Example 15-5 (page 589)
H2O2(aq) 6 H2O(l) + ½ O2(g)
from experiment,
rate = k[H2O2],
k = 7.30 X 10-4 s-1
If [H2O2]0 = 2.32 M, what is [H2O2] at 1200 s?
Solution:
since reaction is first order,
In ([H2O2]t/[H2O2]o) = -kt
In [H2O2]t - In [H2O2]0 = -kt
In [H2O2]t = In [H2O2]0 - kt = In(2.32) - (7.30 X 10-4)(1200)
*** watch units for k and t; also, [H2O2] will be in M
In [H2O2]t = 0.8416 - 0.876 = -0.0344
[H2O2]t = exp(-0.0344) = 0.966 M
(Value with experimental data provided on page 589 is
0.98 M - reasonable agreement)
Example 15-6 (page 591) - based on same reaction
H2O2(aq) 6 H2O(l) + ½ O2(g)
from experiment,
rate = k[H2O2],
k = 7.30 X 10-4 s-1
What percent of H2O2 is decomposed in the first 500.0 s
after the reaction begins?
In ([H2O2]t/[H2O2]o) = -kt
= - (7.30 X 10-4 s-1)(500.0 s)
= -0.365
[H2O2]t/[H2O2]o = exp(-0.365) = 0.694
[H2O2]t = 0.694[H2O2]o
in other words, 69.4% of the initial H2O2 still remains,
or, 100.0 - 69.4 = 30.6% has been consumed
Practice Ex. B: At what time is the [H2O2] b decomposed?
(i.e. [H2O2]t = a[H2O2]o)
In ([H2O2]t/[H2O2]o) = -kt
In (a[H2O2]0/[H2O2]o) = -(7.30 X 10-4 s-1) t
In(a) = -(7.30 X 10-4 s-1) t
-1.0986 = -(7.30 X 10-4 s-1) t
t = 1500 s
Practice Example 15-7B (page 593)
(CH2)2O(g) 6 CH4(g) + CO(g)
t½ = 56.3 min, k = 2.05 X 10-4 s-1
If the initial pressure of (CH2)2O(g) is 782 mmHg, after
30.0 h
a) What is the partial pressure of (CH2)2O(g)?
Method 1 - integrated rate law
In ((PA)t/(PA)o) = -kt
In (PA)t - In (PA)0 = -kt
In (PA)t = In (PA)0 - kt = In(782) - (2.05 X 10-4)(30X60X60)
= -15.478
(PA)t = exp(-15.478) = 1.90 X 10-7 mmHg
Method 2 - half-lives
# of half-lives = (30.0hX60min/h)/(56.3min) = 31.97 . 32
(i.e. approximately an integer number of half-lives)
Therefore,
(PA)t = (½)32X782 mmHg = 1.82 X 10-7 mmHg
(Method 1 is more accurate)
b) What is the total pressure in the reaction vessel?
Ptotal = Pethylene-oxide + Pmethane + Pcarbon-monoxide
. 2 X 782 = 1562 mmHg
Example 15-8 (page 596)
For the reaction A 6 products
a) establish the order of reaction
(i.e. make suitable plots)
1) [A] versus t
(not linear - not zero order)
2) In[A] versus t
(not linear - not first order)
3) 1/[A] versus t
(linear! - must be second order)
b) What is the rate constant?
- slope of line in graph 3 gives k
k = {(4.00 - 1.00)L/mol}/25 min
= 0.12 M-1min-1
c) What is the half-life if [A]0 = 1.00 M?
t½ = 1/k [A]0 = 8.3 min
Note: see Example B
reaction is first order:
t½ = (In 2)/k, k = (In 2)/(100 s) = 6.93 X 10-3 s-1