Session #23: Homework Solutions
Problem #1
The decomposition of hydrogen peroxide, H2O2, can be represented by the following
reaction:
2H2O2(aq) → 2H2O(l) + O2(g)
The table below reports data taken at room temperature (300 K).
Table 1. Decomposition of H2O2(aq) at 300 K.
concH2O2 (mol/liter)
time (seconds)
2.32
2.01
1.72
1.49
0.98
0.62
0.25
0
200
400
600
1200
1800
3000
(a) Show that the reaction is first order.
(b) Calculate the value of the half-life of this reaction.
(c) Suppose that the initial concentration of H2O2 were 3.5 M. How long would it take at
300 K to reduce the concentration of H2O2 to 25% of its initial value?
Solution
(a) To show that the reaction is first order, try fitting the logarithm of concentration
versus time. Least-squares analysis gives:
ln c = 0.831 – 7.21×10–4 t
with a correlation coefficient of 0.998.
(b) The half-life is given by t1/2 =
ln 2
0.693
= 961 s
=
k
7.21 × 10-4
(c) To decrease the concentration to 25% of initial value would take 2 half-lives, since
after t1/2 the concentration would be 50% and after 2t1/2 it would be 50% of 50%. So
the answer is 2×961 s = 1922 s.
Problem #2
A chemical reaction which has an activation energy of 167.0 kJ/mole is to proceed at
T = 450 K with a very constant rate; the rate is allowed to vary at most by ±1%.
How constant must the temperature be to achieve this required rate stability? (For
T>> Δ T, T1 x T2 = T2)
Solution
EA ⎛ 1 1 ⎞
⎜⎜ T − T ⎟⎟
⎝ 2 1⎠
k1
R
= 1.02 = e
k2
ln 1.02 =
ln 1.02 =
ΔT =
EA ⎛ 1
1 ⎞ EA ⎛ T1 − T2 ⎞ EA ⎛ ΔT ⎞
−
⎜
⎟=
⎜
⎟=
⎜
⎟
R ⎝ T2 T1 ⎠ R ⎝ T1 × T2 ⎠
R ⎝ T2 ⎠
EA ΔT
RT 2
R × T2 × ln 1.02
= 2.0 × 10-1 K = ± 0.1 K
EA
The required temperature stability is beyond the capabilities of conventional
thermostats.
Problem #3
Determine the diffusivity D of lithium (Li) in silicon (Si) at 1200°C, knowing that
D1100°C =10–5 cm2/s and D695°C = 10–6 cm2/s.
Solution
D1
10-6
=
= 10-1 = e
D2
10-5
EA =
EA ⎛ 1
1 ⎞
−
R ⎜⎝ 968 1373 ⎟⎠
R ln 10
= 62.8 kJ/mole
1
1
−
968 1373
−
D1100
=e
D1200
EA ⎛ 1
1 ⎞
−
R ⎜⎝ 1373 1473 ⎟⎠
-5
D1200 = 10
×e
EA ⎛ 1
1 ⎞
−
R ⎜⎝ 1373 1473 ⎟⎠
= 1.45 × 10−5 cm2 / sec
Problem #4
For a chemical reaction, the concentrations of reactant as a function of time are
given below for 25°C and for 50°C.
at 25°C
time (h)
.00
3.15
10.00
13.50
26.00
37.30
at 50°C
conc. (mole/L)
0.1039
0.0896
0.0639
0.0529
0.0270
0.0142
time (min)
0
9
18
54
105
180
conc. (mole/L)
0.1056
0.0961
0.0856
0.0536
0.0270
0.0089
(a) Indicate schematically (in two different graphic presentations) how you could prove,
given concentration data at certain times, that a reaction is of first order.
(b) Determine, from graphic presentations, the rate constants (k) for the given reaction
at 25°C and 50°C.
(c) Determine the half-life (t1/2) for the reaction at 50°C.
(d) Determine the half-life (t1/2) for the reaction at 70°C.
(e) What is the time required for the reaction at 25°C to be completed to the extent of
42%?
Solution
(a) For the first order reactions:
∧
co
straight
line
ln c
co/2
>
t
t1/2 = constant (time)
(b) at 25oC, k = 0.0533 hr–1
255C
±2
±3
k = 0.0533 hr±1
ln c
±4
±5
0
10
20
30
40
t (hr)
at 50oC, k = 0.0138 min–1
505C
±2
±3
k = 0.0138 min±1
ln c
±4
±5
0
(c) t1/2 =
40
80
120
160
200
t (min)
ln 2
ln 2
=
= 50.2 min
k
+ 0.0138
(d) k = Ae
E
− A
RT
We need to determine A (and EA):
−
k25
=e
k50
EA
R
⎛ 1
1 ⎞
−
⎜
⎟
⎝ T25 T50 ⎠
k25 = 0.0533 hr-1 ×
ln
k25 ⎛ EA
= ⎜−
k50 ⎝ R
1 hr
= 8.88 × 10-4 min-1
60 min
⎞⎛ 1
1 ⎞
−
⎟
⎟⎜
⎠ ⎝ T25 T50 ⎠
⎛ 8.88 × 10-4 ⎞
⎛k ⎞
-8.3 ln ⎜
-R ln ⎜ 25 ⎟
⎜ 1.38 × 10-2 ⎟⎟
k50 ⎠
⎝
⎝
⎠ = 87.7 × 103 J/mole
EA =
=
1 ⎞
⎛ 1
⎛ 1
1 ⎞
−
⎜ 298 − 323 ⎟
⎜
⎟
⎝
⎠
⎝ T25 T50 ⎠
o
At 25 C: k =
A =k
E
− A
Ae RT
EA
e RT
= (8.88 × 10-4 )
⎛ 8.77×104 ⎞
⎜
⎟
⎜ 8.31×298 ⎟
⎠
e⎝
A = 2.22 × 1012 m-1
So, at 70oC (343 K):
k =A
E
- A
e RT
= (2.22 × 1012 )
⎛ 8.77×104 ⎞
⎟
-⎜
⎜ 8.31×343 ⎟
⎠
e⎝
= 9.28 × 10-2 min-1
and (finally!):
t1/2 =
ln 2
ln 2
=
= 7.47 min
k
9.28 × 10-2 min-1
(e) If the reaction is 42% complete, then 58% of the reactants remain. Therefore, c =
0.58 co.
c = coe–kt → ln
c
= -kt
co
⎛
c ⎞
c
ln ⎜ 0.58 o ⎟
co ⎠
co
= - ⎝
= 10.22 hr = 10 hr, 13 min, 12.1 sec
k
0.0533 hr-1
ln
t =Problem #5
In a chemical reaction the concentration of a rate-determining component is
measured (in moles) at one minute intervals from zero to 5 minutes. The data are:
1.0 x 10–2, 0.683 x 10–2, 0.518 x 10–2, 0.418 x 10–2, 0.350 x 10–2, and 0.301 x 10–2.
(a) Determine the order (n) of this reaction.
(b) Determine the rate constant (k).
(c) Determine the half-life of this reaction.
Solution
(a)
In c
400
–4.5
In c
–5.0
300
–
5.5
–6.0
1/c
200
time (min)
0
1
2
3
4
5
100
The graph indicates that the ln c vs time plot yields almost a straight line, but the
1/c plot does yield a straight line. This identifies the reaction as a 2nd order reaction.
(b) The rate constant (k) is given by the slope of the 1/c vs time plot:
k =
332-146
1
= 46.5
5-1
mole × min
(c) The half life of a second-order reaction (given in class) can be simply obtained from
the second order rate equation. It is:
τ1/2 =
1
1
=
= 2.15 min
k × co
46.5 × 10-2
MIT OpenCourseWare
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3.091SC Introduction to Solid State Chemistry
Fall 2009
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