/mol-rxn)(0 kJ/mol) ][(1 mol Na(s)/mol-rxn)(0 kJ/mol)+(1 mol H2O(l)/mol-rxn)(285.83 kJ/mol)] ] =183.32 kJ/mol-rxn

ΔrS°=nS°(products)nS°(reactants)=[[(1 mol NaOH(aq)/mol-rxn)S°[NaOH(aq)]+(0

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