CHEM 102 Discussion Class 3 (Chapter 13)
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Integrated Rate Laws and t
/12
:
[A] t
− [A]
0
= − k ⋅ t t
1/ 2
=
[A]
0 k ⋅ t
0 order
€ ln
[A]
0
[A] t
1
= ln[A]
0
−
1
[A] t
[A]
0
− ln[A] t
= k ⋅ t
= k ⋅ t t
1/ 2
= t
1/ 2
1 k ⋅ [A]
0
=
0.693
k
1st order
2nd order
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Using Integrated Rate Laws – Calculating Quantities
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(1) A zero order reaction starts with an initial concentration of reactant of 1.00 M and has a rate constant of 1.65x10
-5 M/s. What is the concentration of the reactant after (a) 1 s (b) 100 s and (c)
100 minutes?
This problem asks for quantities after a certain amount of time – all problems like this use integrated rate laws. We must rearrange the zero order rate law for [A] t time t. Note 100 minutes = 6000 s.
, the concentration at
[A] t
− [A]
0
[A] t
= − k ⋅ t
= − k
⋅ t
+
[A]
0
Therefore
(a)
[A]
1 s
= − k ⋅ t + [A]
0
= ( − 1.65x10
− 5
M / s × 1 s) + 1.00 M
= 0.999 M
(b)
[A]
100 s
= − k ⋅ t + [A]
0
=
(
−
1.65x10
− 5
= 0.998 M
M / s
×
100 s)
+
1.00 M
(c)
[A]
6000 s
= − k ⋅ t + [A]
0
=
(
−
1.65x10
− 5
= 0.901 M
M / s
×
6000 s)
+
1.00 M
(2) A first order reaction ABC → AB + C has a rate constant of 0.0125 /s. If the initial concentration of ABC is 0.500 M, what is the concentration of ABC after 5 minutes?
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First note we must turn 5 minutes into seconds (300 s). Next we must rearrange the integrated first order rate law to solve for [A] t ln[A]
0
− ln[A] t
− ln[A] t ln[A] t
[A] t
= k ⋅ t
= k ⋅ t − ln[A]
0
= −
(k
⋅ t
− ln[A]
= e − (k ⋅ t − ln[A]
0
)
0
=
) e − ((0.0125 / s × 300 s) − ln(0.500 M)
= e − 4.44
= 0.0118 M
(3) The reaction NO
2
(g) + NO
2
(g) → N
2
O
4
(g) is second order with a rate constant of 0.044 /M·s at a particular temperature and pressure. If the initial concentration of NO
2
(g) is 0.100 M, what is the concentration of NO
2
after 20 minutes?
Note that 20 minutes = 1200 s. We must rearrange the second order integrated rate law for
[NO
2
] t
. Actually, it’s easier to rearrange for 1/[A] t
first.
1
[A] t
1
−
[A]
0
1
[A] t
1
[NO
2
]
1200 s
=
= k k ⋅
⋅ t t
+
1
[A]
0
= (0.044 / M ⋅ s × 1200 s) +
[NO
2
]
1200 s
=
62.8 / M
1
=
62.8 M
= 0.016 M
1
0.100 M
(4) A liquid evaporates following zero order kinetics with k = 6.75x10
-3 take for 10.0 g of the liquid to evaporate?
g/s. How long does it
This problem asks us to find t but gives us no initial (or final) concentrations. The equations still work for amounts (g) instead of concentrations. All we know is that the final amount is 10.0 g less than the initial amount. We could use a pair of values, for example [Liquid]
0
= 100.0 g and
[Liquid] t
= 90.0 g.
[A] t
− [A]
0
= − k ⋅ t
[A] t
− [A]
0
− k
= t t =
90.0 g − 100.0 g
− 6.75x10
− 3 g / s
=
− 10.0 g
− 6.75x10
− 3 g / s
= 1480 s
Notice that it would make no difference whether we chose an initial amount of 100.0 g or 1000.0
g or any other number because we’d be left with the -10.0 g as the numerator regardless! All we

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€ care about is the difference! Another way of saying this is that the rate of evaporation for a zero order process doesn’t depend on the initial amount, as you know.
(5) A reaction X
2
Y → 2 X + Y has a rate law Rate = 0.104 /M·s·[X
2
Y] 2 . If the initial concentration of products is zero, how long does it take for the concentration of X to reach 0.100
M starting with an initial concentration of X
2
Y of 0.200 M
We can see from the stoichiometry that when X
2
Y decomposes it makes two X molecules. This means, in order to make 0.100 M of X, we would need to use 0.050 M of X to find out how long it takes for the initial concentration of X
2
2
Y. Hence, we need
Y to fall by 0.050 M, that is, using the conditions [X
2
Y]
0
= 0.200 M and [X
2
Y] t
= 0.150 M. This is a second order reaction so we need to use the second order integrated rate law.
1
[A] t
1
[A] t
1
−
[A]
0
1
− k
[A]
0
=
= t k
⋅ t t
=
1
0.150 M
−
1
0.200 M
0.104 / M ⋅ s
=
6.667 / M − 5.000 / M
0.104 / M ⋅ s
=
16.0 s
(6) The elimination of alcohol in the bloodstream by humans is a zero order reaction. The legal maximum for driving is 0.080 g/100 mL of blood, corresponding to a concentration of 17.3 mM.
The rate constant varies for individuals but an average is about 0.015g/mL·hour (3.25 mM/hour).
If an individual has a blood alcohol level of twice the legal limit for driving, how long will it be before the person can legally drive?
This one can be solved in a similar way to (4). It’s easiest to work in the mM (millimolar) units but identical results are obtained using the mg/mL units. Twice the legal limit would be (2 x
17.3 mM = 34.6 mM).
[A] t
− [A]
0
= − k ⋅ t
[A] t
− [A]
0
− k
= t t
=
17.3 mM
−
34.6 mM
−
3.25 mM / hour
=
−
17.3 mM
−
3.25 mM / hour
=
5.32 hours
These numbers are approximately correct! It takes about 5 hours to ‘sober up’ enough to drive if you are at twice the legal blood alcohol content.
Half Life
(7) The half life of a particular reaction decreases with increasing concentration. What kind of kinetics, zero, first or second order, is characteristic of this behavior?

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For zero order kinetics, the half-life gets shorter as the reaction proceeds t
1/2
= [A]
0
/2·k. The equation shows that as [A]
0
increases, t
1/2
increases. In other words, the half-life is longer for the higher initial concentrations. The reactions cannot be zero order.
For first order kinetics, the half-life doesn’t change as the reaction proceeds t
1/2 equation shows that the half life is constant and changing [A]
0
has no effect.
= 0.693/k. The
For second order kinetics, the half-life gets longer as the reaction proceeds t
1/2
= 1/k·[A]
0
. This t equation shows that, because of the inverse relationship between t
1/2
and [A]
0
, as [A]
0
increases,
1/2
decreases. This is the behavior observed so the reaction must be second order.
(8) The concentration changes from 0.196 M to 0.144 M during the first 30 s of a first order reaction. What is the half-life?
At first sight, there doesn’t seem to be enough information here. To find t
1/2
we need k. But there is enough information to calculate k here.
ln
[A]
0
[A] t ln
[A]
0
[A] t t
= k
⋅ t
= k k
= ln
0.196 M
0.144 M
30 s
= ln1.361
30 s
=
0.0103 / s
Now we can find t
1/2
: t
1/ 2
=
0.693
k
=
0.693
0.0103 / s
= 67.3 s
Arrhenius Equation
(9) A graph of lnk versus 1/T for a particular reaction has a slope of -6013 /K and a y-axis intercept of 24.10. (a) What is E reaction at 300 K?
a
and A for this reaction? (b) What is the rate constant for this
(a) The linearized version of the Arrhenius equation is obtained by taking natural logarithms of both sides of the Arrhenius equation: k = A ⋅ e − E a
/ R ⋅ T ln k
= −
E a
R
⋅
1
T
+ ln A
This shows a graph of lnk versus 1/T would have a slope m = –E a
/R and an intercept b = lnA.
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€ m
= −
E a
R
E a
= − m
⋅
R
= −
(
−
6013 / K)
×
8.314 J / K
⋅ mol
=
49990 J / mol (50 kJ / mol) b
= ln A
A = e b
= e
24.10
= 2.92x10
10
(b) The rate constant k is k
=
A
⋅ e − E a
/ R ⋅ T
=
2.92x10
10
= 2.92x10
10
× e − 20.04
= 57.8
/ s
× e − 50000 J / mol /(8.314 J / K ⋅ mol × 300 K)
One curious thing to note here is that we cannot calculate the units for A because an antilog has no units. This carries through to (b) because we cannot therefore predict the units of k in this case. This is actually good news: the Arrhenius method works for zero, first or second order kinetics and each one has different units for k. We can’t use the Arrhenius plot to predict the reaction order unfortunately.
Reaction mechanisms
(10) I propose that a reaction has three elementary steps:
(1) H
2
(g) + O
2
(g) → H
2
O
2
(g) Slow
(2) H
2
O
2
(g) → H
2
O(g) + O(g) Fast
(3) O(g) + H
2
(g) → H
2
O(g) Very fast
(a) Which step is the rate determining step? (b) What is the overall reaction? (c) Identify any catalysts or intermediates. (d) Write down the expected rate law for this reaction.
(a) The slowest step is the rate-determining step, reaction (1).
(b) The overall reaction involves canceling the H
2
O
2
(it is on the right in (1) and left in (2)) and
O (it is on the right in (2) and left in (3)) to leave 2 H
2
(g) + O
2
(g) → 2 H
2
O(g). Note it is balanced automatically if the three elementary steps are balanced.
(c) H
2
O
2
and O are intermediates. There are no catalysts.
(d) The slowest step determines the rate law and so we look at (1). It is a bimolecular reaction A
+ B and so is expected to have the form including both reactants as first order: Rate = k·[H
2
]·[O
2
]. We would have guessed Rate = k·[H
2
] 2 ·[O
2
] if we’d looked at the overall equation!