   Chapter 18, Problem 26PS

Chapter
Section
Textbook Problem

Is the reaction Si(s) + 2 H2(g) → SiH4(g) spontaneous under standard conditions at 298.15 K? Answer this question by calculating ΔS°(system), ΔS°(surroundings), and ΔS°(universe). (Define reactants and products as the system.)

Interpretation Introduction

Interpretation:

It should be identified that whether the given reaction is spontaneous at 298.15K by calculating  ΔS(system), ΔS(surroundings) and ΔS(universe) also the reactants and products should be defined as system.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

ΔS(universe)= ΔS(system)+ΔS(surroundings)

The ΔS(universe) should be greater than zero for a spontaneous process.

The  ΔS(system) can be calculated by the following expression,

ΔS(system)=ΔrS°=nS°(products)nS°(reactants)

The ΔS(surroundings) can be calculated by the following expression,

ΔS(surroundings)=ΔrHT

Here, ΔrHo is the enthalpy change for the reaction.

Explanation

Given: Si(s) + 2H2(g)SiH4(g)

The standard entropy of SiH4(g) is 204.65 J/Kmol.

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

The standard entropy of Si(s) is 18.82 J/Kmol.

The standard enthalpy of SiH4(g) is 34.31 kJ/mol.

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

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

The ΔS(universe) for the reaction Si(s) + 2H2(g)SiH4(g) is calculated below.

The balanced chemical equation is:

Si(s) + 2H2(g)SiH4(g)

Calculation for ΔSo(system):

The  ΔS(system) can be calculated by the following expression,

ΔS(system)=ΔrS°=nS°(products)nS°(reactants)ΔS(system)=[(1 mol SiH4(g)/mol-rxn)S°[SiH4(g)](1 mol Si(s)/mol-rxn)S°[Si(s)]+(2 mol H2(g)/mol-rxn)S°[H2(g)]]

Substituting the respective entropy values,

ΔS(system)=[(1 mol SiH4(g)/mol-rxn)(204.65 J/Kmol)(1 mol Si(s)/mol-rxn)(18.82 J/Kmol)+(2 mol H2(g)/mol-rxn)(130.7 J/Kmol)]=75

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