Diacetone Rate=k[DAA]a[OH-]bRate= k'[DAA] where k’=k[OH-]b as [OH-]Diacetone Rate=k[DAA]a[OH-]bRate= k'[DAA] where k’=k[OH-]b as [OH-]

Diacetone alcohol dissociates in the presence of hydroxide ions with the following reaction:Figure 1: Cleavage of diacetone alcohol to acetone (source: Kinetic study of base catalysed diacetone alcohol depolymerisation, David Hewett)  The reaction involves the formation of two products from one reactant.The progress of the reaction was followed carrying out the reaction in a dilatometer and recording the change of volume of the reaction as a function of time. The volume of 1 molecule of diacetone alcohol is not the same as the volume of 2 molecules of acetone and as a result, the total volume of the reaction solution increases as the reaction proceeds.  The temperature and the concentration of diacetone alcohol have been varied to see what effect they could have on the reaction. The rate equation law and the activation enthalpy will be used to identify the rate determining step and find the mechanism of the reaction.Several reactions were carried out varying the concentration of NaOH and the temperature at which they were done. By changing the temperature, the concentration of reactants and including a catalyst can increase the number of collisions per unit of time. By rising the temperature, more collisions occur in the reaction per unit of time1 as the molecules are excited and are moving at a greater speed. Rising the concentration of the reactants, which is the number of molecules per unit of volume, will expand the number of collisions per unit of time, thus increasing the rate reaction1. The concentration of the molecules involved in the rate determining step is responsible for the rate of the reaction. We also used sodium hydroxide as a catalisis which provides an alternative reaction with a lower activation energy. There are more successful collisions per unit of time and the reaction rate increases. In this lab, five different experiments were carried out:For each of them, we read the concentration, which was in this case the height within the capillarity, along the time t. We then made several tables (see Tables…..in Appendix) with the headings time t/s ; reading x/mm ; reading x’/mm ; x’-x/mm and ln(x’-x) to realise the different plots of the experiences as seen on Figure 2 and 3, to determine the different rate constants.The rate law is found to represent a two reactant system in which the reaction is first order. The reaction is catalysed by the presence of OH-. The rate of our system is: Rate=kDAAaOH-bRate= k’DAA where k’=kOH-b as OH- being constant during the reaction. OH- can be put into the rate law and hence become 1st order. At t=0, the value of the constant is to be lnDAA?. So k’ can be found easily by plotting a graph of lnDAA against t, k’ being obtained by the gradient of the graph (-k’). This will help us calculate the activation enthalpy of the reaction later. In Figure 1, we have first plotted the three lnDAA at 0,1 M against t, varying the temperature to 25?C, 30?C and 35?C. From Figure 1,  we could determine the first order rate constants k’, which corresponds to the slope of the curve. Figure 2: Plot of ln(x’-x) against t  for the reaction using 0.1 M, 0.25 M and 0.075 NaOH at 25?C We determined the values of the rate constants k’:Table 2: Rate constants of 1st order – k’Figure 2: Plot of  ln(x’-x) against t for the reaction using 0.1 M NaOH at 30?C and 35?C,From these values, we were able to determine the 2nd order rate constant kOH, which is related to the 1st order rate constant, k’, by the expressionTable 3: Values of the second order rate constants –  kOHConcentrationTemperatureRate constant kOHSo the average value of kOHis 0.012.We approximated t1/2 to 20 minutes. From that, we designed an experiment using 35ml of 0.075 M, taking readings every 2 minutes for 20 minutes, leaving a break of 25 minutes before re-reading values every 2 minutes until 65 minutes. We then plotted our values in Figure 1, obtaining a rate constant of k’=0.0006 at 25?C. The energy to form the rate determining transition state compound, that is, the enthalpy of activation, can be calculated. The activation enthalpy, AE, of a reaction is related to the rate constant k’ by the Arrhenius equation where A is a pre-exponential factor which can be considered independent of the temperature and concentration, R is the gas constant and T the temperature in Kelvin. Using the law of logarithms, this may also be written in the form: The rate law includes the reactant species which are present and involved in the rate determining step.The activation enthalpy represents the sum of all the bond enthalpies (those which are broken) in the rate determining step. The rate law includes the concentration of all the molecules which are involved in the rate determining step, as the rate at which this takes place is responsible for the rate of the reaction. The rate law represents a two reactant system, both diacetone alcohol and hydroxide ions are involved in the rate determining step and the reaction is first order with respect to both. As both reactants are first order, one molecule of diacetone alcohol collides and reacts with one hydroxide ion. The rate determining step involves the attack of a diacetone alcohol molecule by one hydroxide ion, so the mechanism is supported by both the calculated activation enthalpy and the rate law. The order of the reaction with respect to both diacetone alcohol and hydroxide ions was detemrined to be first order. The experimental rate law was as followsRate= kDAAaOH-bThe effect of temperature on the rate constant was investigated and from this, the activation enthalphy of the reaction was calculated to be…..Both the rate law and activation enthalpy obtained supported the proposed reaction mechanism. The activation enthalpy