   # If Δ r H° = +467.9 kJ/mol-rxn and Δ r S° = +560.7 J/K · mol-rxn for the following reaction 2 Fe 2 O 3 (s) + 3 C(graphite) → 4 Fe(s) + 3 CO 2 (g) then, under standard conditions, this reaction will be spontaneous (a) at all temperatures (b) at higher temperatures (c) at lower temperatures (d) at no temperature ### Chemistry & Chemical Reactivity

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

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18.5, Problem 3RC
Textbook Problem
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## If ΔrH° = +467.9 kJ/mol-rxn and ΔrS° = +560.7 J/K · mol-rxn for the following reaction2 Fe2O3(s) + 3 C(graphite) → 4 Fe(s) + 3 CO2(g)then, under standard conditions, this reaction will be spontaneous (a) at all temperatures (b) at higher temperatures (c) at lower temperatures (d) at no temperature

Interpretation Introduction

Interpretation: the spontaneity of the given reaction is to be identified from the given options.

Concept introduction:

In Spontaneous process, energy transfer from higher concentration to lower, or the energy is dispersed.

Entropy (S) is the quantity to measure the dispersal of energy in the spontaneous processes.

The equation for finding ΔS(system) is,

ΔrS=nS(products)nS(reactants)

ΔrH(system) is the change in enthalpy is calculated by sum of the product of stoichiometric coefficients and molar enthalpies of products minus sum of the product of stoichiometric coefficients and molar enthalpies of reactants. That is,

ΔrH(system)=nΔfH(products)nΔfH(reactants)

Equation for ΔS(surrounding) is,

ΔS(surrounding)=ΔfH(system)/T

### Explanation of Solution

For the reaction,

Total 5 moles (2+3) of reactants reacted to form 7 (4+3) moles of products. So entropy is favored.

ΔS(system) is 560J/Kmol ,

ΔrH(system) is 467.9kJ/mol

Equation for ΔS(surroundi

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