1. Choose the most correct option.
A. A sample of pure water, whatever the source always contains .......... by mass of oxygen and 11.1 % by mass of hydrogen .
a. 88.8 b.18 c.80 d.16
B. Which of the following compounds can NOT demonstrate the law of multiple proportions?
Answers
Answer:
64 kJ mol
−
1
Explanation:
Your tool of choice here will be the Arrhenius equation, which looks like this
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
k
=
A
⋅
exp
(
−
E
a
R
T
)
a
a
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−
, where
k
- the rate constant for a given reaction
A
- the pre-exponential factor, specific to a given reaction
E
a
- the activation energy of the reaction
R
- the universal gas constant, useful here as
8.314
J mol
−
1
K
−
1
T
- the absolute temperature at which the reaction takes place
As you can see, the Arrhenius equation establishes a relationship between the rate constant and the absolute temperature at which the reaction takes place.
In other words, this equation allows you to determine how a change in temperature affects the rate of the reaction.
Now, convert the two temperatures from degrees Celsius to Kelvin by using the conversion factor
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
T
[
K
]
=
t
[
∘
C
]
+
273.15
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−
You will thus have
T
1
=
100
∘
C
+
273.15
=
373.15 K
T
2
=
150
∘
C
+
273.15
=
423.15 K
So, you know that at
T
1
, the rate constant for your reaction takes the form
k
1
=
A
⋅
exp
(
−
E
a
R
⋅
T
1
)
Likewise, at
T
2
the rate constant takes the form
k
2
=
A
⋅
exp
(
−
E
a
R
⋅
T
2
)
Divide these two equations to get rid of the pre-exponential factor,
A
k
1
k
2
=
A
⋅
exp
(
−
E
a
R
⋅
T
1
)
A
⋅
exp
(
−
E
a
R
⋅
T
2
)
This will be equivalent to - using the properties of exponents
k
1
k
2
=
exp
[
E
a
R
⋅
(
1
T
2
−
1
T
1
)
]
Take the natural log of both sides of the equation to get
ln
(
k
1
k
2
)
=
E
a
R
⋅
(
1
T
2
−
1
T
1
)
Rearrange to solve for
E
a
, the activation energy of the reaction
∣
∣
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
E
a
=
R
⋅
ln
(
k
1
k
2
)
1
T
2
−
1
T
1
a
a
∣
∣
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−
Finally, plug in your values to get
E
a
=
8.314 J mol
−
1
K
−
1
⋅
ln
(
1.3
⋅
10
−
4
M
−
1
s
−
1
1.5
⋅
10
−
3
M
−
1
s
−
1
)
(
1
423.15
−
1
373.15
)
K
−
1
E
a
=
64,212.3 J mol
−
1