Ch.16 kinetics problems
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Overall, the document discusses two chemical reactions and their step-by-step mechanisms. The first reaction of 2A + B gives the products C + BC through the elementary steps of A + A giving C and C + B giving BC. The second reaction of ClO- + I- giving Cl- + IO- is proposed to occur through three steps involving H2O, OH-, HIO, and HClO. However, the proposed slow step rate law is found to be inconsistent with the mechanism, as HClO should be treated as an intermediate rather than a reactant. Replacing it gives a rate law that does not match observations.
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![November 26, 2010
66. A + B gives AB Elementary step
A + A gives C
C + B gives BC
Overall 2A + B gives C + BC
72. a) A + B + C gives D
b) X and Y
c) step 1 is bimolecular rate = k[A][B]
step 2 is bimolecular rate = k[X][C]
step 3 is unimolecular rate = k[Y]
d) yes
e) yes it's possible](https://image.slidesharecdn.com/ch-16kineticsproblems-101126100755-phpapp02/85/Ch-16-kinetics-problems-1-320.jpg)
![November 26, 2010
73. a. ClO- + I- gives Cl- + IO-
b.H2O, OH-, HIO, HClO
c. step 1 is bimolecular rate = k[ClO-][H2O]
step 2 is bimolecular rate = k[ I-][HClO]
step 3 is bimolecular rate = k[OH-][HIO]
d. slow step rate law is rate =k[ I-][HClO] this is not
consistent with the proposed rate law. The HClO is an
intermediate and should be replaced. From step 1, leaving out
water, [HClO] = [ClO-]/[OH-]. Replacing [HClO] in the slow step
rate law gives:
rate = k[I-][ClO-]/[OH-]. This is not the observed rate law.](https://image.slidesharecdn.com/ch-16kineticsproblems-101126100755-phpapp02/85/Ch-16-kinetics-problems-2-320.jpg)
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Ch.16 kinetics problems
- 1. November 26, 2010 66. A + B gives AB Elementary step A + A gives C C + B gives BC Overall 2A + B gives C + BC 72. a) A + B + C gives D b) X and Y c) step 1 is bimolecular rate = k[A][B] step 2 is bimolecular rate = k[X][C] step 3 is unimolecular rate = k[Y] d) yes e) yes it's possible
- 2. November 26, 2010 73. a. ClO- + I- gives Cl- + IO- b.H2O, OH-, HIO, HClO c. step 1 is bimolecular rate = k[ClO-][H2O] step 2 is bimolecular rate = k[ I-][HClO] step 3 is bimolecular rate = k[OH-][HIO] d. slow step rate law is rate =k[ I-][HClO] this is not consistent with the proposed rate law. The HClO is an intermediate and should be replaced. From step 1, leaving out water, [HClO] = [ClO-]/[OH-]. Replacing [HClO] in the slow step rate law gives: rate = k[I-][ClO-]/[OH-]. This is not the observed rate law.