Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN: 9781337398909
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
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Chapter 10, Problem 10.84PAE

Methane can be produced from CO and H2.The process might be done in two steps, as shown below, with each step carried out in a separate reaction vessel within the production plant.

Reaction 1 C O ( g ) + 2H 2 ( g ) CH 3 OH ( l )

Δ S ° = -532 J/K

Reaction 2 CH 3 OH ( l ) CH 4 ( g ) + 1 2 O 2 ( g )

   Δ S ° = +162 J/K

NOTE: You should be able to work this problem without using any additional tabulated data.

(a) Calculate Δ H ° for reaction 1.

(b) Calculate Δ G f ° for CO(g).

(c) Calculate S° for O2(g).

(d) At what temperatures is reaction 1 spontaneous?

(e) Suggest a reason why these two steps would need to be carried out separately.

    Substance    Δ H f ° (kJ mol-1)    Δ G f ° (kJ mol-1)    S ° (J mol-1 K-1)
    CO(s) -110.5 197.674
    CH3OH(
       l
    )
    -238.7 -166.4 126.8
    CH4(g) -74.8 186.2

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

Thermodynamic entropy, enthalpy and Gibb’s free energy as asked must be calculated for the reactions given.

Concept Introduction:

Enthalpy change for a process is determined as:

ΔHo=iviΔHfo(product)ijvjΔHfo(reactant)j

where v is for stoichiometric coefficients

Similarly entropy change for a process is determined as:

ΔSo=iviSo(product)ijvjSo(reactant)j

where v is for stoichiometric coefficients and So is for absolute entropies of reactants and products..

Gibb’s free energy is a state function which predicts whether a process is spontaneous or not at conditions of constant pressure and temperature. Gibb’s free energy change for a process at constant temperature is defined as:

ΔGo=ΔHoTΔSo

where ΔS is the total change in the entropy of the system and ΔH is the total change in the enthalpy of the system. For a spontaneous process ΔG must be negative.

Similar to enthalpy and entropy, standard Gibb’s free energy change can be calculated as:

ΔGo=iviΔGfo(product)ijvjΔGfo(reactant)j

Also ΔHfo and ΔGfo for elements in their standard states is zero.

Answer to Problem 10.84PAE

Solution:

a)ΔHo for reaction 1 is 128.2 kJ mol1.

b)ΔGfo for CO(g) is 137.1 kJ mol1.

c)So for O2(g) is 205.2 J K1 mol1.

d) Hence reaction 1 is spontaneous at lower temperatures.

e) Since reaction 1 is spontaneous at lower temperatures and reaction 2 is spontaneous at higher temperatures, they must be carried out separately.

a)

Explanation of Solution

Given reaction 1 is:

CO(g)+2 H2(g)CH3OH(l)

Calculate enthalpy change the above reaction that is ΔHo, enthalpy of formation for hydrogen would be zero.

ΔHo=ΔHfo[CH3OH(l)]ΔHfo[CO(g)]2 ΔHfo[H2(g)]=238.7 kJ mol1(110.5  kJ mol 1)0=128.2 kJ mol1

b)

Using the value of enthalpy change of reaction 1 calculated in part a) and the given entropy change calculate the Gibb’s free energy change for reaction 1. Since we are using standard values the temperature will be 298 K.

ΔGo=ΔHoTΔSo=128.2 kJ mol1(298 K×332  J K 1  mol 1)=128.2×103 J mol1+98936 J mol1=29.3×103 J mol1=29.3 kJ mol1

Now use the formula described above in the concept introduction to calculate ΔGfo for CO(g).

ΔGo=ΔGfo[CH3OH(l)]ΔGfo[CO(g)]2 ΔGfo[H2(g)]29.3 kJ mol1=166.4 kJ mol1ΔGfo[CO(g)]0ΔGfo[CO(g)]=137.1 kJ mol1

c)

Given reaction 2 is:

CH3OH(l)CH4(g)+1/2 O2(g)

The entropy change for the above reaction is ΔSo=+162 J K1 mol1, calculate So for O2(g).

ΔSo=12So[O2(g)]+So[CH4(g)]So[CH3OH(l)]+162 J K1 mol1=12So[O2(g)]+186.2 J K1 mol1126.8 J K1 mol112So[O2(g)]=102.6 J K1 mol1So[O2(g)]=205.2 J K1 mol1

d)

Gibb’s free energy is a state function which predicts whether a process is spontaneous or not at conditions of constant pressure and temperature. Gibb’s free energy change for a process at constant temperature is defined as:

ΔGo=ΔHoTΔSo

Where ΔS is the total change in the entropy of the system and ΔH is the total change in the enthalpy of the system. For a spontaneous process ΔG must be negative.

Given reaction 1 is:

CO(g)+2 H2(g)CH3OH(l)

The entropy change for the above reaction is ΔSo=332 J K1 mol1. The value of enthalpy change of reaction 1 calculated in part a) is 128.2 kJ mol1. Since both ΔH<0 and ΔS<0 Gibb’s free energy depends on temperature. At higher temperatures the term TΔS dominates, so the process will be spontaneous at lower temperatures where ΔH dominates.

Hence reaction 1 is spontaneous at lower temperatures.

e)

As explained above in part c) reaction 1 is spontaneous at lower temperatures. Similarly given reaction 2 is:

CH3OH(l)CH4(g)+1/2 O2(g)

The entropy change for the above reaction is ΔSo=+162 J K1 mol1

. Calculate enthalpy change for this reaction:

ΔHo=12ΔHfo[O2(g)]+ΔHfo[CH4(g)]ΔHfo[CH3OH(l)]=0+(74.8  kJ mol 1)(238.7  kJ mol 1)=164 kJ mol1

Gibb’s free energy is a state function which predicts whether a process is spontaneous or not at conditions of constant pressure and temperature. Gibb’s free energy change for a process at constant temperature is defined as:

ΔGo=ΔHoTΔSo

Where ΔS is the total change in the entropy of the system and ΔH is the total change in the enthalpy of the system. For a spontaneous process ΔG must be negative.

Now since both ΔH>0 and ΔS>0 Gibb’s free energy again depends on temperature. At higher temperatures the term TΔS dominates, so the process will be spontaneous at higher temperatures where ΔS dominates.

Since reaction 1 is spontaneous at lower temperatures and reaction 2 is spontaneous at higher temperatures, they must be carried out separately.

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Chapter 10 Solutions

Chemistry for Engineering Students

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