   Chapter 18, Problem 63GQ

Chapter
Section
Textbook Problem

The following reaction is reactant-favored at equilibrium at room temperature.COCl2(g) → CO(g) + Cl2(g)Will raising or lowering the temperature make it product-favored?

Interpretation Introduction

Interpretation:

It should be identified that increasing or decreasing the temperature for the given reaction will makes it as product favoured.

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 reaction can be made product-favoured or reactant-favoured depending upon the magnitude and sign of ΔrS°.

Explanation

The ΔrS° for the decomposition of COCl2(g) is calculated below.

The Appendix L referred for the values of standard entropies.

The standard entropy of COCl2(g) is 283.53 J/Kmol.

The standard entropy of CO(g) is 197.674 J/Kmol.

The standard entropy of Cl2(g) is 223.08 J/Kmol.

The balanced chemical equation for the evaporation of ethanol is:

COCl2(g)CO(g)+Cl2(g)

The expression for the standard entropy change is,

ΔrS°=nS°(products)nS°(reactants)=[[(1 mol CO(g)/mol-rxn)S°[CO(g)]+(1 mol Cl2(g)/mol-rxn)S°[Cl2(g)

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