CHAPTER 23. Chemical Thermodynamicsmcqua ?· CHAPTER 23. Chemical Thermodynamics ... =−5.03 ×105…

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<ul><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>CHAPTER 23. Chemical Thermodynamics</p><p>23-1. (a) H2O(l) H2O(s) (0C, 1 bar)Because ice is more ordered structurally than liquid water, ice has less positional disorder thanliquid water. Thus, when compared at the same temperature and pressure, the entropy of solidwater is less than the entropy of liquid water, and so Ssys &lt; 0.</p><p>(b) Cu(s)(1 bar, 500C) Cu(s)(25C, 1 bar)A decrease in temperature at fixed pressure decreases the energy and therefore decreases thethermal disorder of the Cu(s). The decrease in thermal disorder means that the entropydecreases, and thus, Ssys &lt; 0.</p><p>(c) 4 Al(s) + 3 O2(g) 2 Al2O3(s) (25C, 1 bar)The reaction described by this equation involves the conversion of seven moles of reactants(three moles of which are gas) into two moles of solid, which corresponds to a large decrease inpositional disorder; consequently Ssys &lt; 0.</p><p>23-2. Sublimation means the direct conversion of a solid to a gas without passing through the liquidphase.</p><p>(1) solid gas (sublimation) S sys = S subFor the two-stage process</p><p>(2) solid liquid gaswe have S sys = S fus + S vap. But the initial and final states are the same as in (a) and theentropy is a state function; therefore,</p><p>S sub = S fus + S vapFurther, because S fus &gt; 0 and S vap &gt; 0, S fus + S vap &gt; S vap. Thus, S sub &gt; S vap.</p><p>23-3. The order is mainly based upon the number of atoms because the masses of the oxygen andnitrogen atoms are about the same; thus we have</p><p>NO(g) &lt; N2O(g) &lt; NO2(g) &lt; N2O4(g) &lt; N2O5(g)</p><p>Note that N2O(g) and NO2(g) both have three atoms per molecule, but N2O(g) has a lowermolecular mass than NO2(g), and thus we predict that it has the lower entropy of the pair.</p><p>75</p></li><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>76 GENERAL CHEMISTRY, FOURTH EDITION | McQuarrie, Rock, and Gallogly</p><p>23-4. Because Hrxn &gt; 0 and Srxn &lt; 0, the reaction is not spontaneous under any conditions.</p><p>Plants use energy from the sun (photosynthesis) to drive the reaction and produce glucose. Anonspontaneous reaction will only form products upon the input from an external source ofthe energy needed to drive the reaction.</p><p>23-5. The concentrations of Ag+(aq ) and Cl(aq ) before the addition of 5.0 mL of 0.10 MAgNO3(aq ) are calculated from the Ksp expression</p><p>Ksp = 1.8 1010 M2 = [Ag+] [Cl]</p><p>But [Ag+] = [Cl] and thus</p><p>[Ag+] = [Cl] =</p><p>1.8 1010 M2 = 1.3 105 M</p><p>The values of [Ag+]0 and [Cl]0 immediately after the addition of of 5.0 mL of 0.10 MAgNO3(aq ) are</p><p>[Ag+]0 = (5.0 mL)(0.10 M) + (100.0 mL)(1.3 105 M)</p><p>105.0 mL= 4.8 103 M</p><p>[Cl]0 = (100.0 mL)(1.3 105 M)</p><p>105.0 mL= 1.2 105 M</p><p>Thus the value of Qc is</p><p>Qc = [Ag+]0[Cl]0 = (4.8 103 M)(1.2 105 M) = 5.8 108 M2</p><p>The value of G rxn is calculated using Equation 23.16.</p><p>G rxn = RT lnQcKc</p><p>= (8.3145 JK1 mol1)(298 K) ln(</p><p>5.8 108 M21.8 1010 M2</p><p>)= +14 kJmol1</p><p>The positive value of G rxn is consistent with the common-ion effect that states that addition ofa common ion to a solution decreases the solubility of an ionic solid.</p><p>23-6. K = Kp/Qp = 1.90 bar/1.00 bar = 1.90 (unitless)23-7. The value of G rxn can be calculated from the value of K by using Equation 23.19.</p><p>G rxn = RT ln K = (8.3145 JK1 mol1)(298 K) ln(1.54 1088)= 5.03 105 Jmol1 = 503 kJmol1</p><p>The reaction as written is spontaneous at 25C under standard conditions.</p><p>23-8. For the reaction described by the equation</p><p>Ag+(aq , 2.0 106 M) + Cl(aq , 2.0 106 M) AgCl(s)</p><p>we have</p><p>Q = QcQc</p><p>= 1([Ag+]0/M)([Cl]0/M)</p><p>= 1(2.0 106)2 = 2.5 10</p><p>11</p><p>University Science Books, 2011. All rights reserved. www.uscibooks.com</p></li><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>Chapter 23: Chemical Thermodynamics 77</p><p>and because Kc = 1/Ksp,</p><p>K = KcK c</p><p>= 11.80 1010 = 5.6 10</p><p>9</p><p>Thus, using Equation 23.22, we obtain for G rxn</p><p>G rxn = RT ln QK = (8.3145 JK1 mol1)(298 K) ln</p><p>(2.5 10115.6 109</p><p>)= +9.41 kJmol1</p><p>The positive value of G rxn tells us that no AgCl(s) can form under the given conditions of[Ag+]0 and [Cl]0.</p><p>23-9. We have that</p><p>G rxn = G f [CO(g)] + G f [Cl2(g)] G f [COCl2(g)]= (1)(137.2 kJmol1) + (1)(0 kJmol1) (1)(205.9 kJmol1)= 68.7 kJmol1 = 6.87 104 Jmol1</p><p>Using Equation 23.19,</p><p>ln K = Grxn</p><p>RT= 6.87 10</p><p>4 Jmol1(8.3145 JK1 mol1)(298 K) = 27.7</p><p>K = e 27.7 = 9.1 1013</p><p>23-10. Start with (from Example 23-10)</p><p>G rxn = 29.4 kJmol1 + (8.3145 JK1 mol1)(298 K) ln 1PCO P</p><p>2H2</p><p>= 2.94 104 Jmol1 + 2.48 103 Jmol1 ln 1P 3</p><p>where P = PCO = PH2. We want G rxn &lt; 0, so we want</p><p>2.48 103 Jmol1 ln 1P 3</p><p>&lt; 2.94 104 Jmol1</p><p>or</p><p>ln1P 3</p><p>(</p><p>11.4 105</p><p>)1/3= 0.019 bar.</p><p>23-11. Given that Hrxn = +55.8 kJmol1 for the reaction described by the equation</p><p>2 H2O(l) H3O+(aq ) + OH(aq )</p></li><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>78 GENERAL CHEMISTRY, FOURTH EDITION | McQuarrie, Rock, and Gallogly</p><p>and that K = 1.0 1014 at 25C, we use Equation 23.28 to compute K at 37C (human bodytemperature)</p><p>lnK</p><p>1.0 1014 =(</p><p>55.8 103 Jmol18.3145 JK1 mol1</p><p>) [310 K 298 K</p><p>(310 K)(298 K)</p><p>]= 0.87</p><p>and</p><p>K = (1.0 1014)e 0.87 = 2.4 1014 at 37CTherefore,</p><p>[H3O+] = [OH] = K 1/2 = 1.55 107 Mand</p><p>pH = log([H3O+]/M) = 6.81Because the reaction described by the equation is endothermic, we predict from Le Chteliersprinciple that an increase in temperature will shift the reaction to the right, increasing theconcentration of H3O+(aq ) and decreasing the pH. Thus, our results are consistent with LeChteliers principle.</p><p>University Science Books, 2011. All rights reserved. www.uscibooks.com</p></li><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>CHAPTER 24. Oxidation-Reduction Reactions</p><p>24-1. (a) Applying Rule 7, we assign each oxygen atom an oxidation state of +2. From Rule 2 with xthe oxidation state of each chromium atom, we have</p><p>2x + (7)(2) = 2 or x = +6Thus, the oxidation states of Cr = +6 and O = 2.(b) Applying Rule 4, we assign each fluorine atom an oxidation state of 1. From Rule 2 with xthe oxidation state of the hydrogen atom, we have</p><p>x + (2)(1) = 1 or x = +1Thus, the oxidation states are H = +1 and F = 1.(c) Applying Rule 6, we assign each hydrogen atom an oxidation state of +1. From Rule 2 withx the oxidation state of the nitrogen atom, we have</p><p>x + (4)(+1) = +1 or x = 3Thus, the oxidation states are H = +1 and N = 3.(d) Applying Rule 2 with x the oxidation state of each iodine atom, we have</p><p>3x = 1 or x = 1/3</p><p>Thus, the oxidation state is I = 1/3.24-2. (a) The chlorine atom is more electronegative than the bromine atom, so, according to</p><p>Rule 8b, we assign both electrons of the covalent bond to the chlorine atom</p><p>Br Cl</p><p>A neutral bromine atom has seven valence electrons and has been assigned six in theformula, so</p><p>oxidation state of Br in BrCl = 7 6 = +1</p><p>A neutral chlorine atom has seven valence electrons and has been assigned eight in theformula, so</p><p>oxidation state of Cl in BrCl = 7 8 = 1</p><p>79</p></li><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>80 GENERAL CHEMISTRY, FOURTH EDITION | McQuarrie, Rock, and Gallogly</p><p>(b) In this case, the bromine atom is more electronegative than the iodine atom, so we have</p><p>Br I</p><p>Thus, we have an oxidation state of +1 for I and 1 for Br.(c) The Lewis formula for ICl2 is</p><p>Cl I Cl</p><p>The chlorine atom is more electronegative than the iodine atom, so write</p><p>Cl I Cl</p><p>A neutral iodine atom has seven valence electrons and has been assigned six in the formula, so</p><p>oxidation state of I in ICl2 = 7 6 = +1</p><p>A neutral chlorine atom has seven valence electrons and each has been assigned eight in theformula, so</p><p>oxidation state of Cl in ICl2 = 7 8 = 1</p><p>Notice that the sum of the oxidation states of the atoms equals the overall charge, as it must foran ion.(d) The Lewis formula of BrCl3 is</p><p>Cl Br Cl</p><p>Cl</p><p>The chlorine atom is more electronegative than the bromine atom, so write</p><p>Br</p><p>Cl</p><p>ClCl</p><p>Thus, the oxidation state of Br is +3 and that of Cl is 1.24-3. The oxidation state of each Mn atom in a Mn2O7 molecule is (Rule 7 and Rule 2) given by</p><p>2x + (7)(2) = 0, or x = +7. Therefore, there is a total of 2 7 + 7 6 = 56 valence electronsin Mn2O7 . The Lewis formula is given in the text.</p><p>24-4. The oxidation state of iodine changes from 1 in an I ion to 1/3 in an I3 ion. Thus, I is thereducing agent and thus the species oxidized. The oxidation state of oxygen changes from 0 inan O2 molecule to 2 in a OH ion. Thus, O2(aq ) is the oxidizing agent and the speciesreduced.</p><p>24-5. The oxidation state of copper changes from +1 in a Cu+ ion to both +2 in a Cu2+ ion and to 0in Cu(s). Thus, the half-reaction equations are</p><p>Cu+(aq ) Cu2+(aq ) + e (oxidation)Cu+(aq ) + e Cu(s) (reduction)</p><p>University Science Books, 2011. All rights reserved. www.uscibooks.com</p></li><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>Chapter 24: Oxidation-Reduction Reactions 81</p><p>In the overall reaction, the Cu+ ion is both oxidized (to Cu2+) and reduced (to Cu). Such selfoxidation-reduction reactions are called disproportionation reactions.</p><p>24-6. The H+(aq ) over the reaction arrow indicates that we have an acidic solution. The unbalancedhalf-reaction equations are</p><p>Br BrO3 (oxidation)MnO4 Mn2+ (reduction)</p><p>The two half-reactions are balanced with respect to the elements other than H and O. Tobalance each half reaction with respect to the oxygen atoms, we add three H2O molecules tothe left side of the oxidation half reaction and four H2O molecules to the right side of thereduction half reaction.</p><p>3 H2O + Br BrO3 (oxidation)MnO4 Mn2+ + 4 H2O (reduction)</p><p>To balance each half reaction with respect to the hydrogen atoms, we add six H+ ions to theright side of the oxidation half reaction and eight H+ ions to the left side of the reduction halfreaction.</p><p>3 H2O + Br BrO3 + 6 H+ (oxidation)8 H+ + MnO4 Mn2+ + 4 H2O (reduction)</p><p>Next we balance the half-reactions with respect to charge by adding six electrons to the rightside of the oxidation half reaction and five electrons to the left side of the reduction halfreaction.</p><p>3 H2O + Br BrO3 + 6 H+ + 6 e (oxidation)8 H+ + MnO4 + 5 e Mn2+ + 4 H2O (reduction)</p><p>Multipling the oxidation half-reaction equation by 5 and the reduction half-reaction equationby 6 yields</p><p>15 H2O + 5 Br 5 BrO3 + 30 H+ + 30 e (oxidation)48 H+ + 6 MnO4 + 30 e 6 Mn2+ + 24 H2O (reduction)</p><p>Adding the above two equations yields</p><p>15 H2O + 5 Br + 48 H+ + 6 MnO4 5 BrO3 + 30 H+ + 6 Mn2+ + 24 H2O</p><p>Cancelling like terms and indicating phases yield the final balanced equation.</p><p>5 Br(aq ) + 18 H+(aq ) + 6 MnO4 (aq ) 5 BrO3 (aq ) + 6 Mn2+(aq ) + 9 H2O(l)</p><p>24-7. The oxidation state of oxygen in H2O2 is 1. When H2O2 acts as a reducing agent, theoxidation state of oxygen increases. The oxidation state of oxygen in O2 is zero, which is greaterthan 1. Thus, we have</p><p>H2O2 O2</p></li><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>82 GENERAL CHEMISTRY, FOURTH EDITION | McQuarrie, Rock, and Gallogly</p><p>This equation is balanced with respect to all elements except hydrogen. Because we have anacidic aqueous solution, we balance the hydrogen atoms by adding 2 H+ to the right side of theequation</p><p>H2O2 O2 + 2 H+</p><p>Finally, we add 2 e to the right to balance the charge.</p><p>H2O2 O2 + 2 H+ + 2 e</p><p>The balanced final half-reaction equation is</p><p>H2O2(aq ) O2(g) + 2 H+(aq ) + 2 e</p><p>24-8. The oxidation state of the iron atom changes from +2 in Fe(OH)2(s) to +3 in Fe(OH)3(s) andthus iron is oxidized. Because iron is oxidized, the other reactant, O2(g), must be reduced.Thus we have</p><p>Fe(OH)2 Fe(OH)3 (oxidation)O2 (reduction)</p><p>Because the oxidation half-reaction equation involves a metal hydroxide, we skip steps 3 and 4and balance it directly by adding an OH to the left and and an electron on the right</p><p>OH + Fe(OH)2 Fe(OH)3 + e</p><p>Because the solution is alkaline, the reduction half reaction is balanced according to steps 3and 4 by adding 2 H2O to the left side and 4 OH to the right side</p><p>2 H2O + O2 4 OH</p><p>The charge is balanced by adding 4 e to the left side</p><p>2 H2O + O2 + 4 e 4 OH (reduction)</p><p>We can balance the electrons by multiplying the oxidation half-reaction equation by 4 and thenadding the result to the reduction half-reaction equation to obtain</p><p>4 OH + 4 Fe(OH)2 + 2 H2O + O2 4 Fe(OH)3 + 4 OH</p><p>Cancellation of like terms and designation of phases yield</p><p>4 Fe(OH)2(s) + 2 H2O(l) + O2(g) 4 Fe(OH)3(s)</p><p>24-9. (a) The oxidation and reduction half-reaction equations are</p><p>CH3CH2OH CH3COOH (oxidation)Cr2O</p><p>27 Cr3+ (reduction)</p><p>University Science Books, 2011. All rights reserved. www.uscibooks.com</p></li><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>Chapter 24: Oxidation-Reduction Reactions 83</p><p>The balancing proceeds as follows:</p><p>(i) elements other than O and H</p><p>CH3CH2OH CH3COOH (oxidation)Cr2O</p><p>27 2 Cr3+ (reduction)</p><p>(ii) oxygen</p><p>H2O + CH3CH2OH CH3COOH (oxidation)Cr2O</p><p>27 2 Cr3+ + 7 H2O (reduction)</p><p>(iii) hydrogen and charge</p><p>H2O + CH3CH2OH CH3COOH + 4 H+ + 4 e (oxidation)14 H+ + Cr2O27 + 6 e 2 Cr3+ + 7 H2O (reduction)</p><p>(iv) multiplying the oxidation half-reaction equation by 3 and the reduction half-reactionequation by 2 (to balance the electrons) and adding the resulting equations yield</p><p>3 H2O + 3 CH3CH2OH + 28 H+ + 2 Cr2O27 3 CH3COOH + 12 H+ + 4 Cr3+ + 14 H2O(v) cancellation of like terms and designating phases yields the final balanced equation</p><p>3 CH3CH2OH(aq ) + 16 H+(aq ) + 2 Cr2O27 (aq ) 3 CH3COOH(aq ) + 4 Cr3+(aq ) + 11 H2O(l)</p><p>(b) The number of millimoles of Cr2O27 (aq ) consumed is given by</p><p>millimoles of Cr2O27 consumed = (10.0 mL)(0.0100 M)</p><p>(10%100%</p><p>)= 0.0100 mmol</p><p>The number of millimoles of CH3CH2OH(aq ) oxidized is</p><p>millimoles of CH3CH2OH = (0.0100 mmol Cr2O27 )3 mol CH3CH2OH</p><p>2 mol Cr2O27</p><p> = 0.0150 mmol</p><p>The number of milligrams of ethanol is</p><p>mass of CH3CH2OH = (0.0150 103 mol)(46.07 gmol1) = 6.9 104 g = 0.69 mg</p></li><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>CHAPTER 25. Electrochemistry</p><p>25-1. A current of 0.50 amperes corresponds to a current flow of 0.50 coulombs per second. Thenumber of electrons that flow through a cross section in one minute is</p><p>flow of electrons = (0.50 Cs1)(</p><p>1 electron1.602 1019 C</p><p>) (60 s</p><p>1 min</p><p>)= 1.9 1020 electronsmin1</p><p>25-2. Electrons are produced at the cadmium electrode</p><p>Cd(s) Cd2+(aq ) + 2 e</p><p>Electrons are consumed at the indium electrode.</p><p>In3+(aq ) + 3 e In(s)</p><p>For the description of the electrochemical cell, see the text.</p><p>25-3. Oxidation takes place in a basic solution at the left electrode</p><p>Zn(s) ZnO(s) (oxidation)</p><p>To balance this equation, we use the following steps:</p><p>Zn(s) ZnO(s) + H2O(l)2 OH(aq ) + Zn(s) ZnO(s) + H2O(l)2 OH(aq ) + Zn(s) ZnO(s) + H2O(l) + 2 e</p><p>Reduction takes place in a basic solution at the right electrode.</p><p>HgO(s) Hg(l) (reduction)</p><p>To balance this equation, we use the following steps:</p><p>H2O(l) + HgO(s) Hg(l)H2O(l) + HgO(s) Hg(l) + 2 OH(aq )</p><p>H2O(l) + HgO(s) + 2 e Hg(l) + 2 OH(aq )</p><p>84</p></li><li><p>Practice-Problem-6660020 book December 22, 2010 8:37</p><p>Chapter 25: Electrochemistry 85</p><p>The overall equation for the reaction is the sum of the oxidation half-reaction equation and thereduction half-reaction equation.</p><p>Zn(s) + HgO(s) ZnO(s) + Hg(l)</p><p>25-4. See the text.</p><p>25-5. The cell will continue to discharge until Q = K , that is until chemical equilibrium is attained.When Q = K , ln(Q/K ) = 1, and the cell voltage E = 0, from Equation 25.7.The value of K for the equation tha...</p></li></ul>

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