Aspirin, C
9
H
8
O
4
, slowly decomposes at room temperature by reacting with water in the atmosphere to produce acetic acid, HC
2
H
3
O
2
, and 2-hydro
x
ybenzoic acid, C
7
H
6
O
3
(this is why old bottles of aspirinoften smell like vinegar):C
9
H
8
O
4
+ H
2
O
p
HC
2
H
3
O
2
+ C
7
H
6
O
3
Concentration and rate data for this reaction are given below
.
[C
9
H
8
O
4
] (M) [H
2
O] (M) Rate (M/s)
0
.
0
1
00 0
.
02002
.
4
v
1
0
±
1
3
0
.
0
1
00 0
.
08009
.
6
v
1
0
±
1
3
0
.
0200 0
.
02004
.
8
v
1
0
±
1
3
W
rite the rate law for this reaction and calculate k (be sure to include the correct units)
To write the rate law for the reaction, we need to determine the order of reaction with respect to each reactant. This can be done by analyzing the concentration and rate data provided.
Looking at the given data, we can observe that the concentration of water ([H2O]) remains constant throughout, while the concentration of aspirin ([C9H8O4]) varies. This suggests that the decomposition of aspirin is only dependent on the concentration of aspirin itself.
Based on the data, we can see that as the concentration of aspirin doubles (from 0.0100 M to 0.0200 M), the rate of the reaction also doubles (from 2.4 M/s to 4.8 M/s). This indicates that the reaction is first order with respect to aspirin.
Therefore, the rate law for this reaction can be written as:
Rate = k[C9H8O4]^1
Now, to calculate the rate constant (k), we can choose any data point and substitute the values into the rate law equation. Let's use the first data point (0.0100 M aspirin and 0.0200 M water) for this calculation.
2.4 M/s = k * (0.0100 M)^1
Rearranging the equation, we get:
k = (2.4 M/s) / (0.0100 M)
k = 240 M^-1s^-1
Therefore, the rate constant (k) for this reaction is 240 M^-1s^-1.