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Results - iodine & propanone method 2
As I said earlier, it is quite difficult to estimate the times accurately. I have done so to the nearest 10 seconds, but you may well disagree with me!
|
Run |
time/s |
rate/arbitrary units |
|
A |
100 |
0.040 |
|
B |
200 |
0.020 |
|
C |
200 |
0.020 |
|
D |
50 |
0.040 |
You should note that we used distilled water to make the total volume of solution constant in all four experiments. This means that the concentration of each reactant is proportional to the volume of solution we used.
|
Experiment |
Run A |
Run B |
Run C |
Run D |
|
Volume of 2 M HCl/cm3 |
20 |
10 |
20 |
20 |
|
Volume of 2 M propanone/cm3 |
8 |
8 |
4 |
8 |
|
Volume of water/cm3 |
0 |
10 |
4 |
2 |
|
Volume of 0.01 M iodine/cm3 |
4 |
4 |
4 |
2 |
|
Rate |
0.040 |
0.020 |
0.020 |
0.040 |
If we compare run A with run B, we find that the only difference is that the concentration of hydrochloric acid is twice as big in run A. We also find that the rate is twice as big. This means that the order of reaction with respect to the hydrochloric acid is first order. If it were zero order the rate would have been unchanged, and if it were second order the rate would have gone up by a factor of 4 (2 × 2).
Next we compare run A with run C. Here we find that the only difference is that the concentration of propanone is twice as big in run A. We also find that the rate is twice as big. This means that the order of reaction with respect to the propanone is first order. The reason being the same as that for the hydrochloric acid.
Finally we compare run A with run D. Here we find that the only difference is that the concentration of iodine is twice as big in run A, but the rate is unchanged. This means that the order of reaction with respect to the iodine is zero order.
Putting these three bits of information together we obtain the rate equation:
|
Rate = k[CH3COCH3][H+] |