Spontaneity and Energy Transfer Lecture Material Exercise \(\PageIndex{1a}\) Which of the following is a spontaneous process? ice melting at 25°C heat flowing from a hot to a cold object an iron tool rusting 1 only 2 only 3 only 1 and 2 only 1, 2, and 3 Answer e. 1, 2, and 3 Dispersal of Energy - Entropy Entropy - A Microscopic Understanding Entropy Measurements and Values Exercise \(\PageIndex{4a}\) Predict the entropy change in the reaction, \(2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)\) increase decrease same not enough information Answer b. decrease Exercise \(\PageIndex{4b}\) Predict the entropy change in the reaction, \(Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)\) increase decrease same 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. Exercise \(\PageIndex{4c}\) Which one of the following decreases of the entropy of the system? There is only one correct answer. dissolving NaCl in water sublimation of benzene dissolving oxygen in water 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 Exercise \(\PageIndex{4d}\) Which one has the highest absolute entropy at 25 0 C? H 2 O(l) He(g) C(s) NH 3 (g) Answer d. NH 3 (g), then He(g), as the ammonia is more complex and has more microstates available. Exercise \(\PageIndex{4e}\) Arrange the following reactions in order of increasing ∆S° rxn values: \(H_{2}(g)+Cl_{2}(g)\rightarrow 2HCl(g)\) \(3H_{2}(g)+N_{2}(g)\rightarrow 2NH_{3}(g) \) \((NH_{4})_{2}Cr_{2}O_{7}(s)\rightarrow Cr_{2}O_{3}(s)+4H_{2}O(l)+N_{2}(g)\) 1 < 3 < 2 1 < 2 < 3 2 < 1 < 3 2 < 3 < 1 3 < 1 < 2 Answer c. 2 < 1 < 3 Exercise \(\PageIndex{4f}\) Calculate the change of entropy for the reaction at 25 0 C, where S f 0 for Al(s), O 2 (g) and Al 2 O 3 (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 \[\Delta S^{0}=\left ( 2*51.0 \right )-\left ( 4*28.32 \right )-\left ( 3*205.0 \right )=-626.3J/K\nonumber\] Second Law and Gibbs Free Energy Spontaneity and Temperature Exercise \(\PageIndex{5.1a}\) For a reaction to be spontaneous at all temperatures, what should be the signs of ∆H° and ∆S°? +,+ +,- -,+ -,- Answer c. -,+ 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. +,+ +,- -,+ -,- Answer a. +,+ Exercise \(\PageIndex{5.1c}\) When ammonium sulfate dissolves in water, the temperature of the solution decreased. Predict the sign of ∆H and ∆S? +,- +,+ -,+ -,- Answer b. +,+ Exercise \(\PageIndex{5.1d}\) If a reaction is endothermic and nonspontaneous at 25°C, then it can never be spontaneous can become spontaneous by adding a catalyst may be spontaneous at higher temperatures may be spontaneous at lower temperatures is exothermic and spontaneous at high temperatures Answer c. may be spontaneous at higher temperatures Exercise \(\PageIndex{5.1e}\) If a reaction is exothermic and nonspontaneous at 25°C and 1 atm of pressure, it may be spontaneous at higher temperatures spontaneous at lower temperatures endothermic at lowest temperatures endothermic at higher temperatures nonspontaneous at all temperatures Answer b. spontaneous at lower temperatures Exercise \(\PageIndex{5.1f}\) For the following process: \(Br_{2}(l)\rightarrow 2Br(g)\) ∆H is + and ∆S is + for the reaction ∆H is - and ∆S is - for the reaction ∆H is + and ∆S is - for the reaction ∆H is - and ∆S is + for the reaction ∆G is + for all temperatures Answer a. ∆H is + and ∆S is + for the reaction 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 – – – – + + + + + 0 + – + – – Answer c. + + + Exercise \(\PageIndex{5.1h}\) The reaction \(Br_{2}(l)\rightarrow 2Br(g)\) is spontaneous at 1,600 °C. We can conclude that ∆H is + and ∆S is + for the reaction ∆H is - and ∆S is - for the reaction ∆H is + and ∆S is - for the reaction ∆H is - and ∆S is + for the reaction ∆G is + for all temperatures Answer a. ∆H is + and ∆S is + for the reaction Phase Transitions 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, \(\Delta H-T\Delta S=0\) \[\Delta H=T\Delta S\nonumber\] \[T=\frac{\Delta H}{\Delta S}=\frac{12.36}{0.0092}=1343K\nonumber\] Exercise \(\PageIndex{5.2b}\) What is the state of gold if the temperature is 1200°C? solid liquid gas a mixture of solid and liquid Answer b. liquid Exercise \(\PageIndex{5.2c}\) What is the state of gold if the temperature is 900°C? solid liquid gas a mixture of liquid and gas Answer a. solid 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 \[\Delta H=T\Delta S\nonumber\] \[\Delta S=\frac{\Delta H}{T}=\frac{6.01}{273.15}=22\,J/mol\nonumber\] 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 \[\Delta H=T\Delta S\nonumber\] \[\Delta S=\frac{\Delta H}{T}=\frac{40.7}{273.15+100}=109\,J/mol\nonumber\] Driving Chemical Reactions 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? reactant product both none of the above Answer a. reactant 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 \[Q=\left ( P_{HCl} \right )^{2}\nonumber\] 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 \[K=\left ( P_{HCl} \right )^{2}=e^{\frac{-\Delta G^{0}}{RT}}=e^{\frac{-69.27kJ/mol}{(0.008314kJ/mol \cdot K)(298K)}}=7.2*10^{-13}\nonumber\] \[\left ( P_{HCl} \right )^{2}>7.2*10^{-13}\nonumber \] \[P_{HCl}>\sqrt{7.2*10^{-13}}\nonumber\] \[P_{HCl}>8.5*10^{-7}\nonumber\] 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 \[\Delta G=\Delta G^{0}+RTlnQ\nonumber\] \[\Delta G=69.27\,kJ/mol+\left ( 0.008315kJ/(mol*K)*298.15K*ln(5.2*10^{-4})^{2} \right )\nonumber\] \[\Delta G=69.27\,kJ/mol+\left ( -37.49\,kJ/mol \right )=31.78\,kJ/mol\nonumber\] Exercise \(\PageIndex{5.3e}\) What is the ∆G and does the reaction proceed towards products or reactants? 0 1 2 3 2NO(g) + Cl2(g) ⟶ 2NOCl (g) 4.9*10-6M 3.1*10-3M NaN 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 Exercise \(\PageIndex{5.3f}\) What is the ∆G and does the reaction proceed towards products or reactants? 0 1 2 3 2NOCl (g) ⟶ 2NO(g) + Cl2(g) 3.1*10-1M NaN 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 Exercise \(\PageIndex{6.1a}\) Balance the following equations Calculate the change of the Gibbs free energy for the reaction at 25°C, where standard free energy of formation of C 2 H 4 (g), H 2 O(g), C 2 H 5 OH(l) are 68, -229, -175 kJ/mol respectively. \(C_{2}H_{4}(g)+H_{2}O(g)\rightarrow C_{2}H_{5}OH(l)\) What is the change of entropy if S 0 of C 2 H 4 (g), H 2 O(g), C 2 H 5 OH(l) are 219, 189, and 161 J/K*mol. 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 \[\Delta H^{0}=\Delta G^{0}+T\Delta S^{0}\nonumber\] \[\Delta H^{0}=-14+298*\left ( -0.247 \right )=-87.606\,kJ/mol\nonumber\] 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. \(MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\) Find the ΔH rxn for the reaction above. Find the ΔS rxn for the reaction above. 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. \[MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\nonumber\] \[\Delta G^{0}_{rxn}=\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}\nonumber\] \[\Delta G^{0}_{rxn}=141.03\,kJ/mol-298.15K(0.24067\,kJ/mol*K)=69.27\,kJ/mol\nonumber\] 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. \(2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\) Find the ΔH rxn for the reaction above. Find the ΔS rxn for the reaction above. 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. \[2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\nonumber\] \[\Delta G^{0}_{rxn}=\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}\nonumber\] \[\Delta G^{0}_{rxn}=75.54\,kJ/mol-298.15K(0.1162kJ/(mol*K))=40.89\,kJ/mol\nonumber\] 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. \(2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\) Find the ΔH rxn for the reaction above. Find the ΔS rxn for the reaction above. 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. \[2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\nonumber\] \[\Delta G^{0}_{rxn}=\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}\nonumber\] \[\Delta G^{0}_{rxn}=-75.54\,kJ/mol-298.15K(-0.1162kJ/(mol*K))=-40.89\,kJ/mol\nonumber\] Gibbs Free Energy and K 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 \[\Delta G^{0}=-RTlnK=-8.314*298*ln{0.48}=1818.46\,J/mol\nonumber\] Exercise \(\PageIndex{6.2b}\) Using the information below, calculate the value of K at 25°C for the reaction of C(s) and O 2 (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 ) \[C(s)+O_{2}(g)\rightarrow \cancel{CO_{2}(g)}\,\,\,\,\,\Delta G^{0}=-394.4\,kJ/mol\nonumber\] \[+\cancel{CO_{2}(g)}\rightarrow CO(g)+\frac{1}{2}O_{2}\,\,\,\,\,\Delta G^{0}=257.2\,kJ/mol\nonumber\] \[C(s)+\frac{1}{2}O_{2}(g)\rightarrow CO(g)\,\,\,\,\,\Delta G^{0}=-137.2\,kJ/mol\] \[K=e^{-\frac{\Delta G^{0}}{RT}}=e^{-\frac{-137.2}{0.008314*298}}=1.12*10^{24}\] 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 \[\Delta G^{0}=-RTlnK=-8.314*298*ln{0.48}=1818.46\,J/mol\nonumber\] Exercise \(\PageIndex{6.2d}\) Using the reaction and the giving the following data to answer the following questions. \(Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)\) 0 1 2 NaN ∆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 Determine the value for ΔS°. Determine the value for ΔH°. Determine the value for ΔG°. 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 \[K=e^{-\frac{\Delta G^{0}}{RT}}=e^{-\frac{-55.5}{0.008314*268}}=5.35*10^{9}\nonumber\] Exercise \(\PageIndex{6.2e}\) Find the K for the reaction at the following temperatures. \(MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\) Temperature at 25°C? Temperature at 20°C? 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. \[MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\nonumber\] \[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\] \[\Delta H^{0}_{rxn}=\left [ -601.8\,kJ/mol+2(-92.30\,kJ/mol) \right ]-\left [ -641.6\,kJ/mol+-285.83\,kJ/mol \right ]=141.03\,kJ/mol\nonumber\] \[Delta S^{0}_{rxn}=\sum \Delta S_{f}^{0}n(products)-\sum \Delta S_{f}^{0}m(reactants)\nonumber\] \[\Delta S^{0}_{rxn}=\left [ 26.8\,J/(mol*K)+2(186.69\,J/(mol*K))\right ]+\left [ 89.6\,J/(mol*K)+69.91\,J/(mol*K) \right ]=240.67\,J/(mol*K)\nonumber\] \[K=e^{\frac{\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}}{-RT}}\nonumber\] \[K=e^{\frac{141.03\,kJ/mol-298.15K(0.24067\,kJ/(mol*K))}{-0.008314(298)}}=2.769*10^{-13}\nonumber\] Answer c. \[MgCl_{2}(s)+H_{2}O(l)\rightarrow MgO(s)+2HCl(g)\nonumber\] \[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\] \[\Delta H^{0}_{rxn}=\left [ -601.8\,kJ/mol+2(-92.30\,kJ/mol) \right ]-\left [ -641.6\,kJ/mol+-285.83\,kJ/mol \right ]=141.03\,kJ/mol\nonumber\] \[Delta S^{0}_{rxn}=\sum \Delta S_{f}^{0}n(products)-\sum \Delta S_{f}^{0}m(reactants)\nonumber\] \[\Delta S^{0}_{rxn}=\left [ 26.8\,J/(mol*K)+2(186.69\,J/(mol*K))\right ]+\left [ 89.6\,J/(mol*K)+69.91\,J/(mol*K) \right ]=240.67\,J/(mol*K)\nonumber\] \[K=e^{\frac{\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}}{-RT}}\nonumber\] \[K=e^{\frac{141.03\,kJ/mol-373.15K(0.24067\,kJ/(mol*K))}{-0.008314(373.15)}}=6.756*10^{-8}\nonumber\] Exercise \(\PageIndex{6.2f}\) Find the K for the reaction at the following temperatures. \(2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\) Temperature at 25°C? Temperature at 15°C? 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. \[2NOCl(g)\rightarrow 2NO(g)+Cl_{2}(g)\nonumber\] \[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\] \[\Delta H^{0}_{rxn}=\left [ 2(90.37\,kJ/mol)+0\,kJ/mol \right ]-\left [ 2(52.6\,kJ/mol)\right ]=75.54\,kJ/mol\nonumber\] \[Delta S^{0}_{rxn}=\sum \Delta S_{f}^{0}n(products)-\sum \Delta S_{f}^{0}m(reactants)\nonumber\] \[\Delta S^{0}_{rxn}=\left [ 2(210.62\,J/(mol*K))+222.96\,J/(mol*K)\right ]+\left [ 2(264\,J/(mol*K)) \right ]=116.2\,J/(mol*K)\nonumber\] \[K=e^{\frac{\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}}{-RT}}\nonumber\] \[K=e^{\frac{75.54\,kJ/mol-423.15K(0.1162\,kJ/(mol*K))}{-0.008314(423.15)}}=5.558*10^{-4}\nonumber\] Exercise \(\PageIndex{6.2g}\) Find the K for the reaction at the following temperatures. \(2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\) Temperature at 25°C? Temperature at 10°C? 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. \[2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)\nonumber\] \[\Delta H^{0}_{rxn}=\sum \Delta H_{f}^{0}n(products)-\sum \Delta H_{f}^{0}m(reactants)\nonumber\] \[\Delta H^{0}_{rxn}=\left[ 2(52.6\,kJ/mol) \right ]-\left [ 2(90.37\,kJ/mol)+0\,kJ/mol \right ]=-75.54\,kJ/mol\nonumber\] \[Delta S^{0}_{rxn}=\sum \Delta S_{f}^{0}n(products)-\sum \Delta S_{f}^{0}m(reactants)\nonumber\] \[\Delta S^{0}_{rxn}=\left [ 2(264\,J/(mol*K)) \right ]-\left [ 2(210.62\,J/(mol*K))+222.96\,J/(mol*K)\right ]=-116.2\,J/(mol*K)\nonumber\] \[K=e^{\frac{\Delta H^{0}_{rxn}-T\Delta S^{0}_{rxn}}{-RT}}\nonumber\] \[K=e^{\frac{-75.54\,kJ/mol-473.15K(-0.1162\,kJ/(mol*K))}{-0.008314(473.15)}}=1.861*10^{2}\nonumber\] General Questions Exercise \(\PageIndex{7a}\) Which of the following must have a negative value for an exothermic process? enthalpy change entropy change free energy change electrode cell potential equilibrium constant Answer a. enthalpy change Exercise \(\PageIndex{7b}\) Which of the following is true when one mole of solid naphthalene (mothballs) sublimes to gas? The entropy increases The entropy decreases The enthalpy increases The enthalpy decreases 1 only 2 only 1 and 3 only 2 and 3 only 1 and 4 only Answer c. 1 and 3 only Exercise \(\PageIndex{7c}\) The total entropy of a system and its surroundings always increases for a spontaneous process. This is a statement of the law of constant composition the first law of thermodynamics the second law of thermodynamics the third law of thermodynamics the law of conservation of matter Answer c. the second law of thermodynamics 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 \[\Delta S=\frac{\Delta H}{T_{b}}=\frac{23400\,J/mol}{240.15K}=97.5\,J/(mol*K)\nonumber\] Exercise \(\PageIndex{7e}\) At the boiling point of benzene, C 6 H 6 , ∆H vap = 30.78 kJ/mol, ∆S vap = 87.15 J/(mol*K). Determine the normal boiling temperature in degrees Celsius for C 6 H 6 . 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 \(\PageIndex{7f}\) Arrange the following in order of increasing entropy: CH 4 (g), C(s), Li(s), Na(s) Li, C, CH 4 , Na C, Li, Na, CH 4 Na, Li, C, CH 4 CH 4 , Li, Na, C Na, Li, CH 4 , C Answer b. C, Li, Na, CH 4 Exercise \(\PageIndex{7g}\) Arrange the following in order of increasing entropy, S°: Hg(l), Hg(s), C 6 H 6 (l), CH 3 OH(g) Hg(s), CH 3 OH(g), C 6 H 6 (l), Hg(l) CH 3 OH(g), Hg(s), Hg(l), C 6 H 6 (l) Hg(l), Hg(s), C 6 H 6 (l), CH 3 OH(g) Hg(s), Hg(l), C 6 H 6 (l), CH 3 OH(g) Hg(s), Hg(l), CH 3 OH(l), C 6 H 6 (l) Answer d. Hg(s), Hg(l), C 6 H 6 (l), CH 3 OH(g) Exercise \(\PageIndex{7h}\) Which of the following processes would be expected to have a positive ∆S value? \(2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(g)\) \(I_{2}(g)\rightarrow I_{2}(s)\) \(2ClBr(g)\rightarrow Cl_{2}(g)+Br_{2}(g)\) \(NH_{4}HS(s)\rightarrow NH_{3}(g)+H_{2}S(g)\) \(2NO(g)+O_{2}(g)\rightarrow 2NO_{2}(g)\) Answer d. \(NH_{4}HS(s)\rightarrow NH_{3}(g)+H_{2}S(g)\) Exercise \(\PageIndex{7i}\) Which of the following compounds has the highest entropy in J/(mol*K) at 298K? CH 3 OH(l) CO(g) SiO 2 (s) H 2 O(l) CaCO 3 (s) Answer b. CO(g) Exercise \(\PageIndex{7j}\) Which of the following processes would be expected to have a ∆S value very close to zero? \(H_{2}O(s)\rightarrow H_{2}O(l)\) \(2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(g)\) \(H_{2}O(s)\rightarrow H_{2}O(g)\) \(N_{2}(g)+O_{2}(g)\rightarrow 2NO(g)\) \(OF_{2}(g)+H_{2}O(g)\rightarrow O_{2}(g)+2HF(g)\) Answer d. \(N_{2}(g)+O_{2}(g)\rightarrow 2NO(g)\) Exercise \(\PageIndex{7k}\) Calculate ∆S° for the decomposition of ozone from oxygen. \(2O_{3}(g)\rightarrow 3O_{2}(g)\) S 0 =205 J/(mol*K) for O 2 (g) and 239 for O 3 (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 \(\PageIndex{7l}\) 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. \[Fe_{3}O_{4}(s)+CO(g)\rightarrow Fe(s)+Fe_{2}O_{3}(g)+CO_{2}(g)\,\,\,\,\,\Delta G^{0}=16.1kJ/mol\nonumber\] Exercise \(\PageIndex{7m}\) For the reaction \(3C(s)+4H_{2}(g)\rightleftharpoons C_{3}H_{8}(g)\), ∆S 0 = -269 J/(mol*K) ∆H 0 = -103.8 kJ/mol Calculate the equilibrium constant at 25°C for the reaction above. Answer \[\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}\nonumber\] \[\Delta G^{0}=(-1.038*10^{5}\,J/mol)-298K(-269\,J/(mol*K))=-2.3638*10^{4}\,J/mol\nonumber\] \[\Delta G^{0}=-RTlnK\nonumber\] \[-2.3638*10^{4}=-(8.314)(298)lnK\nonumber\] \[-2.3638*10^{4}=-2.4776*10^{3}lnK\nonumber\] \[lnK=9.5408\nonumber\] \[K=e^{9.5408}=1.4*10^{4}\nonumber\]