Domenica DiStasio April 2 2019 Experiment 4 Chemical Kinetics Effect of Temperature Arrhenius Equation Objective The objective of this lab is to determine the effect that temperature has on the speed of a chemical reaction We are going to look at how the activation energy in the Arrhenius Equation changes as the temperature of H2C2O4 increases decreases Procedure 1 Using a buret place 20mL of 0 5M H2C2O4 in an Erlenmeyer Flask and 10mL of 0 02M KMnO4 in a 15cm test tube 2 Place both vessels in your water bath and let them sit there for five minutes 3 Set up your second trial at this temperature 4 Mix the reactants and record the time it takes for the solution again to go yellow brown Swirl the contents regularly 5 Repeat the procedure with another sample at this temperature 6 Using the average time determine k at this temperature 7 Repeat this procedure for the other suggested temperatures 8 When complete determine the Activation Energy by plotting ln k v 1 T Remember temperature must be in Kelvin Repeat this 4 times at temperatures 0 10 20 and 30 degrees Only do 1 trial for the temp at 0 degrees Data Collection Table 1 Temperature and concentration trials and their reaction times KMnO4 H2C2O4 mL mL 10 mL 10 mL Temp Reaction Time Trial 2 s Average Time C Reaction Time Trial 1 s 20mL 0 769 s 769 s 20 mL 10 410 s 470 s 0 s 440 s 10 mL 20 mL 20 320 s 299 s 309 5 s temp was at 23 at end 10 mL 20 mL 30 113 s 105 s 109 s Calculations 1 Determining the Rate Temperature at 0oC KMnO4 t Rate KMnO4 H2C2O4 KMnO4 021 mol 1000 ml 10mL 1000 mL 30mL 007 M R k KMnO4 H2C2O4 R KMnO4 t 007 M 769 s 9 1 10 6 mol L sec 2 Calculation on Determining k Temperature at 0oC KMnO4 021 mol 1000 ml 10mL 1000 mL 30mL 007 M H2C2O4 5 mol 1000 mL 20 mL 1000 mL 30 mL 333 M R k KMnO4 H2C2O4 K R KMnO4 H2C2O4 9 1 10 6 mol L sec 007 M 333M 00391 M sec Table 2 Calculations of k for each temperature trial o T C T K 1 T 1 K k Ln k 0 273 15 00366 00391 5 55 10 283 15 00353 00683 4 99 20 293 15 00341 00970 4 64 30 303 15 00330 0276 3 59 Discussion The activation energy tells you the minimum amount of energy required for a reaction to occur The lower the activation energy the faster the reaction In this experiment the reaction s activation should be lower than the typical bond energy because the temperature is increasing with each reaction which means the activation energy should be decreasing This makes sense regarding our graph because as the temperature increases the activation energy is decreasing hence the negative slope Figure 1 Scatter plot with trend line showing the relationship between Ln k vs 1 T Kelvin The slope is y 5150 5x 13 205 Ea Equation Slope 5150 5x 13 205 Ea R 5150 5x Ea 5150 5 8 314 JmolK 42821 1000 42 821 kJmol Ea 42 821 kJmol
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