Introduction: The study of kinetics is compulsory for our understanding chemical reactions. The speed of reaction often determines its value in which these chemical processes are precise to the greatest efficiency. The effect on the rate of the reaction due to the concentration of the reactants is shown by the order of the reaction. It can be expressed by the following equation:  ate [A] [B]  R = kx y    K is the rate constant, x is the order respect to A and y is the order respect to B. The order of reaction must be determined experimentally. This demonstrates the relationship between the concentration and the rate of reaction.  Activation energy is the minimum energy that the molecules need in order to for the reaction to proceed. The Molecules in the reaction that have the energy will lead to effective collisions. A low activation energy of the reaction means it can easily react. Whereas a high activation energy means there are more molecules that may not have the required energy to create effective collisions. For endothermic reaction, the activation energy will be positive and for exothermic it will be negative. The EA can be determined experimentally by performing the experiment at different temperatures. This will be observed in part 2 of the experiment when examining the influence on temperature of the reaction with catalase and with iodide.The activation energy can be calculated from the arrhenius equation which shows the effect of change in the temperature on the rate of reaction.  Aek =  (−Ea/RT )  K is the rate constant, A is the arrhenius constant or the pre-exponential factor, Ea is the activation energy in joules, R is the constant 8.31451 and T is the temperature in kelvin(Logan,1999). The equation can be rearranged to create  , so that thenk nA a/RTl = l − E  natural log can be plotted against 1/T and the slope of the graph represents:
m = -ea / r

The rate can be increased in several ways. It can be increased by increasing temperature, increasing the concentration of the reactants, increasing the pressure of the gas, increasing the surface area of reactants and finally using the catalyst. A catalyst reduces the activation energy which becomes essential in biological context. In human body there many enzymes which act as                             a catalysts for certain reactions. This is important because without these enzymes to catalyze it                             would be too slow which can lead to serious consequences to the human body.  Catalase is an enzyme that maintains the proper functioning and maintenance of the human body.                             When carbohydrates and fats are metabolized, the produce . catalyze prevents cell                2O2H        oxidative damage by speeding up the decomposition of into water and oxygen which are                2O2H            harmless.(Alfonso-Prieto, Biarnés, Vidossich, & Roviro, 2009). This can be expressed in the                       following equation:  H O H O22 2 22+ O2  Without catalyze, H2O2 would alter the cell membrane which will lead to cell death. To prevent the catalyze are required which are found in the liver, kidney and red blood cells. The rate of reaction for catalase can be described by using the following equation where x is the partial order in relation to H2O2 and y is the partial order to the catalyse. The equation can be expressed as:  ate [H O ] [catalyst]R = k2 2x y  The purpose of this experiment is to determine the partial order respect to hydrogen peroxide in the present of the catalyst and then to determine the activation energy for the catalysed decomposition of hydrogen peroxide. A rate law is based on the reactants in an experiment. The k, x and y(components of rate law) must be found experimentally. The order of the reaction do not correspond with the stoichiometry of the reaction. Sometimes they are the same but that is the consequence of mechanism. When the data is plotted on the graph, the slope is the reaction order. Using the graphs obtained of pressure and time, the slope of the graph will represent the value of k.

Calculations:  Partial order with respect to hydrogen peroxide  Rate is founded by the amount of O2 produced  2O2(aq)  −−> H2O(l)  O2(g)H −   +     The rate law for the H2O2 and the catalase can be found from this equation ate [A]  [B]R = kx y ate  k[H2O2] [catalase]R =  x y When constants are removed: ate  C[H2O2]R =  x   determine the dilution of hydrogen peroxide: V VC1 1= C2 2 C1 is the initial concentration V1 is the initial volume  C2 is the final concentration  V2 is the final volume  C1=0.75% V1= 1 mL C2= ? V2= 2 mL 0.75)(1) (2)( = C2 .375%

discussion:

Part b was to compare the efficiency of both catalysts and which of these catalysts is better the hydrogen peroxide or potassium iodide. Before the experiment was started the first thing to do was to extract the catalase from the spinach. This was done by blending it in a blender and then the liquid from the blender was strained with a towel so you could remove all the solid from liquid. The catalyst was placed in ice bath to keep it stable. The importance of placing it in ice bath is to make sure the enzyme does not denature. They won't be able to function properly if the temperature is too high.

How X is calculated of the partial order: To find the the partial the rate was changed to rate = K[H2O2]x  so it could be easier to solve. K is some constant and it doesn't really if you find it or not. The rate of reaction was found experimentally where the concentrations remained but the temperatures differed. The reason why the volume and concentration remained constant was so that they weren't playing a factor on the rate of the equation. If the concentration and volume changed along with temperature it would have been impossible to find the rate. All of this was recorded in the labquest where it was pressure vs time. The slope of the graph gave the initial rate of reaction. To find the partial order the ln rate has to be plotted and the slope of that graph will give us the order of the graph. A error that could have occured when experimentally finding x was that the tube that was sealed could have let some air in which would have gave inaccurate result. The partial order when calculated should have been 1 but it was originally 1.389 which was rounded to 1. This is probably because the some air escaped most likely.   How activation energy was calculated: In order to find the activation energy of the iodide-catalyzed reaction with hydrogen peroxide, a plot of ln(k) vs. 1/T must be created. By taking the ln of each rate value from the best graph of fit, and plotting it against the reciprocal of the corresponding temperature it will give us the slope. The slope will give -EaR and by using this value the activation energy can be calculated. It just has to be multiplied by R (8.314J/(mol*K). The slope just had to be multiplied by the constant, R, 8.314J/(mol times kelvin) and that would give the activation energy. In these two experiments, temperature is the independent variable and pressure is the dependent variable. Pressure differed due to the temperature of water not the other way. To improve this lab more trials can be done with different temperature which will give more accurate results because the data is different. The activation energy for the catalase was   whereas the2.210KJ /mol  4 activation energy for potassium iodide was  . this is telling us that the catalase is a6.550 KJ /mol  4 better catalyst than the potassium iodide because it has low activation energy.  We predicted that the potassium iodide wouldn't be that affective as the catalase since the catalase is actually found in animals and humans. Based on our data the catalase is better catalyst than potassium iodide because it has lower activation energy than potassium iodide. The catalase is an enzyme that is found in our liver. Its job is to break down hydrogen peroxide into oxygen and water. Knowing this it is good that activation energy is low because that means it won't take that long to function for the enzyme to repair in our body. We also observed that the greater the concentration is, the greater the rate is and the greater the temperature, the greater the rate is. Conclusively the rate is is proportional to temperature and the concentration which is shown in our experiment. Some error that could have caused inaccurate results is that tube that sealed erlenmeyer flask was opened which allowed air to escape which resulted in giving inaccurate result. The temperature was very hard to be controlled if it wasn't controlled properly it would have given an inaccurate result. Another error that could have happened is putting the flask in the water and the glass vial won't fall which would result in having a greater temperature but the reaction officially hasn't started.