Propanone reacts with Iodine slowly at room temperature. This reaction can be faster if we add Hydrochloric acid (HCl, a catalyst).

I will use different volumes of HCl by adding water, thus decreasing its concentration. I will use the following volumes:

• 4 cm3 of HCl
• 3.5 cm3 of HCl  + 0.5 cm3 of water (in order to lower the concentration)
• 3 cm3 of HCl  + 1 cm3 of water
• 2.5 cm3 of HCl  + 1.5 cm3 of water
• 2 cm3 of HCl  + 2 cm3 of water
• 1.5 cm3 of HCl  + 2.5 cm3 of water
• 1 cm3 of HCl  + 3 cm3 of water

APPARATUS

• One measuring cylinder
• Three small test tubes
• 1 bigger test tube (where the reaction is going to take place)
• Test tube rack for the small test tubes
• Another test tube rack for the big test tubes
• 1 stopwatch

The following are the steps of my experiment:

1. First do a trial operation to decide on the measurements of different solutions.
2. Using a measuring cylinder I will measure out 1 cm3 of Iodine, and put it in one of the test tubes.
3. Measure out 4 cm3 of Propanone, and put in another test tube.
4. Measure 4 cm3 of HCL and put in the third test tube.
5. Mix the HCl and Propanone together (no reaction should take place).
6. Hold the stopwatch in one hand and the Iodine in another, or get someone to help you.
7. Press the START button on the stopwatch as soon as you poor in the iodine.
8. Wait until the colour of the Iodine completely disappears, and then press the STOP button on the stopwatch.
9. Record the time at the end of each experiment.
10. Repeat the experiment using different concentrations of HCl.

Obtaining Evidence:

After my trial operation I decided to use 4 cm3 of Propanone, and only 1 cm3 of Iodine, because a bigger volume of Iodine means more time for the reaction, which I couldn’t afford. My volume of HCl will start form 4 cm3 and go down, and its concentration is 1 mol dm-3.

RESULTS:

These were my results on the first day:

        

Since the same concentration of Propanone and Iodine has been used in all my experiments, I did not include them in the table.

These results were about what I predicted. I have noticed the following based on my results:

1. Time increases by decreasing the concentration of Hydrochloric acid.

2. If we decrease the concentration to a half, the time doubles. For example, a concentration of 4 cm3 of HCl takes 294 seconds, if this volume decreases to 2 cm3 of HCl and 2 cm3 of water (which will decrease the concentration), the time increases to 563 seconds, which in other words is 4:54 min to 9:23 min respectively. Therefore the time almost doubled as the concentration is halved.

The following were my results on the second day:

These results were noticeably different than those of the first day.

I noticed the following:

1. Although the times are different, they follow the same rule. The time almost doubles as the concentration of the HCl is halved. Take 4 cm3 as an example, it takes 430 seconds, but if it was halved to 2 cm3, then the time almost doubles.
2. The difference in the time between the two days is probably due to the change in temperature, since the two experiments are one week apart.

I decided to take the average of both readings as my final result. I will do this by adding the value on table 1 to that on table 2 and the dividing by 2. For instance, for 4 cm3 the following will happen: 430 + 294 = 724, then we divide this value by 2. So, 724 / 2 = 362 seconds, which is 6 minutes and 2 seconds.

My final results are as follows:

Having these results, I can now use the following formula to draw a graph of concentration against time, but first I have to calculate the concentration.

Concentration = Original Concentration  x  (Volume of Acid Used / Total       Volume)

I can also draw a graph of concentration against 1/Time. All my calculations of 1/Time where given to 3 significant figures.

Analysis:

Looking at my graph it is obvious that there is a relationship between the concentration of the acid and the time it takes for the reaction to finish. The more the concentration of the acid the quicker the reaction will be, but still as the concentration goes up to 1 mol dm-3 the graph starts to flatten. In the Time against Concentration graph the curve starts at the top moving downwards as the concentration increases. Therefore we can say that time is inversely proportional to concentration.

 

Prediction- “The more concentration of the acid used the faster the particles will collide, and therefore the quicker the rate of the reaction will be, which means that atoms collide more often in the presence of a catalyst, such as Hydrochloric acid, and there for increasing the rate of the reaction”.

This prediction was actually true, looking back at the graph I could see that the curve goes down as the concentration of the acid increases, which means that the more the concentration of the acid the less time it takes for the reaction to happen. I also found a rate for the reaction the reaction almost doubles as the concentration is halved.

These results were taken from the my final results:

Looking at these results we can see that it is multiplied by approximately 1.8, and therefore, the time almost doubles as the concentration is halved.

Evaluation:

In both my graph (Time against Concentration and 1/Time against Concentration) I had a reading, which was slightly away from the curve. This reading was the 0.875mol dm-3 probably because I couldn’t tell exactly when the solution was completely colourless, but even thought this could be a reason after redoing this specific reading I got almost the same reading of 426 seconds which is 17 seconds different from the other reading, which was 443. This slight difference made the curve look much better.

Apart from this reading, all other readings were on the curve. The results of the first trial where so much different than those of the second because the two trials where one week apart, and therefore a change of the atmospheric temperature might have happened. But when I averaged the two trials I got reasonable results.

If I could find a better way to decide when the solution becomes colourless it would help me obtain better, more accurate results.

I can also work on keeping the temperature constant so that the trials on different days do not affect my results. This can be done if I put the solutions with there test tubes in a beaker full of water, and measure the temperature of the water to make sure the reaction is carried in constant temperature. In case the water gets colder the beaker could be heated until it returns back to normal temperature. If it gets hotter ice cubes could be added to the water in the beaker to make it colder.  

In order for me to obtain more accurate results I can re-do the experiment, but this time checking another variable, such as the temperature. If I was to do that I will have to keep the concentration of the solutions constant this time, and control the temperature using: a Bunsen burner, ice cubes, and most importantly a thermometer that is going to be used to record my readings. When heating the solutions I cannot expose the test tubes to direct fire, they have to be put in a beaker full of water, otherwise I wont be able to take readings, as the thermometer will break. I can then compare these results to mine and see if increasing the temperature also speeds up the rate of a reaction, as I expect to do or not.