18: Entropy and Free Energy

Spontaneity and Energy Transfer

Lecture Material

Exercise \(\PageIndex{1a}\)

Which of the following is a spontaneous process?

1. ice melting at 25°C
2. heat flowing from a hot to a cold object
3. an iron tool rusting

 

1. 1 only
2. 2 only
3. 3 only
4. 1 and 2 only
5. 1, 2, and 3

Answer

e. 1, 2, and 3

 

Exercise \(\PageIndex{1}\)

For which one of the following reactions does the entropy of the system INCREASE?

a.       NH₃(g) + HCl(g) --> NH₄Cl(s)

b.      Ag⁺(aq) + Cl^-(aq) --> AgCl(s)

c.       2H₂(g) + O₂(g) --> 2H₂O(g)

d.      NH₃(g) + H₂O(l) --> NH₄⁺(aq) + OH^-(aq)

e.       2H₂O₂(l) --> 2H₂O(l) + O₂(g)

Answer

      2H₂O₂(l) --> 2H₂O(l) + O₂(g)

Entropy Measurements and Values

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Exercise \(\PageIndex{4a}\)

Predict the entropy change in the reaction,

\(2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)\)

1. increase
2. decrease
3. same
4. not enough information

Answer

b. decrease

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Exercise \(\PageIndex{4b}\)

Predict the entropy change in the reaction,

\(Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)\)

 

1. increase
2. decrease
3. same
4. not enough information

Answer

b. decrease, although you might say not enough information, as we do not have a bearing on the solvent orientation around the ions (the solvent will undergo an increase in entropy), but if you look at the thermodynamic tables, you see there is a decrease in entropy. 

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Exercise \(\PageIndex{4c}\)

Which one of the following decreases of the entropy of the system? There is only one correct answer.

1. dissolving NaCl in water
2. sublimation of benzene
3. dissolving oxygen in water
4. boiling of alcohol

Answer

c. dissolving oxygen in water, once again, the salts can be tricky, but clearly a gas dissolving in a liquid shows a decrease

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Exercise \(\PageIndex{4d}\)

Which one has the highest absolute entropy at 25⁰C?

1. H₂O(l)
2. He(g)
3. C(s)
4. NH₃(g)

Answer

d. NH₃(g), then He(g), as the ammonia is more complex and has more microstates available.  

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Exercise \(\PageIndex{4e}\)

Arrange the following reactions in order of increasing ∆S°_rxn values:

1. \(H_{2}(g)+Cl_{2}(g)\rightarrow 2HCl(g)\)
2. \(3H_{2}(g)+N_{2}(g)\rightarrow 2NH_{3}(g) \)
3. \((NH_{4})_{2}Cr_{2}O_{7}(s)\rightarrow Cr_{2}O_{3}(s)+4H_{2}O(l)+N_{2}(g)\)

 

1. 1 < 3 < 2
2. 1 < 2 < 3
3. 2 < 1 < 3
4. 2 < 3 < 1
5. 3 < 1 < 2

Answer

c. 2 < 1 < 3

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Exercise \(\PageIndex{4f}\)

Calculate the change of entropy for the reaction at 25⁰C, where S_f⁰ for Al(s), O₂(g) and Al₂O₃(s) are 28.32, 205.0 and 51.0 J/K*mol respectively.

\(4Al(s)+3O_{2}(g)\rightarrow 2Al_{2}O_{3}(s)\)

Answer

\[ΔS°=\sum nS^\circ_{298}(\ce{products})−\sum mS^\circ_{298}(\ce{reactants}) \nonumber\]

see eq.18.4.9

\[\Delta S^{0}=\left ( 2*51.0 \right )-\left ( 4*28.32 \right )-\left ( 3*205.0 \right )=-626.3J/K\nonumber\]

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Exercise \(\PageIndex{1}\)

Which of the following produces an INCREASE in entropy of the system?

a.   H2O(l) --> H2O(s)                                                      b.  2O2(g) + 2SO(g) --> 2SO3(g)
c.  2CH3OH(g) + 3O2(g) --> 2CO2(g) + 4H2O(l)                 d.  I2(s) --> I2(l)                                        e.  None of the above.

 

 

Answer

   d.  I2(s) --> I2(l)   

 

 

Second Law and Gibbs Free Energy

Spontaneity and Temperature

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Exercise \(\PageIndex{5.1a}\)

For a reaction to be spontaneous at all temperatures, what should be the signs of ∆H° and ∆S°?

1. +,+
2. +,-
3. -,+
4. -,-

Answer

c. -,+

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Exercise \(\PageIndex{5.1b}\)

A reaction that is not spontaneous at low temperatures becomes spontaneous at high temperatures.  Predict the sign of ∆H and ∆S.

1. +,+
2. +,-
3. -,+
4. -,-

Answer

a. +,+

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Exercise \(\PageIndex{5.1c}\)

When ammonium sulfate dissolves in water, the temperature of the solution decreased.  Predict the sign of ∆H and ∆S?

1. +,-
2. +,+
3. -,+
4. -,-

Answer

b. +,+

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Exercise \(\PageIndex{5.1d}\)

If a reaction is endothermic and nonspontaneous at 25°C, then it

1. can never be spontaneous
2. can become spontaneous by adding a catalyst
3. may be spontaneous at higher temperatures
4. may be spontaneous at lower temperatures
5. is exothermic and spontaneous at high temperatures

Answer

c. may be spontaneous at higher temperatures

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Exercise \(\PageIndex{5.1e}\)

If a reaction is exothermic and nonspontaneous at 25°C and 1 atm of pressure, it may be

1. spontaneous at higher temperatures
2. spontaneous at lower temperatures
3. endothermic at lowest temperatures
4. endothermic at higher temperatures
5. nonspontaneous at all temperatures

Answer

b. spontaneous at lower temperatures

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Exercise \(\PageIndex{5.1f}\)

For the following process:

\(Br_{2}(l)\rightarrow 2Br(g)\)

1. ∆H is + and ∆S is + for the reaction
2. ∆H is - and ∆S is - for the reaction
3. ∆H is + and ∆S is - for the reaction
4. ∆H is - and ∆S is + for the reaction
5. ∆G is + for all temperatures

Answer

a. ∆H is + and ∆S is + for the reaction

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Exercise \(\PageIndex{5.1g}\)

The normal boiling point of ammonia is 33°C. For the process

\(NH_{3}(l)\rightarrow NH_{3}(g)\)

at -40°C, the signs of ∆H, ∆S, and ∆G would be

   ∆G          ∆H          ∆S

1. –           –             –
2. –           +             +
3. +          +             +
4. 0           +             –
5. +           –             –

Answer

c.       +          +             +

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Exercise \(\PageIndex{5.1h}\)

The reaction \(Br_{2}(l)\rightarrow 2Br(g)\) is spontaneous at 1,600 °C. We can conclude that

1. ∆H is + and ∆S is + for the reaction
2. ∆H is - and ∆S is - for the reaction
3. ∆H is + and ∆S is - for the reaction
4. ∆H is - and ∆S is + for the reaction
5. ∆G is + for all temperatures

Answer

a. ∆H is + and ∆S is + for the reaction

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Exercise \(\PageIndex{5.1i}\)

For a particular chemical reaction, H^o is positive and S^o is positive. Which of the following statements about the spontaneity of the reaction under standard conditions is TRUE?

a)      The reaction can become spontaneous by lowering the temperature
b)     The reaction can become spontaneous by raising the temperature
c)      The reaction will be spontaneous at all temperatures
d)      The reaction is not spontaneous at any temperature.
e)      None of the above.

Answer

b)     The reaction can become spontaneous by raising the temperature

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Exercise \(\PageIndex{5.1j}\)

Which of the following reactions is unfavorable at low temperatures but becomes favorable as the temperature increases?

a.  2 CO(g) + O₂(g) --> CO₂(g); ∆H = -566 kJ; ∆S = -173 J/K
b.  2 H₂O(g) --> 2 H₂(g) + O₂(g); ∆H = 484 kJ; ∆S = 90.0 J/K
c.  2 N₂O(g) --> 2 N₂(g) + O₂(g); ∆H = -164 kJ; ∆S = 149 J/K
d.  PbCl₂(s) --> Pb²⁺(aq) + O₂(g); ∆H = 23.4 kJ; ∆S = -12.5 J/K
e.  none of the above

Answer

b.  2 H₂O(g) --> 2 H₂(g) + O₂(g); ∆H = 484 kJ; ∆S = 90.0 J/K

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Exercise \(\PageIndex{5.1k}\)

Which of the following reactions is unfavorable at low temperatures but becomes favorable as the temperature increases?

a.       2 CO(g) + O₂(g) --> CO₂(g); H^o = -566 kJ; S^o = -173 J/K

b.      2 H₂O(g) --> 2 H₂(g) + O₂(g); H^o = 484 kJ; S^o = 90.0 J/K

c.       2 N₂O(g) --> 2 N₂(g) + O₂(g); H^o = -164 kJ; S^o = 149 J/K

d.      PbCl₂(s) --> Pb²⁺(aq) + O₂(g); H^o = 23.4 kJ; S^o = -12.5 J/K

Answer

b.      2 H₂O(g) --> 2 H₂(g) + O₂(g); H^o = 484 kJ; S^o = 90.0 J/K 

Exercise \(\PageIndex{5.1l}\)

For a particular chemical reaction, H^o is positive and S^o is negative. Which of the following statements about the spontaneity of the reaction under standard conditions is TRUE?

1. 1. The reaction will be spontaneous only if the magnitude of H^o is large enough to overcome the unfavorable entropy change.
   2. The reaction will be spontaneous only if the magnitude of S^o is large enough to overcome the unfavorable enthalpy change.
   3. The reaction will be spontaneous regardless of the magnitudes of H^o and S^o.
   4. The reaction cannot be spontaneous.
   5. The reaction will be spontaneous only of G^o is positive.

Answer

The reaction cannot be spontaneous.

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Phase Transitions

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Exercise \(\PageIndex{5.2a}\)

What is the melting point of gold, Au, at 100kPa if ΔH°_fusion=12.36kJ/mol, and ΔS°_fusion = 9.2J/mol-K?

Answer

\[\Delta G=\Delta H-T\Delta S\nonumber\]

At the equilibrium, \(\Delta G=0\)

Therefore, \(\Delta H-T\Delta S=0\)

\[\Delta H=T\Delta S\nonumber\]

\[T=\frac{\Delta H}{\Delta S}=\frac{12.36}{0.0092}=1343K\nonumber\]

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Exercise \(\PageIndex{5.2b}\)

What is the state of gold if the temperature is 1200°C?

1. solid
2. liquid
3. gas
4. a mixture of solid and liquid

Answer

b. liquid

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Exercise \(\PageIndex{5.2c}\)

What is the state of gold if the temperature is 900°C?

1. solid
2. liquid
3. gas
4. a mixture of liquid and gas

Answer

a. solid

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Exercise \(\PageIndex{5.2d}\)

Knowing the freezing point of water, what is the ΔS°_fusion of water if ΔH°_fusion=6.01kJ/mol?

Answer

\[\Delta G=\Delta H-T\Delta S\nonumber\]

\(\Delta H-T\Delta S=0\) at equilibrium

\[\Delta H=T\Delta S\nonumber\]

\[\Delta S=\frac{\Delta H}{T}=\frac{6.01}{273.15}=22\,J/mol\nonumber\]

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Exercise \(\PageIndex{5.2e}\)

Knowing the normal boiling point of water, what is the ΔS°_vap of water if ΔH°_vap=40.7kJ/mol?

Answer

\[\Delta G=\Delta H-T\Delta S\nonumber\]

\(\Delta H-T\Delta S=0\) at equilibrium

\[\Delta H=T\Delta S\nonumber\]

\[\Delta S=\frac{\Delta H}{T}=\frac{40.7}{273.15+100}=109\,J/mol\nonumber\]

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Exercise \(\PageIndex{5.2f}\)

Methylene chloride, CH₂Cl₂, is a common organic solvent. Using the thermodynamic data below, calculate the boiling point (°C) for CH₂Cl₂.

Compound	ΔHf∘​ (kJ mol⁻¹)	S∘(J mol⁻¹ K⁻¹)
CH₂Cl₂(l)	−121.5	178.0
CH₂Cl₂(g)	−92.5	270.1

Answer

At the boiling point, the liquid and vapor are in equilibrium:
\[ \Delta G_{\text{vap}} = 0 = \Delta H_{\text{vap}} - T_b\,\Delta S_{\text{vap}} \]
\[ T_b = \frac{\Delta H_{\text{vap}}}{\Delta S_{\text{vap}}} \]

Enthalpy of vaporization:
\[ \Delta H_{\text{vap}} = \Delta H_f^\circ(\text{g}) - \Delta H_f^\circ(\text{l})
= (-92.5) - (-121.5) = 29.0\ \text{kJ mol}^{-1} \]
\[ \Delta H_{\text{vap}} = 29.0 \times 10^{3}\ \text{J mol}^{-1} \]

Entropy of vaporization:
\[ \Delta S_{\text{vap}} = S^\circ(\text{g}) - S^\circ(\text{l})
= 270.1 - 178.0 = 92.1\ \text{J mol}^{-1}\text{K}^{-1} \]

Solve for \(T_b\):
\[ T_b = \frac{29.0 \times 10^{3}}{92.1} = 315.0\ \text{K} \]
\[ T_b = 315.0 - 273.15 = 41.9^\circ\text{C} \]

Final Answer: \(\boxed{T_b = 41.9^\circ\text{C}}\)
 

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Exercise \(\PageIndex{5.2g}\)

 The melting point of tungsten (used in some light bulbs) is the second highest among the elements (only that of carbon is higher). If the enthalpy of fusion is 35.23 kJ/mol and the entropy of fusion is 9.65 J/mol K, calculate the melting point (in K) of tungsten.

Answer

\[
\Delta G = \Delta H - T\Delta S
\]

At the melting point, solid and liquid tungsten are at equilibrium, so

\[
\Delta G = 0
\]

Therefore,

\[
0 = \Delta H_{\text{fus}} - T\Delta S_{\text{fus}}
\]

Solving for \(T\),

\[
T = \frac{\Delta H_{\text{fus}}}{\Delta S_{\text{fus}}}
\]

Convert \(\Delta H_{\text{fus}}\) from kJ/mol to J/mol so the units match:

\[
35.23 \, \text{kJ/mol} \times \frac{1000 \, \text{J}}{1 \, \text{kJ}}
= 35230 \, \text{J/mol}
\]

Now substitute:

\[
T = \frac{35230 \, \text{J/mol}}{9.65 \, \text{J/mol K}}
\]

\[
T = 3650.8 \, \text{K}
\]

With three significant figures,

\[
\boxed{T = 3.65 \times 10^3 \, \text{K}}
\]

or

\[
\boxed{T = 3650 \, \text{K}}
\]

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Driving Chemical Reactions

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Exercise \(\PageIndex{5.3a}\)

\(MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\,\,\,\,\,\Delta G^{0}=69.27\,kJ/mol\)

Does the reaction favor products or reactants?

1. reactant
2. product
3. both
4. none of the above

Answer

a. reactant

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Exercise \(\PageIndex{5.3b}\)

\(MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\,\,\,\,\,\Delta G^{0}=69.27\,kJ/mol\)

What is the expression for Q?

Answer

\[Q=\left ( P_{HCl} \right )^{2}\nonumber\]

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Exercise \(\PageIndex{5.3c}\)

\(MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\,\,\,\,\,\Delta G^{0}=69.27\,kJ/mol\)

At 298K,  what minimal pressures would the reaction proceed to the left?

Answer

\[K=\left ( P_{HCl} \right )^{2}=e^{\frac{-\Delta G^{0}}{RT}}=e^{\frac{-69.27kJ/mol}{(0.008314kJ/mol \cdot K)(298K)}}=7.2*10^{-13}\nonumber\]

\[\left ( P_{HCl} \right )^{2}>7.2*10^{-13}\nonumber \]

\[P_{HCl}>\sqrt{7.2*10^{-13}}\nonumber\]

\[P_{HCl}>8.5*10^{-7}\nonumber\]

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Exercise \(\PageIndex{5.3d}\)

\(MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\,\,\,\,\,\Delta G^{0}=69.27\,kJ/mol\)

What is the ∆G for P_HCl = 5.2*10^-4?

Answer

\[\Delta G=\Delta G^{0}+RTlnQ\nonumber\]

\[\Delta G=69.27\,kJ/mol+\left ( 0.008315kJ/(mol*K)*298.15K*ln(5.2*10^{-4})^{2} \right )\nonumber\]

\[\Delta G=69.27\,kJ/mol+\left ( -37.49\,kJ/mol \right )=31.78\,kJ/mol\nonumber\]

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Exercise \(\PageIndex{5.3e}\)

What is the ∆G and does the reaction proceed towards products or reactants?

2NO(g)     +	Cl2(g)	⟶	2NOCl (g)
4.9*10^-6M	3.1*10^-3M		2.4*10^-1M

Answer

\[ 2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g) \nonumber\]

\[\Delta G = \Delta G^o + RTlnQ \nonumber \]

\[\Delta G^{0}_{rxn}=\sum n\Delta G^{0}_{f}(products)-\sum m\Delta G^{0}_{f}(reactants)\nonumber\]

\[\Delta G^{0}_{rxn}=\left [ 2(66.3\,kJ/mol) \right ] - \left [ 2(86.71\,kJ/mol)+0\,kJ/mol \right ]=-40.82\,kJ/mol\nonumber\]

\[Q=\frac{\left ( 2.4*10^{-1} \right )^{2}}{\left ( 4.9*10^{-6} \right )^{5}\left ( 3.1*10^{-3} \right )}=7.7*10^{11}\nonumber\]

\[\Delta G=-40.82\,kJ/mol+(0.008315kJ/(mol*K)*298.15K*ln(7.7*10^{11}))=27.1 kJ/mol\nonumber\]

So it will produce more reactants

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Exercise \(\PageIndex{5.3f}\)

What is the ∆G and does the reaction proceed towards products or reactants?

2NOCl (g)	⟶	2NO(g)     +	Cl2(g)
3.1*10^-1M		2.4*10^-6M	4.9*10^-3M

Answer

\[2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\nonumber\]

\[\Delta G = \Delta G^o + RTlnQ \nonumber \]

\[\Delta G^{0}_{rxn}=\sum n\Delta G^{0}_{f}(products)-\sum m\Delta G^{0}_{f}(reactants)\nonumber\]

\[\Delta G^{0}_{rxn}=\left [ 2(86.71\,kJ/mol)+0\,kJ/mol \right ]-\left [ 2(66.3\,kJ/mol) \right ]=40.82\,kJ/mol\nonumber\]

\[Q=\frac{\left ( 2.4*10^{-6} \right )^{2}\left ( 4.9*10^{-3} \right )}{\left ( 3.1*10^{-1} \right )^{2}}=2.9*10^{-13}\nonumber\]

\[\Delta G=40.82\,kJ/mol+(0.008315\,kJ/(mol*K)*298.15K*ln(2.9*10^{-13}))=-30.4\,kJ/mol\nonumber\]

So it will produce more products.

 

Gibbs Free Energy

Standard State Calculations

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Exercise \(\PageIndex{6.1a}\)

Balance the following equations

1. Calculate the change of the Gibbs free energy for the reaction at 25°C, where standard free energy of formation of C₂H₄(g), H₂O(g), C₂H₅OH(l) are 68, -229, -175 kJ/mol respectively.

   \(C_{2}H_{4}(g)+H_{2}O(g)\rightarrow C_{2}H_{5}OH(l)\)

2. What is the change of entropy if S⁰ of C₂H₄(g), H₂O(g), C₂H₅OH(l) are 219, 189, and 161 J/K*mol. 
3. Using the results above, what is the change of enthalpy at 25°C for the same reaction?

Answer a

\[\Delta G^{0}=-175-\left [ \left ( -229 \right )+68 \right ]=-14kJ/mol\nonumber\]

Answer b

\[\Delta S^{0}=161-189-219=-247\,J/(mol*K)\nonumber\]

Answer c

\[\Delta H^{0}=\Delta G^{0}+T\Delta S^{0}\nonumber\]

\[\Delta H^{0}=-14+298*\left ( -0.247 \right )=-87.606\,kJ/mol\nonumber\]

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Exercise \(\PageIndex{6.1b}\)

Using a table of Standard State Thermodynamic Values (there is one in the references tab of LibreTexts), answer the questions below. Your answer may differ slightly due to different tables in different texts.

\(MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\)

1. Find the ΔH_rxn for the reaction above.
2. Find the ΔS_rxn for the reaction above.
3. Find the ΔG_rxn for the reaction above.

Answer a.

\[MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\nonumber\]

\[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta H^{0}_{rxn}=\left [ -601.8\,kJ/mol+2(-92.30\,kJ/mol) \right ]-\left [ -641.6\,kJ/mol+-285.83\,kJ/mol \right ]=141.03\,kJ/mol\nonumber\]

Answer b.

\[MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\nonumber\]

\[\Delta S^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta S^{0}_{rxn}=\left [ 26.8\,J/(mol*K)+2(186.69\,J/(mol*K)) \right ]-\left [ 89.6\,J/(mol*K)+69.91\,J/(mol*K) \right ]=240.67\,J/(mol*K)\nonumber\]

Answer c.

\[MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\nonumber\]

\[\Delta G^{0}_{rxn}=\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}\nonumber\]

\[\Delta G^{0}_{rxn}=141.03\,kJ/mol-298.15K(0.24067\,kJ/mol*K)=69.27\,kJ/mol\nonumber\]

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Exercise \(\PageIndex{6.1c}\)

Using a table of Standard State Thermodynamic Values (there is one in the references tab of LibreTexts), answer the questions below. Your answer may differ slightly due to different tables in different texts.

\(2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\)

1. Find the ΔH_rxn for the reaction above.
2. Find the ΔS_rxn for the reaction above.
3. Find the ΔG_rxn for the reaction above.

Answer a.

\[2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\nonumber\]

\[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta H^{0}_{rxn}=\left [ 2(90.37\,kJ/mol)+0\,kJ/mol \right ]-\left [ 2(52.6\,kJ/mol) \right ]=75.54\,kJ/mol\nonumber\]

Answer b.

\[2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\nonumber\]

\[Delta S^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta S^{0}_{rxn}=\left [ 2(210.62\,J/(mol*K))+222.96\,J/(mol*K) \right ]-\left [ 2(264\,J/(mol*K)) \right ]=116.2\,J/(mol*K)\nonumber\]

Answer c.

\[2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\nonumber\]

\[\Delta G^{0}_{rxn}=\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}\nonumber\]

\[\Delta G^{0}_{rxn}=75.54\,kJ/mol-298.15K(0.1162kJ/(mol*K))=40.89\,kJ/mol\nonumber\]

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Exercise \(\PageIndex{6.1d}\)

Using a table of Standard State Thermodynamic Values (there is one in the references tab of LibreTexts), answer the questions below. Your answer may differ slightly due to different tables in different texts.

\(2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\)

1. Find the ΔH_rxn for the reaction above.
2. Find the ΔS_rxn for the reaction above.
3. Find the ΔG_rxn for the reaction above.

Answer a.

\[2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\nonumber\]

\[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\nonumber\]

\[\Delta H^{0}_{rxn}=\left [ 2(52.6\,kJ/mol) \right ]-\left [ 2(90.37\,kJ/mol)+0\,kJ/mol \right ]=-75.54\,kJ/mol\nonumber\nonumber\]

Answer b.

\[2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\nonumber\]

\[Delta S^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta S^{0}_{rxn}=\left [ 2(264\,J/(mol*K))\right ]+\left [ 2(210.62\,J/(mol*K))+222.96\,J/(mol*K) \right ]=-116.2\,J/(mol*K)\nonumber\]

Answer c.

\[2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\nonumber\]

\[\Delta G^{0}_{rxn}=\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}\nonumber\]

\[\Delta G^{0}_{rxn}=-75.54\,kJ/mol-298.15K(-0.1162kJ/(mol*K))=-40.89\,kJ/mol\nonumber\]

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Exercise \(\PageIndex{6.1e}\)

At elevated temperatures, hydrogen iodide, HI, decomposes to produce hydrogen, H₂, and iodine, I₂, as shown in the equation:     2 HI (g) -> H₂ (g) + I₂ (g)

Using the thermodynamic data below, calculate the standard free energy change (in kJ) for this reaction at 900^o C.

compound	Hf^o, kJ mol^-1	S^o, J mol^-1 K^-1
HI(g)	25.9	206.3
H2(g)	0	131.0
I2(g)	62.4	260.6

Answer

\[
\mathrm{2HI(g) \rightarrow H_2(g) + I_2(g)}
\]

Use

\[
\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ
\]

First calculate \(\Delta H^\circ_{\text{rxn}}\):

\[
\Delta H^\circ_{\text{rxn}}
=
\sum n\Delta H_f^\circ(\text{products})
-
\sum n\Delta H_f^\circ(\text{reactants})
\]

\[
\Delta H^\circ_{\text{rxn}}
=
\left[1(0) + 1(62.4)\right]
-
\left[2(25.9)\right]
\]

\[
\Delta H^\circ_{\text{rxn}}
=
62.4 - 51.8
\]

\[
\Delta H^\circ_{\text{rxn}}
=
10.6 \, \text{kJ}
\]

Now calculate \(\Delta S^\circ_{\text{rxn}}\):

\[
\Delta S^\circ_{\text{rxn}}
=
\sum nS^\circ(\text{products})
-
\sum nS^\circ(\text{reactants})
\]

\[
\Delta S^\circ_{\text{rxn}}
=
\left[1(131.0) + 1(260.6)\right]
-
\left[2(206.3)\right]
\]

\[
\Delta S^\circ_{\text{rxn}}
=
391.6 - 412.6
\]

\[
\Delta S^\circ_{\text{rxn}}
=
-21.0 \, \text{J/K}
\]

Convert entropy to \(\text{kJ/K}\):

\[
-21.0 \, \text{J/K}
\times
\frac{1 \, \text{kJ}}{1000 \, \text{J}}
=
-0.0210 \, \text{kJ/K}
\]

Convert temperature to Kelvin:

\[
T = 900^\circ \text{C} + 273.15 = 1173.15 \, \text{K}
\]

Now substitute into the free energy equation:

\[
\Delta G^\circ
=
10.6 \, \text{kJ}
-
(1173.15 \, \text{K})(-0.0210 \, \text{kJ/K})
\]

\[
\Delta G^\circ
=
10.6 \, \text{kJ}
+
24.6 \, \text{kJ}
\]

\[
\Delta G^\circ
=
35.2 \, \text{kJ}
\]

\[
\boxed{\Delta G^\circ = +35.2 \, \text{kJ}}
\]

The positive value means the reaction is not spontaneous under standard-state conditions at \(900^\circ \text{C}\).

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Gibbs Free Energy and K

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Exercise \(\PageIndex{6.2a}\)

The equilibrium constant for a reaction is 0.48 at 25°C. What is the value of ΔG° at this temperature? R=8.314J/K*mol

Answer

\[\Delta G^{0}=-RTlnK=-8.314*298*ln{0.48}=1818.46\,J/mol\nonumber\]

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Exercise \(\PageIndex{6.2b}\)

Using the information below, calculate the value of K at 25°C for the reaction of C(s) and O₂(g) to form CO.

\[C(s)+O_{2}(g)\rightarrow CO_{2}\,\,\,\,\,\Delta G^{0}=-394.4\,kJ/mol\nonumber\]

\[CO(g)+\frac{1}{2}O_{2}(g)\rightarrow CO_{2}\,\,\,\,\,\Delta G^{0}=-257.2\,kJ/mol\nonumber\]

Answer

Applying Hess's Law (section 5.7.1)
\[C(s)+O_{2}(g)\rightarrow \cancel{CO_{2}(g)}\,\,\,\,\,\Delta G^{0}=-394.4\,kJ/mol\nonumber\]

\[+\cancel{CO_{2}(g)}\rightarrow CO(g)+\frac{1}{2}O_{2}\,\,\,\,\,\Delta G^{0}=257.2\,kJ/mol\nonumber\]

---

\[C(s)+\frac{1}{2}O_{2}(g)\rightarrow CO(g)\,\,\,\,\,\Delta G^{0}=-137.2\,kJ/mol\]

\[K=e^{-\frac{\Delta G^{0}}{RT}}=e^{-\frac{-137.2}{0.008314*298}}=1.12*10^{24}\]

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Exercise \(\PageIndex{6.2c}\)

The equilibrium constant for a reaction is 0.48 at 25°C. What is the value of ΔG° at this temperature? R=8.314J/K*mol

Answer

\[\Delta G^{0}=-RTlnK=-8.314*298*ln{0.48}=1818.46\,J/mol\nonumber\]

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Exercise \(\PageIndex{6.2d}\)

Using the reaction and the giving the following data to answer the following questions.

\(Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)\)

	∆Hf° (kJ/mol)	S° (J/mol*K)
Ag^+ (aq)	105.9	73.93
Cl^- (aq)	-167.2	56.5
AgCl (s)	-127.0	96.11

1. Determine the value for ΔS°.
2. Determine the value for ΔH°.
3. Determine the value for ΔG°.
4. Determine the value for K.

Answer a

\[ \begin{align}\Delta S^{0} & = S^{0}(AgCl(s)) -[ S^{0}(Ag^+(aq)) + S^{0}(Cl^-(aq)) \nonumber \\ &= 96.11-[56.5+73.93] \nonumber \\ &=-34.32\,J/(mol*K)\nonumber\end{align}   \nonumber  \]

Answer b

\[\begin{align}\Delta H^{0} & =\Delta H^{0}(AgCl(s)) -[\Delta H^{0}(Ag^+(aq)) + \Delta H^{0}(Cl^-(aq)) \nonumber \\ \; &= -127.0-[105.9-167.2] \nonumber \\ \; &=-65.7 \,kJ/mol \nonumber \end{align} \nonumber\]

Answer c

\[\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}=-65.7-298*(-0.03432)=-55.5\,kJ/mol\nonumber\]

Answer d

\[K=e^{-\frac{\Delta G^{0}}{RT}}=e^{-\frac{-55.5}{0.008314*298}}=5.35*10^{9}\nonumber\]

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Exercise \(\PageIndex{6.2e}\)

Find the K for the reaction at the following temperatures.

\(MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\)

1. Temperature at 25°C?
2. Temperature at 20°C?
3. Temperature at 100°C?

Answer a.

\[MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\nonumber\]

\[\Delta G^{0}_{rxn}=\sum n\Delta G^{0}_{f}(products)-\sum m\Delta G^{0}_{f}(reactants)\nonumber\]

\[\Delta G^{0}_{rxn}=\left [ -569.6\,kJ/mol+2(-95.27\,kJ/mol) \right ]-\left [ -592.1\,kJ/mol+-237.13\,kJ/mol \right ]=69.09\,kJ/mol\nonumber\]

\[K=e^{\frac{\Delta G^{0}_{rxn}}{-RT}}\nonumber\]

\[K=e^{\frac{69.09\,kJ/mol}{-0.008314(298)}}=7.871*10^{-13}\nonumber\]

Answer b.

\[MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\nonumber\]

\[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta H^{0}_{rxn}=\left [ -601.8\,kJ/mol+2(-92.30\,kJ/mol) \right ]-\left [ -641.6\,kJ/mol+-285.83\,kJ/mol \right ]=141.03\,kJ/mol\nonumber\]

\[Delta S^{0}_{rxn}=\sum \Delta S_{f}^{0}n(products)-\sum \Delta S_{f}^{0}m(reactants)\nonumber\]

\[\Delta S^{0}_{rxn}=\left [ 26.8\,J/(mol*K)+2(186.69\,J/(mol*K))\right ]+\left [ 89.6\,J/(mol*K)+69.91\,J/(mol*K) \right ]=240.67\,J/(mol*K)\nonumber\]

\[K=e^{\frac{\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}}{-RT}}\nonumber\]

\[K=e^{\frac{141.03\,kJ/mol-293.15K(0.24067\,kJ/(mol*K))}{-0.008314(293.15)}}=4.425*10^{-13}\nonumber\]

Answer c.

\[MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\nonumber\]

\[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta H^{0}_{rxn}=\left [ -601.8\,kJ/mol+2(-92.30\,kJ/mol) \right ]-\left [ -641.6\,kJ/mol+-285.83\,kJ/mol \right ]=141.03\,kJ/mol\nonumber\]

\[Delta S^{0}_{rxn}=\sum \Delta S_{f}^{0}n(products)-\sum \Delta S_{f}^{0}m(reactants)\nonumber\]

\[\Delta S^{0}_{rxn}=\left [ 26.8\,J/(mol*K)+2(186.69\,J/(mol*K))\right ]+\left [ 89.6\,J/(mol*K)+69.91\,J/(mol*K) \right ]=240.67\,J/(mol*K)\nonumber\]

\[K=e^{\frac{\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}}{-RT}}\nonumber\]

\[K=e^{\frac{141.03\,kJ/mol-373.15K(0.24067\,kJ/(mol*K))}{-0.008314(373.15)}}=6.756*10^{-8}\nonumber\]

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Exercise \(\PageIndex{6.2f}\)

Find the K for the reaction at the following temperatures.

\(2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\)

1. Temperature at 25°C?
2. Temperature at 15°C?
3. Temperature at 150°C?

Answer a.

\[2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\nonumber\]

\[\Delta G^{0}_{rxn}=\sum n\Delta G^{0}_{f}(products)-\sum m\Delta G^{0}_{f}(reactants)\nonumber\]

\[\Delta G^{0}_{rxn}=\left [ 2(86.71\,kJ/mol)+0\,kJ/mol \right ]-\left [ 2(66.3\,kJ/mol) \right ]=40.82\,kJ/mol\nonumber\]

\[K=e^{\frac{\Delta G^{0}_{rxn}}{-RT}}\nonumber\]

\[K=e^{\frac{40.82\,kJ/mol}{-0.008314(298.15)}}=7.058*10^{-8}\nonumber\]

Answer b.

\[2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\nonumber\]

\[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta H^{0}_{rxn}=\left [ 2(90.37\,kJ/mol)+0\,kJ/mol \right ]-\left[ 2(52.6\,kJ/mol) \right ]=75.54\,kJ/mol\nonumber\]

\[Delta S^{0}_{rxn}=\sum \Delta S_{f}^{0}n(products)-\sum \Delta S_{f}^{0}m(reactants)\nonumber\]

\[\Delta S^{0}_{rxn}=\left [ 2(210.62\,J/(mol*K))+222.96\,J/(mol*K)\right ]-\left [ 2(264\,J/(mol*K)) \right ]=116.2\,J/(mol*K)\nonumber\]

\[K=e^{\frac{\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}}{-RT}}\nonumber\]

\[K=e^{\frac{75.54\,kJ/mol-288.15K(0.1162\,kJ/(mol*K))}{-0.008314(288.15)}}=2.378*10^{-8}\nonumber\]

Answer c.

\[2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\nonumber\]

\[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta H^{0}_{rxn}=\left [ 2(90.37\,kJ/mol)+0\,kJ/mol \right ]-\left [ 2(52.6\,kJ/mol)\right ]=75.54\,kJ/mol\nonumber\]

\[Delta S^{0}_{rxn}=\sum \Delta S_{f}^{0}n(products)-\sum \Delta S_{f}^{0}m(reactants)\nonumber\]

\[\Delta S^{0}_{rxn}=\left [ 2(210.62\,J/(mol*K))+222.96\,J/(mol*K)\right ]+\left [ 2(264\,J/(mol*K)) \right ]=116.2\,J/(mol*K)\nonumber\]

\[K=e^{\frac{\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}}{-RT}}\nonumber\]

\[K=e^{\frac{75.54\,kJ/mol-423.15K(0.1162\,kJ/(mol*K))}{-0.008314(423.15)}}=5.558*10^{-4}\nonumber\]

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Exercise \(\PageIndex{6.2g}\)

Find the K for the reaction at the following temperatures.

\(2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\)

1. Temperature at 25°C?
2. Temperature at 10°C?
3. Temperature at 200°C?

Answer a.

\[2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\nonumber\]

\[\Delta G^{0}_{rxn}=\sum n\Delta G^{0}_{f}(products)-\sum m\Delta G^{0}_{f}(reactants)\nonumber\]

\[\Delta G^{0}_{rxn}=\left [ 2(66.3\,kJ/mol) \right ]-\left [ 2(86.71\,kJ/mol)+0\,kJ/mol \right ]=-40.82\,kJ/mol\nonumber\]

\[K=e^{\frac{\Delta G^{0}_{rxn}}{-RT}}\nonumber\]

\[K=e^{\frac{-40.82\,kJ/mol}{-0.008314(298.15)}}=1.417*10^{7}\nonumber\]

Answer b.

\[2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\nonumber\]

\[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta H^{0}_{rxn}=\left[ 2(52.6\,kJ/mol) \right ]-\left [ 2(90.37\,kJ/mol)+0\,kJ/mol \right ]=-75.54\,kJ/mol\nonumber\]

\[Delta S^{0}_{rxn}=\sum \Delta S_{f}^{0}n(products)-\sum \Delta S_{f}^{0}m(reactants)\nonumber\]

\[\Delta S^{0}_{rxn}=\left [ 2(264\,J/(mol*K)) \right ]-\left [ 2(210.62\,J/(mol*K))+222.96\,J/(mol*K)\right ]=-116.2\,J/(mol*K)\nonumber\]

\[K=e^{\frac{\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}}{-RT}}\nonumber\]

\[K=e^{\frac{-75.54\,kJ/mol-283.15K(-0.1162\,kJ/(mol*K))}{-0.008314(283.15)}}=7.337*10^{7}\nonumber\]

Answer c.

\[2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\nonumber\]

\[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\]

\[\Delta H^{0}_{rxn}=\left[ 2(52.6\,kJ/mol) \right ]-\left [ 2(90.37\,kJ/mol)+0\,kJ/mol \right ]=-75.54\,kJ/mol\nonumber\]

\[Delta S^{0}_{rxn}=\sum \Delta S_{f}^{0}n(products)-\sum \Delta S_{f}^{0}m(reactants)\nonumber\]

\[\Delta S^{0}_{rxn}=\left [ 2(264\,J/(mol*K)) \right ]-\left [ 2(210.62\,J/(mol*K))+222.96\,J/(mol*K)\right ]=-116.2\,J/(mol*K)\nonumber\]

\[K=e^{\frac{\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}}{-RT}}\nonumber\]

\[K=e^{\frac{-75.54\,kJ/mol-473.15K(-0.1162\,kJ/(mol*K))}{-0.008314(473.15)}}=1.861*10^{2}\nonumber\]

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Exercise \(\PageIndex{6.2h}\)

The following reaction has a standard free energy change value of 42.6 kJ/mol at 25^oC:

HB(aq) + H₂O(l) <=> H₃O⁺(aq) + B^-(aq)

The K_a of the acid HB is:

Answer

**Solution**

We are given:

\[
\Delta G^\circ = 42.6\ \text{kJ mol}^{-1} = 42.6 \times 10^3\ \text{J mol}^{-1}
\]
\[
T = 25^\circ \text{C} = 298.15\ \text{K}
\]

The relationship between the standard free energy change and the equilibrium constant is:

\[
\Delta G^\circ = -RT \ln K
\]

Rearranging for \( K \):

\[
K = e^{-\Delta G^\circ / (RT)}
\]

Substituting the known values (\( R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} \)):

\[
K = e^{-\frac{42.6 \times 10^3}{(8.314)(298.15)}}
\]

\[
K = e^{-17.2} = 3.3 \times 10^{-8}
\]

Since this is the equilibrium constant for the **acid dissociation reaction**,

\[
K_a = 3.3 \times 10^{-8}
\]

**Final Answer**

\[
\boxed{K_a = 3.3 \times 10^{-8}}
\]
 

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Exercise \(\PageIndex{6.2i}\)

The following reaction is spontaneous as written:

     Zn(s) + Cu²⁺(aq) --> Zn²⁺(aq) + Cu(s)

Which of the following statements is TRUE?

a.  K_eq > 1 and G^o = 0
b.  K_eq > 1 and G^o < 0
c.  K_eq < 1 and G^o < 0
d.  K_eq > 1 and G^o > 0
e.  K_eq < 1 and G^o = 0

Answer

b.  K_eq > 1 and G^o < 0

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Exercise \(\PageIndex{6.2j}\)

The K_sp for the sparingly soluble salt barium fluoride (BaF₂) at 25^oC is equal to
1.17 x 10^-7. Calculate G^o (in kJ) for the following reaction.

                         BaF₂(s) <==> Ba²⁺(aq) +2F^-(aq)

Answer

 

Given \(K_{sp} = 1.17\times 10^{-7}\) at \(T=298.15\ \text{K}\) for the dissolution
\[
\text{BaF}_2(s) \leftrightharpoons \text{Ba}^{2+}(aq) + 2\,\text{F}^-(aq),
\]
the standard free energy change is
\[
\Delta G^\circ = -RT \ln K_{sp}.
\]

\[
\Delta G^\circ
= -(8.3144626\ \text{J mol}^{-1}\text{K}^{-1})(298.15\ \text{K})\,\ln(1.17\times10^{-7})
= 3.9566860971\times 10^{4}\ \text{J mol}^{-1}
= 39.566860971\ \text{kJ mol}^{-1}.
\]

Final Answer: \(\boxed{\Delta G^\circ \approx 39.6\ \text{kJ mol}^{-1}}\)  (rounded at the end)
 

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Exercise \(\PageIndex{6.2k}\)

\(\Delta\)G^o for the following reaction is +55.60kJ.  What is DG for the at 25^oC for the reaction  if [Ag⁺] = [Cl^-] = 1.3 x 10^-5M

                                    AgCl(s)  < == >  Ag⁺  +  Cl^-

a.  50.9 kJ        b. 111.4kJ        c. –111.4kJ      d. 158kJ           e. –0.18kJ

Answer

Given \( \Delta G^\circ = 55.60\times 10^{3}\ \text{J mol}^{-1} \), \(T=298.15\ \text{K}\), and \([ \text{Ag}^+ ] = [ \text{Cl}^- ] = 1.3\times 10^{-5}\ \text{M}\).

\[
\Delta G = \Delta G^\circ + RT\ln Q,\qquad Q=[\text{Ag}^+][\text{Cl}^-]=\big(1.3\times 10^{-5}\big)^2
\]

\[
\Delta G
= 55.60\times 10^{3}
+ (8.3144626)(298.15)\,\ln\!\big(\,(1.3\times 10^{-5})^2\,\big)
= -1.793\times 10^{2}\ \text{J mol}^{-1}
= -0.179\ \text{kJ mol}^{-1}
\]

Final Answer: \(\boxed{\Delta G \approx -0.18\ \text{kJ mol}^{-1}}\)  (rounded at the end)
 

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Exercise \(\PageIndex{6.2l}\)

1.  Consider the following reaction,

2SO₂(g) + O₂(g) <==> 2SO₃(g)

and the following thermochemical data,

substance	Gf^o, kJ/mol
SO2(g)	-300.194
O2(g)	0.0
SO3(g)	-371.06

Calculate the value of the equilibrium constant for this reaction at 25^oC.

Answer

\textbf{Given reaction:}
\[
2\,\mathrm{SO_2(g)} + \mathrm{O_2(g)} \;\rightleftharpoons\;
2\,\mathrm{SO_3(g)}
\]

\[
\begin{array}{c|c}
\text{substance} & \Delta G_f^\circ\;(\mathrm{kJ\,mol^{-1}}) \\\hline
\mathrm{SO_2(g)} & -300.194 \\
\mathrm{O_2(g)}  & 0.000 \\
\mathrm{SO_3(g)} & -371.06
\end{array}
\]

\textbf{Step 1: Calculate } \(\Delta G^\circ_{\mathrm{rxn}}\).

\[
\Delta G^\circ_{\mathrm{rxn}}
= \sum \nu\,\Delta G_f^\circ(\text{products})
  - \sum \nu\,\Delta G_f^\circ(\text{reactants})
\]

\[
\Delta G^\circ_{\mathrm{rxn}}
= 2\bigl(\Delta G_f^\circ(\mathrm{SO_3})\bigr)
  - \Bigl[2\bigl(\Delta G_f^\circ(\mathrm{SO_2})\bigr)
           + 1\bigl(\Delta G_f^\circ(\mathrm{O_2})\bigr)\Bigr]
\]

\[
\Delta G^\circ_{\mathrm{rxn}}
= 2(-371.06)
  - \Bigl[2(-300.194) + 0.000\Bigr]\;\mathrm{kJ\,mol^{-1}}
\]

\[
\Delta G^\circ_{\mathrm{rxn}}
= -742.12 - (-600.388)
= -141.732\;\mathrm{kJ\,mol^{-1}}
\]

Convert to joules:
\[
\Delta G^\circ_{\mathrm{rxn}}
= -1.41732\times 10^{5}\;\mathrm{J\,mol^{-1}}
\]

\textbf{Step 2: Relate \(\Delta G^\circ\) to the equilibrium constant \(K\).}

For a reaction at constant \(T\) and \(P\),
\[
\Delta G^\circ = -RT\ln K
\]

Equivalently, solving directly for \(K\),
\[
K = e^{\left(-\frac{\Delta G^\circ}{RT}\right)}
\]

At \(T = 298\;\mathrm{K}\) and \(R = 8.314\;\mathrm{J\,mol^{-1}K^{-1}}\),
\[
K
= e^{\left(
  -\frac{\Delta G^\circ_{\mathrm{rxn}}}{RT}
  \right)}
= e^{\left(
  -\frac{-1.41732\times 10^{5}\;\mathrm{J\,mol^{-1}}}
        {(8.314\;\mathrm{J\,mol^{-1}K^{-1}})(298\;\mathrm{K})}
  \right)}
\]

Now evaluate the exponent:
\[
K = e^{(57.2)} \approx 7.0\times 10^{24}
\]

\[
\boxed{K \approx 7.0\times 10^{24}\ \text{at }25^\circ\mathrm{C}}
\]
 

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Exercise \(\PageIndex{6.2m}\)

The K_sp for the sparingly soluble salt barium chromate (BaCrO₄) at 25^oC is equal to
2.0 x 10^-10. Calculate G^o (in kJ) for the following reaction.

BaCrO₄(s) <==> Ba²⁺(aq) + CrO₄^2-(aq)

.

Answer

\[
\mathrm{BaCrO_4(s) \rightleftharpoons Ba^{2+}(aq) + CrO_4^{2-}(aq)}
\]

For this dissolution reaction,

\[
K = K_{sp}
\]

and the relationship between standard free energy and the equilibrium constant is

\[
\Delta G^\circ = -RT \ln K
\]

Substitute the given value of \(K_{sp}\):

\[
\Delta G^\circ = -(8.314 \, \text{J/mol K})(298.15 \, \text{K})\ln(2.0 \times 10^{-10})
\]

\[
\ln(2.0 \times 10^{-10}) = -22.33
\]

\[
\Delta G^\circ = -(8.314)(298.15)(-22.33)
\]

\[
\Delta G^\circ = 5.53 \times 10^4 \, \text{J/mol}
\]

Convert to kJ/mol:

\[
\Delta G^\circ = 55.3 \, \text{kJ/mol}
\]

\[
\boxed{\Delta G^\circ = +55 \, \text{kJ/mol}}
\]

The positive value means that dissolving solid barium chromate under standard-state conditions is not spontaneous, which is consistent with its very small \(K_{sp}\).

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Exercise \(\PageIndex{6.2n}\)

The following reaction is spontaneous as written:

Zn(s) + Cu²⁺(aq) --> Zn²⁺(aq) + Cu(s)

Which of the following statements is TRUE?

a.  K_eq > 1 and G^o = 0                     b.  K_eq > 1 and G^o < 0
c.  K_eq < 1 and G^o < 0                     d.  K_eq > 1 and G^o > 0
e.  K_eq < 1 and G^o = 0

Answer

     b.  K_eq > 1 and G^o < 0

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Exercise \(\PageIndex{6.2o}\)

Calculate the value of K_a for hydrofluoric acid (HF) at 25^oC if the standard free energy change for the following reaction is equal to +17.9 kJ/mol:

HF (g) + H₂O (l) <=> H₃O⁺ (aq) + F^- (aq)

a.  4.0 x 10^-38               b.  7.3 x 10^-4                c. 9.2 x 10^-1                 d.  9.9 x 10^-1      e.    1.4 x 10³

Answer

\[
\mathrm{HF(g) + H_2O(l) \rightleftharpoons H_3O^+(aq) + F^-(aq)}
\]

The relationship between standard free energy and the equilibrium constant is

\[
\Delta G^\circ = -RT \ln K
\]

For this reaction, the equilibrium constant is the acid ionization constant:

\[
K = K_a
\]

Rearrange the equation to solve for \(K_a\):

\[
\ln K_a = -\frac{\Delta G^\circ}{RT}
\]

Convert \(\Delta G^\circ\) from kJ/mol to J/mol:

\[
17.9 \, \text{kJ/mol} \times \frac{1000 \, \text{J}}{1 \, \text{kJ}}
= 17900 \, \text{J/mol}
\]

Now substitute:

\[
\ln K_a = -\frac{17900 \, \text{J/mol}}
{(8.314 \, \text{J/mol K})(298.15 \, \text{K})}
\]

\[
\ln K_a = -7.22
\]

Therefore,

\[
K_a = e^{-7.22}
\]

\[
K_a = 7.3 \times 10^{-4}
\]

\[
\boxed{K_a = 7.3 \times 10^{-4}}
\]

The positive value of \(\Delta G^\circ\) gives a value of \(K_a < 1\), which is consistent with HF being a weak acid.

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General Questions

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Exercise \(\PageIndex{7a}\)

Which of the following must have a negative value for an exothermic process?

1. enthalpy change
2. entropy change
3. free energy change
4. electrode cell potential
5. equilibrium constant

Answer

a. enthalpy change

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Exercise \(\PageIndex{7b}\)

Which of the following is true when one mole of solid naphthalene (mothballs) sublimes to gas?

1. The entropy increases
2. The entropy decreases
3. The enthalpy increases
4. The enthalpy decreases

 

1. 1 only
2. 2 only
3. 1 and 3 only
4. 2 and 3 only
5. 1 and 4 only

Answer

c. 1 and 3 only

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Exercise \(\PageIndex{7c}\)

The total entropy of a system and its surroundings always increases for a spontaneous process. This is a statement of

1. the law of constant composition
2. the first law of thermodynamics
3. the second law of thermodynamics
4. the third law of thermodynamics
5. the law of conservation of matter

Answer

c. the second law of thermodynamics

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Exercise \(\PageIndex{7d}\)

The heat of vaporization of ammonia is 23.4 kJ/mol. Its boiling point is -33°C. What is the change in entropy for the vaporization of ammonia in J/(mol*K)?

Answer

\[\Delta S=\frac{\Delta H}{T_{b}}=\frac{23400\,J/mol}{240.15K}=97.5\,J/(mol*K)\nonumber\]

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Exercise \(\PageIndex{7e}\)

At the boiling point of benzene, C₆H₆, ∆H_vap = 30.78 kJ/mol, ∆S_vap = 87.15 J/(mol*K). Determine the normal boiling temperature in degrees Celsius for C₆H₆.

Answer

\[T_{b}=\frac{\Delta H_{vap}}{\Delta S_{vap}}=\frac{30780\,J/mol}{87.15\,J/(mol*K)}=353.2K*273=80.18^{0}C\nonumber\]

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Exercise \(\PageIndex{7f}\)

Arrange the following in order of increasing entropy:

CH₄(g), C(s), Li(s), Na(s)

1. Li, C, CH₄, Na
2. C, Li, Na, CH₄
3. Na, Li, C, CH₄
4. CH₄, Li, Na, C
5. Na, Li, CH₄, C

Answer

b. C, Li, Na, CH₄

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Exercise \(\PageIndex{7g}\)

Arrange the following in order of increasing entropy, S°:

Hg(l), Hg(s), C₆H₆(l), CH₃OH(g)

1. Hg(s), CH₃OH(g), C₆H₆(l), Hg(l)
2. CH₃OH(g), Hg(s), Hg(l), C₆H₆(l)
3. Hg(l), Hg(s), C₆H₆(l), CH₃OH(g)
4. Hg(s), Hg(l), C₆H₆(l), CH₃OH(g)
5. Hg(s), Hg(l), CH₃OH(l), C₆H₆(l)

Answer

d. Hg(s), Hg(l), C₆H₆(l), CH₃OH(g)

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Exercise \(\PageIndex{7h}\)

Which of the following processes would be expected to have a positive ∆S value?

1. \(2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(g)\)
2. \(I_{2}(g)\rightarrow I_{2}(s)\)
3. \(2ClBr(g)\rightarrow Cl_{2}(g)+Br_{2}(g)\)
4. \(NH_{4}HS(s)\rightarrow NH_{3}(g)+H_{2}S(g)\)
5. \(2NO(g)+O_{2}(g)\rightarrow 2NO_{2}(g)\)

Answer

d. \(NH_{4}HS(s)\rightarrow NH_{3}(g)+H_{2}S(g)\)

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Exercise \(\PageIndex{7i}\)

Which of the following compounds has the highest entropy in J/(mol*K) at 298K?

1. CH₃OH(l)
2. CO(g)
3. SiO₂(s)
4. H₂O(l)
5. CaCO₃(s)

Answer

b. CO(g)

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Exercise \(\PageIndex{7j}\)

Which of the following processes would be expected to have a ∆S value very close to zero?

1. \(H_{2}O(s)\rightarrow H_{2}O(l)\)
2. \(2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(g)\)
3. \(H_{2}O(s)\rightarrow H_{2}O(g)\)
4. \(N_{2}(g)+O_{2}(g)\rightarrow 2NO(g)\)
5. \(OF_{2}(g)+H_{2}O(g)\rightarrow O_{2}(g)+2HF(g)\)

Answer

d. \(N_{2}(g)+O_{2}(g)\rightarrow 2NO(g)\)

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Exercise \(\PageIndex{7k}\)

Calculate ∆S° for the decomposition of ozone from oxygen.

\(2O_{3}(g)\rightarrow 3O_{2}(g)\)

S⁰=205 J/(mol*K) for O₂(g) and 239 for O₃(g) at 250C

Answer

\[\Delta S^{0}_{rxn}=S^{0}(products)-S^{0}(reactants)\nonumber\]

\[\Delta S^{0}_{rxn}=3(205\,J/(mol*K))-2(239\,J/(mol*K))=137\,J/(mol*K)\nonumber\]

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Exercise \(\PageIndex{7l}\)

Given the following

\(Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)\,\,\,\,\,\Delta G^{0}=-29.4\,kJ/mol\)

\(3Fe_{2}O_{3}(s)+CO(g)\rightarrow 2Fe_{3}O_{4}(s)+CO_{2}(g)\,\,\,\,\,\Delta G^{0}=-61.6\,kJ/mol\)

calculate ∆G° for

\(Fe_{3}O_{4}(s)+CO(g)\rightarrow Fe(s)+Fe_{2}O_{3}(g)+CO_{2}(g)\)

Answer

Divide both reactions by 2, and reverse the second reaction.

\[\frac{1}{2}Fe_{2}O_{3}(s)+\frac{3}{2}CO(g)\rightarrow Fe(s)+\frac{3}{2}CO_{2}(g)\,\,\,\,\,\Delta G^{0}=-14.7\,kJ/mol \nonumber\]

\[Fe_{3}O_{4}(s)+\frac{1}{2}CO_{2}(g)\rightarrow \frac{3}{2}Fe_{2}O_{3}(s)+\frac{1}{2}CO(g)\,\,\,\,\,\Delta G^{0}=30.8\,kJ/mol \nonumber\]

Now add the reactions together.

\[Fe_{3}O_{4}(s)+CO(g)\rightarrow Fe(s)+Fe_{2}O_{3}(g)+CO_{2}(g)\,\,\,\,\,\Delta G^{0}=16.1kJ/mol\nonumber\]

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Exercise \(\PageIndex{7m}\)

For the reaction \(3C(s)+4H_{2}(g)\rightleftharpoons C_{3}H_{8}(g)\),

∆S⁰ = -269 J/(mol*K)

∆H⁰ = -103.8 kJ/mol

Calculate the equilibrium constant at 25°C for the reaction above.

Answer

\[\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}\nonumber\]

\[\Delta G^{0}=(-1.038*10^{5}\,J/mol)-298K(-269\,J/(mol*K))=-2.3638*10^{4}\,J/mol\nonumber\]

\[\Delta G^{0}=-RTlnK\nonumber\]

\[-2.3638*10^{4}=-(8.314)(298)lnK\nonumber\]

\[-2.3638*10^{4}=-2.4776*10^{3}lnK\nonumber\]

\[lnK=9.5408\nonumber\]

\[K=e^{9.5408}=1.4*10^{4}\nonumber\]

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Exercise \(\PageIndex{7n}\)

 Consider the following characteristics of chemical reactions,

1. 1. spontaneity
   2. maximum amount of work that can be done
   3. speed

The value of G for a chemical reaction under a given set of conditions provides information about:

a.   I only               b.  II only        c.  III only       d. I and II       e.  I, II and III

Answer

d. I and II .

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Exercise \(\PageIndex{7o}\)

.  Which of the following is a statement of the third law of thermodynamics.

a)      The entropy of the universe is constant.
b)      For a spontaneous process the entropy of the universe decreases
c)      For a spontaneous process the entropy of the universe increases
d)      The entropy of an ideal crystal at zero Kelvin is zero .
e)      The change in entropy of the universe must be positive for a spontaneous process.

Answer

The entropy of an ideal crystal at zero Kelvin is zero .

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Exercise \(\PageIndex{7p}\)

Which of the following thermodynamic properties CANNOT be experimentally determined?

a.  S^o           b. H^o           c.  G^o               d.  S^o                e.  G^o

Answer

      c.  G^o              

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