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18: Entropy and Free Energy

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    18.2: Dispersal of Energy - Entropy

    Entropy and Second Law of Thermodynamics

    Exercise \(\PageIndex{a}\)

    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

    Exercise \(\PageIndex{b}\)

    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

    Exercise \(\PageIndex{c}\)

    Which one of the following decreases of the entropy of the system?

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

    c. dissolving oxygen in water

    Exercise \(\PageIndex{d}\)

    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)

    Exercise \(\PageIndex{e}\)

    Calculate the change of entropy for the reaction at 250C, where S0 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

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

    \[\Delta S^{0}=\left ( 2*51.0 \right )-\left ( 4*58.32 \right )-\left ( 3*205.0 \right )=-626.3J/K\nonumber\]

     

    18.6: Gibbs Free Energy

    Gibbs Free Energy

    Exercise \(\PageIndex{6.a}\)

    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.

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

    Answer

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

    Exercise \(\PageIndex{6.b}\)

    What is the change of entropy if S0 of C2H4(g), H2O(g), C2H5OH(l) are 219, 189, and 161 J/K*mol. 

    Answer

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

    Exercise \(\PageIndex{6.c}\)

    What is the change of enthalpy at 25°C for the same reaction?

    Answer

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

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

    xercise \(\PageIndex{6.d}\)

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

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

    c. -,+

    Exercise \(\PageIndex{6.e}\)

    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. +,+

    Exercise \(\PageIndex{6.f}\)

    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. +,+

     

    Gibbs Free Energy and K

    Exercise \(\PageIndex{6.g}\)

    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

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

    Exercise \(\PageIndex{6.h}\)

    Using the reaction and the giving the following data, determine the value for ΔS°.

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

     

    ∆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

    Answer

    \[\Delta S^{0}=96.11-56.5-73.93=-34.32\,J/(mol*K)\nonumber\]

    Exercise \(\PageIndex{6.i}\)

    Use the information from 18.6.8, determine the value for ΔH°.

    Answer

    \[\Delta H^{0}=-127.0-105.9+167.2=-65.7\,kJ/mol\nonumber\]

    Exercise \(\PageIndex{6.j}\)

    Use the information from 18.6.8, determine the value for ΔG°.

    Answer

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

    Exercise \(\PageIndex{6.k}\)

    Use the information from 18.6.8, determine the value for K.

    Answer

    \[K=e^{-\frac{\Delta G^{0}}{RT}}=e^{-\frac{-55.5}{0.008314*268}}=5.35*10^{9}\nonumber\]

    Exercise \(\PageIndex{6.l}\)

    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

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

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


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

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

     

    Free Energy and Temperature

    Exercise \(\PageIndex{6.m}\)

    What is the melting point of gold, Au, at 100kPa if ΔH°fusion=12.36kJ/mol, and ΔS°fusion = 0.0092kJ/mol?

    Answer

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

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

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

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

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

    Exercise \(\PageIndex{6.n}\)

    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

    Exercise \(\PageIndex{6.o}\)

    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

    Exercise \(\PageIndex{6.p}\)

    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

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

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

    Exercise \(\PageIndex{6.q}\)

    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

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

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

     

    Enthalpy, Entropy, and Gibbs Free Energy

    Exercise \(\PageIndex{6.r}\)

    Use the Selected Thermodynamic Values appendix in the back of your book to 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 ΔHrxn for the reaction above.
    2. Find the ΔSrxn for the reaction above.
    3. Find the ΔGrxn 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.

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

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

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

    Exercise \(\PageIndex{6.s}\)

    Use the Selected Thermodynamic Values appendix in the back of your book to 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 ΔHrxn for the reaction above.
    2. Find the ΔSrxn for the reaction above.
    3. Find the ΔGrxn 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.

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

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

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

    Exercise \(\PageIndex{6.t}\)

    Use the Selected Thermodynamic Values appendix in the back of your book to 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 ΔHrxn for the reaction above.
    2. Find the ΔSrxn for the reaction above.
    3. Find the ΔGrxn 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.

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

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

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

     

    Finding K

    Exercise \(\PageIndex{6.u}\)

    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.

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

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

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

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

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

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

    \[K=e^{\frac{141.03\,kJ/mol-298.15K(0.24067\,kJ/(mol*K))}{-0.008314(298)}}=2.769*10^{-13}\nonumber\]

    Answer c.

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

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

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

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

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

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

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

    Exercise \(\PageIndex{6.v}\)

    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.

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

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

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

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

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

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

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

    Answer c.

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

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

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

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

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

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

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

    Exercise \(\PageIndex{6.w}\)

    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.

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

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

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

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

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

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

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

    Answer c.

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

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

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

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

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

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

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

     

    Reactant and Product Loading

    \(MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\,\,\,\,\,\Delta G^{0}=69.27\,kJ/mol\)
    Figure \(\PageIndex{1}\): Use the following reaction for questions 18.6.x-18.6.aa

    Exercise \(\PageIndex{6.x}\)

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

    Does the reaction favor products or reactants?

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

    a. reactant

    Exercise \(\PageIndex{6.y}\)

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

    What is the expression for Q?

    Answer

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

    Exercise \(\PageIndex{6.z}\)

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

    At what minimal pressures would the reaction proceed to the left?

    Answer

    \[K=\left ( P_{HCl} \right )^{2}=e^{\frac{-\Delta G^{0}}{RT}}=7.9*10^{-13}\nonumber\]

    \[\left ( P_{HCl} \right )^{2}>7.9*10^{-13}\nonumber \]

    \[P_{HCl}>\sqrt{7.9*10^{-13}}\nonumber\]

    \[P_{HCl}>8.9*10^{-7}\nonumber\]

    Exercise \(\PageIndex{6.aa}\)

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

    What is the ∆G° for PHCl = 5.2*10-4?

    Answer

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

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

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

    Exercise \(\PageIndex{6.ab}\)

    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

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

    \[\Delta G=-40.98\,kJ/mol+(0.008315kJ/(mol*K)*298.15K*ln(7.7*10^{11}))=26.9*\,kJ/mol\nonumber\]

    So it will produce more reactants

    Exercise \(\PageIndex{6.ac}\)

    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

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

    \[\Delta G=40.98\,kJ/mol+(0.008315\,kJ/(mol*K)*298.15K*ln(2.9*10^{-13}))=-30.6\,kJ/mol\nonumber\]

    So it will produce more products.

     

    General Questions

    Exercise \(\PageIndex{a}\)

    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

    Exercise \(\PageIndex{b}\)

    Which of the following is true when one mole of naphthalene 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

    d. 2 and 3 only

    Exercise \(\PageIndex{c}\)

    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

    Exercise \(\PageIndex{d}\)

    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 \(\PageIndex{e}\)

    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

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

    Exercise \(\PageIndex{f}\)

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

    Answer

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

    Exercise \(\PageIndex{g}\)

    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

    Exercise \(\PageIndex{h}\)

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

    Hg(l), Hg(s), C6H6(l), CH3OH(l)

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

    e. Hg(s), Hg(l), CH3OH(l), C6H6(l)

    Exercise \(\PageIndex{i}\)

    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. \(2NO(g)+O_{2}(g)\rightarrow 2NO_{2}\)
    Answer

    c. \(2ClBr(g)\rightarrow Cl_{2}(g)+Br_{2}(g)\)

    Exercise \(\PageIndex{j}\)

    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)

    Exercise \(\PageIndex{k}\)

    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

    Exercise \(\PageIndex{l}\)

    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. \(N_{2}(g)+O_{2}(g)\rightarrow 2NO\)
    5. \(OF_{2}(g)+H_{2}O(g)\rightarrow O_{2}(g)+2HF(g)\)
    Answer

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

    Exercise \(\PageIndex{m}\)

    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

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

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

    Exercise \(\PageIndex{n}\)

    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

    Exercise \(\PageIndex{o}\)

    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.

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

    Exercise \(\PageIndex{p}\)

    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

    Exercise \(\PageIndex{q}\)

    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

    Exercise \(\PageIndex{r}\)

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

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

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

       ∆G          ∆H          ∆S

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

    a. –           –             –

    Exercise \(\PageIndex{s}\)

    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

    Exercise \(\PageIndex{t}\)

    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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