CHM 101L M5 Factors Affecting Reaction Rates Lab Report Complete

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Jan 9, 2024

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Factors Affecting Reaction Rates Student Name David Arlotta Date 11/26/2023

1 Data Activity 1 Data Table 1: Calibration Trial Na 2 S 2 O 3 (drops) Reaction time (sec) 1 8 16 sec 2 9 20 sec 3 9 24 sec 1. How many drops will be used in the remaining experiments? Activity 2 Data Table 2a. Effects of KI (I - ) Concentration Tri al KI (drop s) HCl (drop s) Starc h (drop s) H 2 O (drop s) Na 2 S 2 O 3 (drops ) H 2 O 2 (mL ) Tim e 1 (sec ) Time 2 (sec) Averag e time (T avg ) (sec) Rate (1/T avg ) 1 8 2 4 0 9 0.4 20.3 0 20.77 1/20.7 7=0.0 48 2 8 2 4 0 9 0.4 21.22 3 6 2 4 2 9 0.4 22.7 6 24.62 1/24.6 2=0.0 41 4 6 2 4 2 9 0.4 26.48 5 4 2 4 4 9 0.4 39.2 8 37.2 1/37.2 =0.03 6 4 2 4 4 9 0.4 35.12 7 2 2 4 6 9 0.4 103. 4 100.64 1/100. 64=0. 009 8 2 2 4 6 9 0.4 97.87 Data Table 2b. Effects of H 2 O 2 Concentration © 2016 Carolina Biological Supply Company

2 Tri al KI (drop s) HCl (drop s) Starc h (drop s) H 2 O (mL ) Na 2 S 2 O 3 (drops ) H 2 O 2 (mL ) Tim e 1 (sec ) Tim e 2 (sec ) Average time (T avg ) (sec) Rate (1/T avg ) 1 8 2 4 0 9 0.4 20.4 19.95 1/19.9 5=0.0 5 2 8 2 4 0 9 0.4 19.5 3 8 2 4 0.1 9 0.3 23.8 8 23.11 1/23.1 1=0.0 4 4 8 2 4 0.1 9 0.3 22.3 4 5 8 2 4 0.2 9 0.2 31.4 9 31.7 1/31.7 =0.03 6 8 2 4 0.2 9 0.2 31.9 1 7 8 2 4 0.3 9 0.1 37.7 2 39.01 1/39.0 1=0.0 2 8 8 2 4 0.3 9 0.1 40.3 Data Table 2c. Effects of HCl (H + ) Concentration Tri al KI (drop s) HCl (%) Starc h (drop s) H 2 O (drop s) Na 2 S 2 O 3 (drops ) H 2 O 2 (mL ) Tim e 1 (sec ) Tim e 2 (sec ) Average time (T avg ) (sec) Rate (1/T avg ) 1 8 100 % 4 0 9 0.4 23.9 6 24.64 1/24.6 4=0.0 40 2 8 100 % 4 0 9 0.4 25.3 1 3 8 75% 4 0 9 0.4 29.8 2 27.46 1/27.4 6=0.0 36 4 8 75% 4 0 9 0.4 25.0 9 5 8 50% 4 0 9 0.4 25.7 7 25.54 1/25.5 4=0.0 39 6 8 50% 4 0 9 0.4 25.3 0 7 8 25% 4 0 9 0.4 22.8 9 24.8 1/24.8 =0.04 0 8 8 25% 4 0 9 0.4 26.7 1 Activity 3 © 2016 Carolina Biological Supply Company

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3 Data Table 3. Temperature Effects Water bath trial Temperature of the water bath (°C) Reaction time (sec) Cold water 6 °C 111 sec Room-temperature water 26 °C 21.22 sec Hot water 41 °C 6.84 sec 2. Explain how each of these treatments affected the reaction rate. Describe the effect at a molecular level. a. Concentration There are more reactant particles traveling together as the concentration of reactants increases. Because there will be more collisions, the reaction rate will be faster. The faster the pace of a reaction, the higher the concentration of reactants. b. Temperature The pace of reaction usually increases as the temperature rises. The average kinetic energy of the reactant molecules will grow as the temperature rises. As a result, a higher percentage of molecules will have the minimal energy required for a successful collision. Activity 4 Data Table 4. Catalyst Trial Water (drops) CuSO 4 (drops) Reaction time (sec) 1 4 0 20.34 sec 2 3 1 14.4 sec 3 2 2 11.61 sec 4 1 3 10.5 sec 5 0 4 8 sec 3. How did the addition of copper(II) sulfate affect the reaction rate? Did the amount of catalyst affect the reaction rate? The addition of copper reduced the reaction times and yes the amount of catalyst does affect the reaction time. © 2016 Carolina Biological Supply Company

4 Activity 5: Data Table 5. Orders of Reactants in the Rate Law R = k [I – ] m [H 2 O 2 ] n [H + ] p Calculated Reaction order (X) (e.g. ( concentration 1 concentration 2 ) X = rate 1 rate 2 ) Reactan t Concentrati on Averag e Rate Calculated Reaction order (X) Averag e Reactio n order Reactio n Order (integer ) KI (I  ) 8 drops 0.050 0.9137 m = 1 (8/6)^x= 0,050/0.040= 0,776 KI (I  ) 6 drops 0.040 (6/4)^x= 0.040/0.026= 1.072 KI (I  ) 4 drops 0.026 (4/2)^x= 0.026/0.014= 0.893 KI (I  ) 2 drops 0.014 H 2 O 2 0.4 mL 0.045 0.425 n = 0 (0.4/0.3)^x= 0.045/0.043= 0.159 H 2 O 2 0.3 mL 0.043 (0.3/0.2)^x= 0.043/0.031= 0.805 H 2 O 2 0.2 mL 0.031 0.2/0.1)^x= 0.031/0.025= 0.312 H 2 O 2 0.1 mL 0.025 HCl (H + ) 100% 0.046 0.211 p = 0 (100/75)^x= 0.046/0.041= 0.405 HCl (H + ) 75% 0.041 (75/50)^x= 0.041/0.038= 0.190 HCl (H + ) 50% 0.038 © 2016 Carolina Biological Supply Company

5 Reactan t Concentrati on Averag e Rate Calculated Reaction order (X) Averag e Reactio n order Reactio n Order (integer ) (50/25)^x= 0.038/0.037= 0.038 HCl (H + ) 25% 0.037 Rate law k*[KI]^m*[H202]^n*[HCI]^p Overall order of the reaction 1+0+0=1 4. Use the rate-law expression you determined for Data Table 5 to answer the following questions. a. If the concentration of I – was doubled, how would that affect the reaction rate? The reaction rate would be doubled b. If the concentration of H 2 O 2 was halved, how would that affect the reaction rate? The reaction rate would be cut in half c. If the concentrations of I – and H 2 O 2 were both doubled, how would that affect the reaction rate? The reaction rate would be quadrupled d. If the concentration of H + (HCl) was doubled, how would that affect the reaction rate? The would be no change © 2016 Carolina Biological Supply Company

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