   Chapter 18, Problem 67GQ

Chapter
Section
Textbook Problem

Elemental boron, in the form of thin fibers, can be made by reducing a boron halide with H2.BCl3(g) + 3 2 H2(g) → B(s) + 3 HCl(g)Calculate ΔrH°, ΔrS°, and ΔrG° at 25 °C for this reaction. Is the reaction predicted to be product-favored at equilibrium at 25 °C? If so, is it enthalpy- or entropy-driven? [S° for B(s) is 5.86 J/K · mol.]

Interpretation Introduction

Interpretation:

The value of ΔrH° ,ΔrS° and ΔrG for formation of elemental boron should be calculated under given conditions.

Concept introduction:

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

ΔrG=ΔrHTΔrS

Here, ΔrH is the change in enthalpy and ΔrS is the change in entropy.

Explanation

The value of ΔrH°, ΔrS° and ΔrG for formation of elemental boron calculated below.

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

The standard entropy of B(s) is 5.86 J/Kmol.

The standard entropy of HCl(g) is 186.2 J/Kmol.

The standard entropy of H2(g) is 130.7 J/Kmol.

The standard entropy of BCl3(g) is  290.17 J/Kmol.

The standard enthalpy of B(s) is 0 kJ/mol.

The standard enthalpy of HCl(g) is 92.31 kJ/mol.

The standard enthalpy of H2(g) is 0 kJ/mol.

The standard enthalpy of BCl3(g) is 402.96 kJ/mol.

The balanced chemical equation is:

BCl3(g) + 32H2(g)B(s) + 3HCl(g)

ΔrH°=nΔfH°(products)nΔfH°(reactants)=[[(1 mol B(s)/mol-rxn)ΔfH°[B(s)]+(3 mol HCl(g)/mol-rxn)ΔfH°[HCl(g)]] [(1 mol BCl3(g)/mol-rxn)ΔfH°[BCl3(g)]+(1.5 mol H2(g)/mol-rxn)ΔfH°[H2(g)]]]

Substituting the enthalpy values,

ΔrH°=[[(1 mol B(s)/mol-rxn)(0 kJ/mol)+(3 mol HCl(g)/mol-rxn)(92.31 kJ/mol)] [(1 mol BCl3(g)/mol-rxn)(402.96 kJ/mol)+(1

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