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CHP13A - released the energy absorbe hich means bonds form...

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Unformatted text preview: released the energy absorbe hich means bonds form. t overall, the reaction release t bonds forming he product bond ' S " AE‘nm .1 #800 k} i his side, leases 500 k] H (-800 k]) = 1600 k] must have in order tor a reaction to r the Eavaiue, the higher ' neethe iarge E d v aiue, been addem Lu uct bonds to be just begrmuho — . approach the molecule initially containing X from the side opposite X, no reaction will occur: H H-..\ CH3Y ..... Z/c—x = CH3CH2Y + 013* + X‘ H3C H "/C — X< ----- YCH3 = No reaction 13.65 Faster, because then all reactants havmg sufficient energy would be converted to product. Because of the orientation requirement, only a small fraction of energetically favorable molecules are converted to product. 13.66 A substance that increases the rate of a reaction without being used up in the process. 13.67 By providing an alternate pathway that has a lower activation energy Ea. 13.68 The Zn2+ bonds with the O of the OH group, and the presence of the Zn2+—O bond weak- ens the C—OI—l bond. This weakening makes the C~—OH bond easier to break, thus lowering the activation energy for the reaction and speeding it up. 13.69 Enzymes. 13.70 The reactant molecule that fits into an enzyme's active site. 13.71 The lock-and—key mechanism describes how, in a chemical reaction catalyzed by an enzyme, a substrate (the key) fits into the enzyme’s active site (the lock). The substrate fits only if it has the correct shape, just as a key fits a lock only if the key has the correct shape. 13.72 Key. Scientists made a sulfanilamide molecule that was a modified version of the PABA key. The sulfanilamide key in the enzyme lock produced faulty vitamin B. 13.73 Because they speed up the processes, thereby decreasing production time and making the processes more financially profitable. Catalysts allow biological processes to proceed at the rate necessary to sustain life at normal temperatures. Without catalysts, the reactions would proceed so slowly that death would occur. . 13.74 Because increasing reactant concentration increases the number of collisions between reac- tant molecules. The more collisions, the faster the rate Of reaction. 13.75 Because decreasing the volume of a gas increases the gas concentration. Increasing the reac-. tant concentration this way increases the number of collisions between reactant molecules, 3 thus increasing the .rate of the reaction. 13.76 Because both factors in the product depend on inherent (unchanging) preperties of the ‘ reactants. ‘ 13.77 Increase because it depends partially on the fraction of collisions having energy equal to 01' greater than Ea. As temperature increases, more collisions have sufficient energy and k increases. 1 13.78 Increase because it partially depends on the fraction of collisions having the proper orienta' tion. A catalyst provides a new mechanism for the reaction that allows a larger fraction of . ' the reactant molecules to have the proper orientation, thus increasing k. 13.79 The fraction of collisions having energy equal to or greater than Ba and the fraction of C sions having the proper orientation. 13.80 it depends on the inherent factors of a reaction, not on whether the reaction is exotherm1 E or endothermic. WW 4334 _ COMPLETE SOLUTIONS 111:)!- UL“..— I. LCDIJCLL 1U f1 01“.; tration of A doubles the rate and doubling the concentration druples the rate. For k = 1 and Therefore doubling the concen of B doubles the rate. Doubling both at the same time qua 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 13.91 A kinetics experiment is one in which the reactant concentration and how these changes affect reaction rate is noted. The exponen law are determined from kinetics experiments. eriments in which the concentra- 13.92 You find the order of each reactant by comparing two exp 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 2‘ kIH202]1[I']‘[H+]?. Determine the order with respec reaction rate to double from 1.15 X order with respect to I": rate : k[H202]1[I']1[f-I+]?. Determine the order with respect to H+ from experiments 1 and 4. Doubling [I-I+] has no effect on the rate, making the reaction zero order with respect to H+2 rate = k[H202]1[I—]1[H+]“. 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[H2021[I‘] The overall order of the reaction is 1 + 1 r— 2. eriments in which the concentra- 1333 You find the order of each reactant by comparing two exp 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]?. ect to B from experiments 1 and 3. Doubling [B] causes the 81 M / s to 0.1124 M/s, making the reaction second 5 are changed one at a time ts in an experimental rate Determine the order with resp reaction rate to quadruple from 0.02 order with respect to B: rate 2 MAP [312. ' It, you get the rate law rate = k[A][B]2. The overall order of the reaction is 1 + 2 2 3, and this problem confirms that the exponents 2' in the rate law are not necessarily the same as the coefficients in the balanced equation. 13.94 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 / 5, making the reaction second order with respect to C102: rate = k[ClOZ]2[OH]?. Determine the order with respect to OH‘ from experiment 0.030 M in experiment 2 to 0.090 M in experiment 3 causes 0.002 76 M / s to 0.008 28 M / s, making the reaction first order with respect to k[C102]2[OH1 1. Dropping the exponent l by convention, you get the rate law rate = k[ClOZ]2{OH_l- ”MW 436 COMPLETE SOLUTIONS rail order of the re ts in the rate law are not nec ccur in one step, to orientation. likely. step, and the balancing in the rate law. This is very un e rate—determining the orders e reaction to n all with the proper e slowest elementary step is th 'cients from the rate-determining step are equal to ' ' = two reactant molecules. ed in one step (a practically impossible occur- ate—determining elementary step (because it is the e nation must re _ ' = ‘ ients can be used as orders to yield rate = erall reaction is the same as any multistep reacti rate law for the ov te law for the rate—determining step. se it is extremely unlikely that three molecules will come to r reaction. it does not agree with the experimentall gether in the ori- the ra .101 No, becau entation necessary to e correct because y determined correct mecha— rith the experimental data. '. ms that also agree w vidence, but you cannot ental e reaction is the correct one. ossibly be the rove easr (a) The first—order dependenc _ [(C 3)3C‘Br] is doubled triples when [(CH3)3CBr] is tripled, and so order dependence on [H20] means the rate is not affected by any change in [H20] (b) You can rule out mechanism [I because it does not agree with the experimental evi- dence. Its one-step mechanism means the reaction is first order 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 (cnghcsr 310“” (CH-93f? + Br“ (eraser + H30 5315‘ (camcon + H+ fluids? er—H (CH3)3COH + HBr chanisrn: (21+ CHCIE, (31' + 01:13 F» (:04 C12 + CHC13 ~—a HCl + CCl4 Rate = lelZ] because only the reactants in the rate-determining step appear in the rate law. The rate—law exponent indicating order is the same as the reactant coefficient 13.107 in the rate-determining step—#1 in this case. ...
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