Introduction My essay is focused on chemical kinetics, which is only about 100 years old subfield of chemistry. Kinetics is topic, which investigates the speed of reaction. It is worth to investigate because it gives us knowledge about how reactions happen and about reaction mechanisms. As well as this, chemical kinetics can be used in many different areas: for instance, in pharmacology to work out how fast the drug dissolves or in food industry in order to understand food decomposition. This investigation is focused on the kinetics of iodination of acetone and the reaction used in this experiment is following: CH3COCH3 (aq) + I2 (aq) ⇒ CH3COCH2I (aq)+ HI (aq) This reaction is suitable because it is fast enough to make multiple runs, but …show more content…
The reaction is first order with respect to propanone and acid, and zero order with respect to iodine. This means that if a concentration of propanone or acid is increased, the rate increases as well, in liner……. However, when the concentration of iodine is changed, the rate is not affected. This is due to it’s reaction mechanism. One way the reaction to occur, is explain as following: In the rate equation k is rate constant. The rate constant is independent of the concentrations of substances, but may depend on environmental factors such as temperatures . Therefore, when the concentration of propanone is increased, the value of k stays the same. This property can be used to find k with different concentrations and find the average in order to ensure that the value of k in specific temperature is reliable. In order to calculate activation energy, the rate constant must be calculated in different temperatures, in this particular experiment, rate constant is calculated in following temperatures: 9°C, 22°C, 29°C, 37°C, 45°C. Rate constant can be calculated by dividing the initial rate of the reaction by the concentrations of CH3COCH3 and H+. In this experiment, the units of k are mol−1 dm3 …show more content…
To ensure the constant rate between HCl and propanone, solutions of propanone and HCl were prepared by following next steps: 100 cm3 of 2M propanone was poured in 250 cm3 measuring cylinder (± 1.5 cm3) 50 cm3 of 2M HCl was poured on the top of propanone (±1.5 cm3) The mixture was poured between two 250 ml flasks Flasks were closed with corks The process was repeated for 5 times but every time the propanone was diluted by 10%. The amounts of propanone, distilled water and HCl were following: 1st solution 2nd solution 3rd solution 4th solution 5th solution CH3COCH3 (cm3) (±1.5 cm3) 100 90 80 70 60 H2O (cm3) (±1.5 cm3) 0 10 20 30 40 HCl (cm3) (±1.5 cm3) 50 50 50 50 50 Therefore, the concentrations of propanone were 2M, 1.8M, 1.6M, 1.4M and 1.2M. Half of the solutions were put in fridge to cool down the solutions and others were left to stay in room temperature over night. Afterwards, for temperatures, 29°C, 37°C and 45°C, solutions were put into water bath to keep temperature constant. Measuring the