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14.83)
The balanced chemical equation shows the overall, net change during a chemical
reaction and only tells us what the reactants and products are and how much of
each is involved. It does NOT tell us HOW the reaction occurs.
Most reactions occur as a series of steps (elementary, single step, reactions). The
rate law is initially written in terms of the reactants from the ratedetermining step.
If any intermediates are present (in a multistep mechanism) in this initial rate law
we try to “get rid” of them by algebraic solving for their concentrations in terms of
reactants, products and/or catalyst.
For any elementary process (singlestep process) the rate law can be written
directly from the balanced equation for the step (using the coefficients). This is
ONLY true for an elementary process. All steps in a multistep mechanism are
elementary processes.
For a reaction that occurs in a single step (an elementary process), the rate law can
be written directly from the balanced equation. However, you can NOT tell that a
reaction is an elementary process simply by looking at it. You must either be told
or have experimental evidence (such as the experimental rate law).
Our experimental rate law tells us something about the mechanism. If the
exponents of the substances in the rate law agree with the coefficients in the
balanced equation, then we can theorize that the reaction may be an elementary
process (occurs as a single step). However, it does not mean that it does, it is just
one possibility. If the exponents of the substances in the rate law do NOT agree
with the coefficients in the balanced equation, then we can say the reaction does
NOT occur as an elementary process and must proceed as a series of steps, MUST
have a multistep mechanism. 57
14.85) 58
14.85) (cont.) 59
14.86) 60
14.86) a) (cont.) 61
14.86) a) (cont.) c) Use the rate law to find the rate
[HgCl2] = 0.100 M [C2O42&] = 0.25 M r = k [HgCl2] [C2O42&]2
r = (8.672 x 10&3 M2s1) [0.100 M] [0.25 M]2
r = 5.420 x 10&5 M/s
= 5.4 x 10&5 M/s 62
14.88)
The following are the halflife relationships for the reactions we’ve studied:
Zeroorder
[A]o
t1/2 = 2k 1storder
0.693
t1/2 = k DECREASES
as rx. proceeds
(successive ½lifes
get smaller  ½ previous
e.g. t1/2,1 = 40 s
t1/2,2 = 20 s
t1/2,3 = 10 s ) CONSTANT
as rx. proceeds
(successive ½lifes
are same
e.g. t1/2,1 = 40 s
t1/2,2 = 40 s
t1/2,3 = 40 s ) t1/2 2ndorder
1
= k [A]o INCREASES
as rx. proceeds
(successive ½lifes
get larger  double
e.g. t1/2,1 = 10 s
t1/2,2 = 20 s
t1/2,3 = 40 s ) For a firstorder reaction the halflife is constant, whereas the halflife for zeroorder and 2ndorder are NOT constant. The halflife for zeroorder gets shorter
as the reaction proceeds (successive halflifes are cut in half each time). The halflife for 2ndorder gets longer as the reaction proceeds (successive halflifes
double each time).
For rxn (1) the halflife is constant as the rxn proceeds. This implies the reaction
is firstorder.
For rxn (2) the halflife gets longer as the rxn proceeds. This implies the reaction
is NOT firstorder. It is also NOT zeroorder (halflife gets shorter). Reaction
(2) must be a higher order than firstorder. We can not say for sure whether it is a
2ndorder but it could be.
A reaction that is 3rd order in reactant A would have the rate law, r = k [A]3. This
would then have an integrated rate equation and halflife of the following forms:
1
1
 = k t + [A]t2
[A]02 3
t 1/2 = k [A]02 This halflife increases as the reaction proceeds (gets longer; each successive halflife is 4 times the previous). 63
14.90)
We have two radioactive particles. The decay process for radioactivity is first
order. The rate constants for the radioactive decay are given. We want the halflifes for each, which one decays at a faster rate and the amount present after 3 halflives. The rate law for both is r = rate = k [substance].
Americium241: 241 Am k = 1.6 x 10!3 yr!1 Iodine125: 125 I k = 1.1 x 10!2 day!1 The rate law for each would be:
r = (1.6 x 10!3 yr!1) [Am] r = (1.1 x 10!2 day!1) [I] a) Halflifes:
241 Am: t1/2 = 0.693/(1.6 x 10!3 yr!1) = 433.1 yr = 4.3 x 102 yr 125 I: t1/2 = 0.693/(1.1 x 10!2 day!1) = 63.0 days = 63 days b) Which decays at a faster rate?
For a given sample size, half of the 241Am decays in 433 years, whereas half of the
125
I sample decays in 63 days. It takes less time for the 125I to decay to half of its
initial conc. The 125I decays at a much faster rate. *** continued next page *** 64
14.90) (cont.)
c) How much of a 1.00mg sample of each remains after 3 halflives?
There are two ways to determine this answer.
1) Method 1: really easy
We simply want what remains after 3 halflives. Remember a halflife means
the time it takes to get to ½ of the original concentration. That means 50%,
25%, 12.5%, 6.25%, etc. Since we are going 3 halflives this is 12.5% of the
original. Thus, each will be 12.5% of the original 1.00 mg sample, 0.125 mg.
The real difference is in how long it will take the concentration of each
radioactive substance to decrease to 12.5% (0.125 mg).
241 Am: t = 3 t1/2 = 3 (433.1 yr) = 1299 yr = 1.3 x 103 yr 125 I: t = 3 t1/2 = 3 (63.0 days) = 189 days = 1.9 x 102 days 2) Method 2: use the 1st order integrated rate equation
ln[A]t = !kt + ln[A]0
Want conc. after 3 halflives (see time above).
241 Am: ln[Am]t = ! (1.6 x 10!3 yr!1)(1299 yr) + ln(1.00)
= ! 2.079 + 0.000
[Am]t = e!2.079 = 0.1250 = 0.13 mg 125 I: ln[I]t = ! (1.1 x 10!2 day!1)(189 days) + ln(1.00)
= ! 2.079 + 0.000
[I]t = e!2.079 = 0.1250 = 0.13 mg 65
14.94)
Use the Arrhenius equation that relates k to Ea and T. k = A e&Ea/RT a) where, A = frequency factor
Ea = activation energy
T = temperature of Kelvin Even if Earx1 = Earx2 the k’s do NOT have to be equal. The rate constant, k,
also depends on A, the frequency factor, which is related to the frequency of
collisions and their effectiveness (proper orientation).
So if Arx1 … Arx2 then krx1 … krx2 b) Two similar reactions have the same rate constant at 25°C, but at 35°C one of
the reactions has a higher rate constant than the other. Why? From the twopoint form of the Arrhenius equation one sees that the
temperature dependance of the rate constant, k, depends on the activation
energy, Ea. Reactions with different behaviors for k with respect to
temperature must have different activation energies, Ea. The fact the two
reactions have the same rate constant at 25°C is a coincidence. The reaction
with the higher rate constant at 35°C has the larger activation energy.
Temperature has a bigger effect on reactions with higher activation energies.
Temperature increases the kinetic energy of the particles and the reaction
with the larger Ea can use this increase in KE more effectively. 66
14.96)
The following mechanism has been proposed for the reaction of NO with H2 to
form N2O and H2O.
a) Show the steps in the mechanism will add to give the overall balanced equation.
NO (g) + NO (g) v N2O2 (g) (1) (Slow)
N2O2 (g) + H2 (g) v N2O (g) + H2O (g) (2)
2 NO (g) + N2O2 (g) + H2 (g) v N2O (g) + H2O (g) + N2O2 (g)
Note: The N2O2 appears on both sides of the equation and will cancel out. It
is an intermediate. Overall balanced eqn: 2 NO (g) + H2 (g) v N2O (g) + H2O (g) * continued on next page * 67
14.96) (cont.) 68
14.99) 69
14.99) (cont.) 70
14.102)
Enzymes are often described using the following twostep mechanism:
E+S k1 » ES (fast) k!1 ES v E + P
E+Sv E+P
k2 E = enzyme (catalyst) (slow) S = substrate (reactant) P = product What is the rate law?
Write the initial rate law in terms of the reactants from the slow (ratedet.) step.
r = k2 [ES]
However, ES is in intermediate. We need to try to get rid of it using step 1.
rate of forward step 1: rfor = k1 [E][S]
rate of reverse step 1: rrev = k!1 [ES]
Since step 1 is faster than step 2, step 1 will come to equilibrium.
rate forward rxn = rate everse rxn
rfor = rrev k1 [E][S] = k!1 [ES] k1
[ES] =  [E][S]
k!1
k2 k1
r = k2 [ES] =  [E][S] = k [E][S]
k!1
so r = k [E][S] where k = (k2 k1/k!1) 71
14.105) 72
14.107)
The gas phase reaction of NO and F2 to form NOF and F,
NO + F2 > NOF + F
is believed to be bimolecular and has an activation energy Ea = 6.3 kJ/mol and a
frequency factor A = 6.0 x 108 M!1s!1.
a) Calculate k at 100 °C.
Use the Arrhenius equation (as below or straightline form): k = A e!Ea/RT ln k = ! Ea/(RT) + ln A 3
k = (6.0 x 108 M!1s!1) e!(6.3 x 10 J/mol)/((8.314 J/molCK)(373.15 K)) k = (6.0 x 108 M!1s!1) e!2.0307 = 7.874 x 107 M!1s!1 = 7.9 x 107 M!1s!1
b) Draw the Lewis structures for NO and NOF (ONF, N in middle) 73
14.107) (cont.)
c) NOF (ONF) has 3 things around it (3 regions of electron density) so its
electron domain is trigonal planar. However, it has 2 atoms and 1 lpe! on
the N so its molecular shape (geometry) is bent, with a bond angle of .120°. d) Possible transition state e) Give a possible reason for the low activation energy.
The NO molecule is an oddelectron system with one unpaired e!. That
makes it e! deficient. It is trying to pair up that e!. It is attracted to the e!
rich F2. This attraction between the molecules means they are more likely to
collide than by just random collisions. Also, you aren’t breaking a bond
between the N and O in order to form the NOF (ONF). This unpaired e! on
the N is attracted to the e! in the bond between the F atoms. You are
breaking the bond between the F atoms, which requires energy, but then
forming a bond between the F atom (with its now unpaired e!) and the
unpaired e! on the N, which releases energy (nearly as much as required 74
14.110) ...
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This note was uploaded on 06/11/2011 for the course CHEM 122 taught by Professor Zellmer during the Spring '07 term at Ohio State.
 Spring '07
 Zellmer
 Chemistry, Reaction

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