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MOOREHW9

# MOOREHW9 - Chapter 13 Chemical Kinetics Rates of Reactions...

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Unformatted text preview: Chapter 13: Chemical Kinetics: Rates of Reactions Three factors that can affect the rate of a chemical reaction are described at the beginning of Section 13.3. The}: are concentrations of reactants. the temperature the reaction is run at, and the presence or absence of a catalyst. Other factors may also be involved. Some ofthese factors may not affect the rate of specific reactions. . .stm11-nr‘: (a) 3 x 10-30 (h) 4 x 10-15 (c) 4 x 10-10 {a} 1.9x 10-ﬁ .S'ri'ategy anﬁ’Eryfanatfon: Given the activation energy and several temperatures, determine the fraction of molecules whose energies would be energetic enough to react. —E ERI In Equation 13.3 of Section 13.5, the exponential term. e a ._ is described as the fraction of sufﬁciently energetic molecules. Convert temperatures to Kelvin: °C - 213.15 = Kelvin. An example ofthe calculation is here for answer (a): 100. 3C — 1179.15 = 3—."3 K I —E RI 4:139] k.T.-'mol).-'I:0_UUES‘_4 kJ.’ mol-KJIISB K) ; Fraction: e a = e = 544-0: 3 x 10—30 1" (K) Fraction of sufficiently energetic molecules (a) 3.?3 e410 = s a: 10—30 (b) 4?3 €355 =4 a: 10—16 (c) F3 e-ll-T' =4 x 10-10 (d) 121's e-13-33=1.9 2c 10—5 V” Roasonaba‘eAnru-nr' Check: The fraction of molecules with enough energy to react increases Wll'l'l increasing temperature. 65. Amour: See graphs below Strategy andEipfarraHon: Draw these diagrams using the information described in the solution to Question :5 and in Section 13.4. 5-H and cm are identical for these diagrams. (a) Em...IEE = Emrmmr AH = (.15 k1 moi-1) — (—145 kJ mot-1) = 220. M mol-l I 7‘5 kJs‘mol 320. kl-"ﬁto. i_|i —145 15111101 13301231113 lde'LlCtS Reaction Progress (b) EHEWEE = Emmmd— AH = {:55 kJ mot-1) — (—TD. kJ moi—1] = 135 M 11101—1 65 kJ.-"1nol i 13 S kJs’mol | i | E I | | on. anm] i I _ _ _ _ _ _ _ _ _ _ _ ___l._ ____________ | | reactants products Reaction Progress (:3 Euﬂme = Eaimm— AH = {35 kJ mot-1) — (+10. M moi—1] = 15 kJ 11101—1 85 Iii-"11101 reactants pI'OdllCIS Reaction Progre 33 Chapter 13: Thermodynanncs: Dnectionality of Chemical Reactions 25. Juan-er: (21) Item 2 (to) Item 2 (1:) Item 2 Strategy and Explanation: Use the qualitative guidelines for entropy changes described in Section 18.3. (a) Item 2 has higher entropy since it is identical to item 1 except that its temperature is higher. Molecules at higher temperature have higher entropy. (In) Item 2, dissolved sugar. has higher entropy than item l_. solid sugar. because solute molecules are more random than those in a solid crystal. (c) Item 2, the mixture of water and alcohol together has higher entropy than item 1, water and alcohol separate. Mixing makes the molecules more random. 27. APTS‘H’QF.‘ (a) NoCl (to) P4 (c) FHﬂu'Oﬁoq) Sontag! and Explanation: Use the methods described in Problem—Solving Example 18.2 and the qualitative guidelines for entropy changes described in Section 13.3. (a) Comparing NaCl and CaO, we find that the biggest difference between these two ionic solids is the attractions due to the charges on the ions. According to Coulomb’s law, the Ca2+ and 02' ions have greater interaction than the Na+ and Cl‘ ions, so NaCl has a larger entropy-"mol than CaO. (h) P4 molecules have more atoms than Cl] molecules, so P4 has a larger entropy per mol than (31:. (c) The solid NH4NO3 crystal is more ordered than the aqueous NH4_ and ND; ions, so the aqueous NH4N03 has a larger entropy per mol than the solid NH4N03. 31. Amn-‘er: (a) negative (13) positive {c} negative (d) positive Strategy and Explanation. Use the qualitative guidelines for entropy changes described in Section 18.3. (a) The reaction has more gas-phase reactants (2 mol) than gas—phase products (1 mol), so the entropy change will be negative. (In) The reaction has fewer gas-phase reactants mol) than gas-phase products {3 mol), so the entropy change will be positive. (c) The reaction has more gas-phase reactants (4 mol) than gas—phase products (2 mol), so the entropy change will be negative. (d) The reaction has fewer gas—phase reactants (0 mol] than gas—phase products (1 mol). so the entropy change will be positive. ...
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