The data given below is for the reaction:
Determine (i) order of reaction
(ii) rate constant
(iii) rate law
Answers
Yes, graphing is the easiest way to do this (or this could be visually challenging!). In fact, learning how to use Excel is a very important skill. I've put a relatively short video on how to use Excel for chemistry here (which goes into graphs near the end):
If you JUST put time in minutes on the
x
axis and the concentration of
N
2
O
5
in
M
on the
y
axis in Excel, it would turn out to be not linear.
But with some trial and error...
graphing the second order integrated rate law does not work; while it does give a positive slope, the graph is curved (not linear), and thus is of the wrong order.
graphing the first order integrated rate law works to give a linear graph, confirming it is first order:
I've included the best fit line equation, which is given by:
y
ln
[
N
2
O
5
]
ln of current conc.
=
m
−
k
rate constant
x
t
+
b
ln
[
N
2
O
5
]
0
ln of initial conc.
where
k
is the rate constant,
[
N
2
O
5
]
is the concentration of
N
2
O
5
in
M
, and
[
]
0
means initial concentration. You know that
t
means time in
min
.
From looking at the best fit line equation, here is what you can immediately get from it:
slope
=
−
k
⇒
k
=
−
slope
⇒
k
=
0.0301 min
−
1
y-int.
=
−
4.3894
⇒
[
N
2
O
5
]
0
=
e
y-int.
=
e
−
4.3894
=
0.0124 M
And furthermore, the rate law for this would therefore be (from knowing what the order is):
r
(
t
)
=
k
[
N
2
O
5
]
2
)
At
t
=
10 min
, to get the concentration of
O
2
, we can examine the concentration of
N
2
O
5
remaining,
0.0092 M
, and determine the amount of
O
2
currently made.
This may or may not be a new concept, but let's consider something called an ICE Table (initial, change, equilibrium), which is normally a way to track the changes in concentration in a reaction on its way to "equilibrium", the point when the reaction has the same rate forwards and backwards.
In the ICE Table below, the coefficients in front of
x
correspond to the coefficients in the chemical reaction.
2
N
2
O
5
(
g
)
→
4
NO
2
(
g
)
+
O
2
(
g
)
I
0.0124 M
0 M
0 M
C
−
2
x
+
4
x
+
x
E
0.0092 M
4
x
x
Well, we can track the progress of the reaction using the reaction quotient,
Q
, even though it hasn't reached equilibrium yet. It's the same kind of definition, but we are using the current concentrations instead.
Since we know that the change in concentration of
N
2
O
5
was
2
x
=
0.0032 M
, we know that
x
=
0.0016 M
is the change in concentration of
O
2
.
And since
O
2
started at
0 M
(i.e. the reaction started with only reactant), that is its concentration at
10 min
!
3
)
The initial rate of production of
NO
2
comes from the rate law, since we know the rate constant and the starting concentration of
N
2
O
5
:
r
(
t
)
=
k
[
N
2
O
5
]
r
(
t
)
=
3.735
×
10
−
4
M/min
=
∣
∣
∣
∣
∣
∣
consumption
−
1
2
Initial rate
Δ
[
N
2
O
5
]
Δ
t
∣
∣
∣
∣
∣
∣
=
production
+
1
4
Initial rate
Δ
[
NO
2
]
Δ
t
So, it is:
Δ
[
NO
2
]
Δ
t
=
4
2
×
3.735
×
10
−
4
M/min
=
7.47
×
10
−
4
M/min