CHP13 - released absorbed. . . hich means ositive, w...

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Unformatted text preview: released absorbed. . . hich means ositive, w released as pro Y value #800 k] te r d initially to bre lAErxn : Eabsorhed Ereleas'ed 7' Eabsorbed 13.49 The minimum .i occur. " 3.50 An inverse pr the reaction); t tion). the energy t duct bonds form. abs lls you that, even; , ak reactant bonds, forming the p Energy 1600 k] released N511“ = H800 k} FL__——H_., Form bondsl More energy on this side; overall reaction releases the difference, 800 k]. ~ Ereleased H AEEX“ : 800 k] ‘ (H800 kl) : 1600 k] t of energy reactant molecules must nr—the larger the Ea alue, the low he smaller the E 1 \‘a ue, the higher the rcac have in order for a reaction to ate (and the flower on r the reac- has been addeu to ..._ just enough energy d the product bonds to be just begnuur.U ’11 an . . t f Finish Reaction coordinate : 60 k] — 30 k] : +3014 4 Ereactants Ea :5 Etransition state reactants : if 30 : +70 f‘ The positiVe ABM value makes this an endothermic reaction. lower the activation energy barrier, the faster the reaction limb the energy b Increase the rate. The molecules have enough energy to c the smaller Ea is, ' , the SIOWer the reaction; hwersely. The larger E8 is te because heating up the reaction mix :ules have enough en come does n h tum ' 13'62 Because the Size 0f Ea depends on t Ethicause the reactan strong bo reaction - - have a much larger Ea t1;;e amount of energy 0 break the A reactant bonds, resulting in a temperature (1 a new atom or W0 action of energetically favorable molecules are converted to product. 13.66 A substance that ' and the presence of the ZEN—O bond weak- he C—OH bond easier to break, thus lowering .eding it up. processes more financially profitable. Catalyst. rate necessary to sustain life would proceed so slowly tha 13.74 Because increasing reactant c tant molecules. The more (:01 13.75 Because decreasing the volume of a gas increases tant concentration this way increases the number thus increasing the rate of the reaction. 13.76 Because both factors in the product depend on inherent (unchanging) properties of the, ' reactants. at normal temperatures. t death would occur. oncentration increases lisions, the faster the rate of reaction. g the proper orientai or the reaction that allows a larger fraction Of or endothermic. 434 _ COMPLETE SOLUTIONS 13.91 13.92 13.93 13.94 -“-‘—v m“ 1”...” pm Lou-cu...“ it: man Ulucl vv1ut LCDIJCLI. Lu :1 quu 111m Uiucl‘ With respect K) D. Therefore doubling the concentration of A doubles the rate and doubling the concentration of B doubles the rate. Doubling both at the same time quadruples the rate. For k = 1 and initial concentration of 1 M for A and 3 M for B, for instance, Rate: 1X(1M)X(3M)=3M/s Rate: 1 X (2M) >< (6M) = ‘l2M/s A kinetics experiment is one in which the reactant concentrations are changed one at a time and how these changes affect reaction rate is noted. The exponents in an experimental rate law are determined from kinetics experiments. You find the order of each reactant by comparing two experiments in which the concentra- tion of that reactant is changed while the concentrations Of all other reactants are held constant. Determine the order with respect to H202 from experiments 1 and 2. Doubling [H202] causes the reaction rate to double from 1.15 X 10'6 M/s to 2.30 X 10'6 M/s, making the reaction first order with respect to H202: rate = k[H202]][I—}?[H+]?. Determine the order with respect to I‘ from experiments 1 and 3. Doubling {I‘} causes the reaction rate to double from 1.15 X 10—6 M/s to 2.30 X 10"6 M/s, making the reaction first order with respect to I’: rate r: k[H202]1[I—]1[H+]?. Determine the order with respect to H+ from experiments 1 and 4. Doubling [H+] has no effect on the rate, making the reaction zero order with respect to H+: rate = k[H202]1[I‘]1[H+]0. Because exponent 1 is left unwritten and any zero-order reactant does not appear in the rate law, the law for this reaction is Rate = k[H202][I‘] The overall order of the reaction is 1 + 1 = 2. You find the order of each reactant by comparing two experiments in which the concentra- tion of that reactant is changed while the concentrations of all other reactants are held con- stant. Determine the order with respect to A from experiments 1 and 2. Doubling [A] caus- es the reaction rate to double from 0.0281 M/s to 0.0562 M/s, making the reaction first order with resPect to A: rate : k[A]1[B]?. Determine the order with reapect to B from experiments 1 and 3. Doubling [B] causes the reaction rate to quadruple from 0.0281 M/s to 0.1124 M/s, making the reaction second order with respect to B: rate I k[A]1[B]2. Dropping the exponent 1 by convention, you get the rate law rate = k[A][B]2. The overall order of the reaction is 1 + 2 = 3, and this problem confirms that the exponents in the rate law are not necessarily the same as the coefficients in the balanced equation. You find the order of each reactant by comparing two experiments in which the concentra- tion of that reactant is changed while the concentrations of all other reactants are held con- stant. Determine the order with respect to C102 from experiments 1 and 2. Tripling [C102] from 0.020 M in experiment 2 to 0.060 M in experiment 1 causes the reaction rate to " increase ninefold from 0.002 76 M/s to 0.024 84 M/s, making the reaction second order p with respect to C102: rate : k[C102]2[OH]?. Determine the order with respect to OH' from experiments 2 and 3. Tripling [OH—I from 0.030 M in experiment 2 to 0.090 M in experiment 3 causes the reaction rate to triple {1‘01}I 0.002 76 M / s to 0.008 28 M / 5, making the reaction first order with respect to OH? rate I k[C102]2[OI-I]1. Dropping the exponent 1 by convention, you get the rate law rate = k[C102]2[OH_l- COMPLETE SOLUTIONS ’rms that the is 2 + l r: 3, and this prob y the same as the coefficients in the balanced of the reaction t necessaril are no ome together tion. the reaction to occur in one step, four molecules would have to c ' a tation. This is very u likely. ' the balancing the proper oricn ' the rate—determini All you can ay is nism. There may e other mechanis Incorrect mechanisms can be ruled out by experimental ev prove eaSily that a given possible mechanism for a reaction is the correct one. 13.105 (a) The first—or e dependence on [(Cl-l3)3CBr] means the reaction rate doubles when [(CH3)3CBr] is doubled, triples when [(CH3)3CBrl s tripled, and so forth. The zero- order depe dence on [H20] means the rate is t affected by any change in [H20] (b) You can rule out mechanism 11 because it does not agree with the experimental EVI‘ dence. lts one—step mechanism means the reaction is first rder in both CH3)3CBr and H20. Mechanism '1 is plausible because it agrees with the experimental data and the sum of the indiVidual steps yields the balanced equation: (cringe): 3‘51"!» (CH3)3C+ + Br gcngigcr + ago “‘5‘ (wagon + l-l+ H+ + Br“ iii-LliiBr ; (CH3)3CBr + H10 fi—J—m (CH3)3COH + HBr 'ned by adding the steps in the mechanism: rall balanced e nation is determ (:1 ’9 cl + ct ——» HCl + col}, 13.106 The ove WNW.“ A37 ...
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CHP13 - released absorbed. . . hich means ositive, w...

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