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Unformatted text preview: i/é/ Md 14.3 Analyze/Plan. Given mol A at a series of times in minutes calculate mol B produced, molarity
of A at each time, change in M of A at each 10 min interval, and AM A/s. For this reaction,
mol B produced equals mol A consumed. M ofA or [A] = mol A/0.100 L. The average rate of disappearance of A for each 10 minute interval is A[A : [A10 ‘ [A11 x 1 min s 10 min 60 s
Solve:
Time(min) Mol A (a) Mol B [A] A [A] (b) Rate (A [Al/s)
0 0.065 0.000 0.65
10 0.051 0.014 0.51 0.14 2.3 X104
20 0.042 0.023 0.42 0.09 1.5 x 10~4
30 0.036 0.029 0.36 006 1.0 x 10”4
40 0.031 0.034 0.31 0.05 0.8 x 10‘4
AME (0 029—0014)moI/0100L 1min
(c) = ' ‘ , ' x =1.25)<10“‘=1.3X10‘4M/s
At (30 — 10)mm 603
14.6 Time Time Interval Concentration A M Rate
(min) (min) (M) (M Is)
___/
0.0 1.85
54.0 54.0 1.58 0.27 8.3 x 104
107.0 53.0 1.36 0.22 6.9 x 106
215.0 108 1.02 034 5.3 x 10*"
430.0 215 0.580 0.44 3.4 x 104 14.10 (a) rate = A[HBr]/2At = A[H2]lAt = A[Br2]/At
(b) rate = A[SOZ]/2At = A[02]/At = A[SOa]/2At (0) rate = A[NO]/2At = —A[H2]/2At = A[N2]/At = A[HZO]/2At 14.14 (a) If [A] is doubled, there will be no change in the rate or the rate constant. The overall rate is unchanged because [A] does not appear in the rate law; the rate constant
changes only with a change in temperature. (b) The reaction is zero order in A, second order in B and second order overall. (c) Units of k = —M’—5 = M1 5“ M2 14.15 Analyze/Plan. Follow the logic in Sample Exercise 14.6. Solve:
(a) rate = k[N205] = 4.82 X104 5'1 [N205]
(b) rate = 4.82 x 10‘3 5'1 (0.0240 M) = 1.16 x 10“ M/s
(c) rate = 4.82 x 10‘3 5‘1 (0.0480 M) = 2.31 x 10‘4 M/s
When the concentration of N205 doubles, the rate of the reaction doubles. 14.19 Analyze/Plan. Substitute relative values into the rate law and solve for x. Solve:
(a) ' rate = [A]"; 3 = [3]"; x = 1
(b) 8 = [2]", x = 3
(c) 1 = [3]"; x = 0 (The rate does not depend on [A].) 14.25 Analyze/Plan. Follow the logic in Sample Exercise 4.6 to deduce the rate law. Rearrange the  a rate law to solve for k and deduce units. Calculate a k value for each set of concentrations
and then average the three values. Solve: (a) increasing [NO] by a factor of 2.5 while holding [Bra] constant (experiments 1 and 2)
increases the rate by a factor 6.25 or (2.5)2. Increasing [Br2] by a factor of 2.5 while
holding [NO] constant increases the rate by a factor of 2.5. The rate law for the appearance of NOBr is: rate = AINOBrllAt = k[NO]2[Br2]. 24 Ms ———2———— =1.20x1o4= 1.ZX10“M“2s“
(0.10M) (0.20M) (b) From experiment 1: k1 =
k2 = 150/(0.25)2(0.20) = 1.20 x 104 = 1.2 X 10‘ M‘ 5‘1
k3 = 60/(0.10)2(0.50) = 1.20 x 104 = 1.2 x 10“ M‘2 s‘1
4 = 735/(O.35)2(0.50) = 1.2 x 104 = 1.2 x 104 M"2 s"1
km= (1.2 X104 +1.2 X104 +1.2 X104+1.2 X104)/4 = 1.2 X104 M‘ZS"1
(c) Use the reaction stoichiometry and Equation 14.4 to relate the designated rates. A[NOBr]/2At = A[Br2]/At; the rate of disappearance of Br2 is half the rate of
appearance of NOBr. (d) Note that the data is given in terms of appearance of NOBr. —A[Br2] k[NO]2[Br2] 1 2 x 104
* = : ' 2 _
At ——2 ————2 28 x (0.075 M) x (0.25 M) — 8.4 M/s 14.26 (a) Increasing [820821 by a factor of 1.5 while holding [I‘] constant increases the rate by a factor of 1.5 (Experiments 1 and 2). Doubling [$20.32*] and increasing [1‘] byafactor /" 5
of 1.5 triples the rate (2 x 1.5 = 3, experiments 1 and 3). Thus the reaction is ﬁrst \~/ order in both [320821 and [I‘] ; rate = k [820321 [I‘]. (b) k = rate/[$20821 [I‘l
k1: 2.6 X 10'6 M/S / (0.018 M)(0.036 M) = 4.01 X 104 = 4.0'x 103 M‘1S'1
k2 = 3.9 X 10'6 /(0.027)(0.036) = 4.01 X 10‘1 = 4.01 X 10“3 = 4.0 X 10“3 M‘1 8"
k3 = 7.8 X 10"5 /(0.036)(0.054) = 4.01 X 104 = 4.01 X 10"3 = 4.0 X 10'3 M“1 s“
k4 = 1.4 X 10'5 /(0.050)(0.072) = 3.89 X 10—3 = 3.9 X 104 M”1 5‘1
kavg = 3.98 x 104 = 4.0 x 104 M“ s1 (c) A[82082‘]/At = A[I‘ ]/3At; the rate of disappearance of $20:— is onethird the rate
of disappearance of I“. (d) Note that the data is given in terms of disappearance of 820821
_A — 3A 8 02‘
ﬂ = A = 3(3.93 x 10*3 M1 s‘)(o.o15 M)(0.040 M) = 7.2 x 104 M/s At At Change of Concentration with Time 14.27 (a) [A]o is the molar concentration of reactant A at time zero, the initial concentration of
A. [A]. is the molar concentration of reactant A at time t. t1,2 is the time required to
reduce [A]o by a factor of 2, the time when [A]t = [Ala/2. k is the rate constant for a
particular reaction. k is independent of reactant concentration but varies with reaction temperature.
(b) A graph of n[A] vs time yields a straight line for a ﬁrstorder reaction.
14.28 (a) A graph of 1/[A] vs time yields a straight line for a secondorder reaction. (b) The halflife of a ﬁrstorder reaction is independent of [A]o, t1,2 = 0.693/k. Whereas,
the halflife of a secondorder reaction does depend on[A]o. t1,2 = 1/k[A]o. 0.693/k. Use this relationship to calculate k for a given t1 ,2, and, at a different temperature, t1,2
given k. Solve: 14.29 Analyze/Plan. The halflife of a ﬁrst—order reaction depends only on the rate constant, t“.2 — (a) tm = 2.3 x 105 s; rm = 0.693/k, k = 0.693/t1,2
k = 0593/23 x 105 s = 3.0 x 104 s“ (b) k=2.2 x 10'55“. t1,2=0.693/2.2 x 104s‘=3.15 x104=3.2 x 10‘s ...
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 Winter '07
 Zellmer
 Chemistry

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