   Chapter 18, Problem 75GQ

Chapter
Section
Textbook Problem

A cave in Mexico was recently discovered to have some interesting chemistry. Hydrogen sulfide, H2S, reacts with oxygen in the cave to give sulfuric acid, which drips from the ceiling in droplets with a pH less than 1. The reaction occurring isH2S(g) + 2 O2(g) → H2SO4(ℓ)Calculate ΔrH°, ΔrS°, and ΔrG°. Is the reaction product-favored at equilibrium at 25 °C? Is it enthalpy- or entropy-driven?

Interpretation Introduction

Interpretation:

The values of ΔrH°, ΔrS° and ΔrGo for the formation of sulfuric acid should be calculated and identify that the reaction is enthalpy or entropy driven.

Concept introduction:

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

ΔrGo= ΔrHo-TΔrSo

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

Explanation

The value of ΔrH°, ΔrS° and ΔrGo for the formation of sulfuric acid is calculated below.

Given:

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

H2S(g)+2O2(g)H2SO4(l)ΔfH°(kJ/mol)-20.630-814So(J/K×mol)205.79205.07156.9

ΔrH°=fH°(products)fH°(reactants)=[(1 mol H2SO4(l)/mol-rxn)ΔfH°[H2SO4(l)]-[(1 mol H2S(g)/mol-rxn)ΔfH°[H2S(g)]+(2 mol O2(g)/mol-rxn)ΔfH°[O2(g)]] ]

Substituting the respective values

ΔrH°=[(1 mol H2SO4(l)/mol-rxn)(-814 kJ/mol)-[(1 mol H2S(g)/mol-rxn)(-20

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