Factors Affecting Reaction Rates Lab Report Dr. Z Edits (1)

docx

School

Southern New Hampshire University *

*We aren’t endorsed by this school

Course

101L

Subject

Chemistry

Date

Feb 20, 2024

Type

docx

Pages

7

Uploaded by DeanMoon13043

Report
Factors Affecting Reaction Rates Date: 11/19/2023 Data Activity 1 Data Table 1: Calibration Trial Na 2 S 2 O 3 (drops) Reaction time (sec) 1 8 drops 29 seconds 2 6 drops 21 seconds 3 N/A N/A Question 1: How many drops will be used in the remaining experiments? 6 drops of Na 2 S 2 O 3 will be used in the remaining experiments. Activity 2 Data Table 2a. Effects of KI (I - ) Concentration Trial KI (drops) HCl (drops) Starch (drops) H 2 O (drops) Na 2 S 2 O 3 (drops) H 2 O 2 (mL) Time 1 (sec) Time 2 (sec) Average time (T avg ) (sec) Average Rate (1/T avg ) 1 8 2 4 0 6 0.4 24 (24 + 22)/2 = 23 1/23= .043 2 8 2 4 0 6 0.4 22 3 6 2 4 2 6 0.4 33 (33 +36)/2 =34.5 1/34.5= .029
Trial KI (drops) HCl (drops) Starch (drops) H 2 O (drops) Na 2 S 2 O 3 (drops) H 2 O 2 (mL) Time 1 (sec) Time 2 (sec) Average time (T avg ) (sec) Average Rate (1/T avg ) 4 6 2 4 2 6 0.4 36 5 4 2 4 4 6 0.4 41 (41 + 39)/2 = 40 1/40= .025 6 4 2 4 4 6 0.4 39 7 2 2 4 6 6 0.4 87 (87+83)/2 = 85 1/85= .012 8 2 2 4 6 6 0.4 83 Data Table 2b. Effects of H 2 O 2 Concentration Trial KI (drops) HCl (drops) Starch (drops) H 2 O (mL) Na 2 S 2 O 3 (drops) H 2 O 2 (mL) Time 1 (sec) Time 2 (sec) Average time (T avg ) (sec) Average Rate (1/T avg ) 1 8 2 4 0 6 0.4 24 (24 + 21)/2= 22.5 1/22.5= .044 2 8 2 4 0 6 0.4 21 3 8 2 4 0.1 6 0.3 32 (32+30)/2=31 1/31= .032 4 8 2 4 0.1 6 0.3 30 5 8 2 4 0.2 6 0.2 42 (42 + 48)/2 = 45 1/45= .022 6 8 2 4 0.2 6 0.2 48 7 8 2 4 0.3 6 0.1 30 (30 + 27)/2 =28.5 1/28.5= .035 8 8 2 4 0.3 6 0.1 27 ©2016 2 Carolina Biological Supply Company
Data Table 2c. Effects of HCl (H + ) Concentration Trial KI (drops) HCl (%) Starch (drops) H 2 O (drops) Na 2 S 2 O 3 (drops) H 2 O 2 (mL) Time 1 (sec) Time 2 (sec) Average time (T avg ) (sec) Average Rate (1/T avg ) 1 8 100% 4 0 6 0.4 20 (20+25)/2= 22.5 1/22.5= .044 2 8 100% 4 0 6 0.4 25 3 8 75% 4 0 6 0.4 24 (24+22)/2= 23 1/23= .043 4 8 75% 4 0 6 0.4 22 5 8 50% 4 0 6 0.4 25 (25+28)/2= 26.5 1/26.5= .038 6 8 50% 4 0 6 0.4 28 7 8 25% 4 0 6 0.4 22 (22 + 26)/2 =24 1/24 = .042 8 8 25% 4 0 6 0.4 26 Activity 3 Data Table 3. Temperature Effects Water bath trial Temperature of the water bath (°C) Reaction time (sec) Cold water 3 25 seconds Room-temperature water 20 18 seconds Hot water 56 6 seconds Question 2: Explain how each of these treatments affected the reaction rate. Describe the effect at a molecular level. a. Concentration b. Temperature ©2016 2 Carolina Biological Supply Company
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Activity 4 Data Table 4. Catalyst Trial Water (drops) Diluted CuSO 4 Solution (drops) Reaction time (sec) 1 4 0 22 seconds 2 3 1 10 seconds 3 2 2 9 seconds 4 1 3 8 seconds 5 0 4 6 seconds Question 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 (II) sulfate increased the reaction rate significantly. The greater the amount of catalyst added (paired with a decrease in the amount of water added each trial), the lower the reaction time (in seconds) was each trial. Activity 5: Data Table 5. Orders of Reactants in the Rate Law Determine the actual 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 ) You must show your full calculation work for credit. ©2016 2 Carolina Biological Supply Company
Reactant Concentration Average Rate (1/T avg ) Calculated Reaction order (X) Show work! Average Reaction order Reaction Order (integer) KI (I ) 8 drops .043 (1.369 + .366 + 1.059)/2 = .931 m = 1 1.369 KI (I ) 6 drops .029 .366 KI (I ) 4 drops .025 1.059 KI (I ) 2 drops .012 You must show your full calculation work for credit. Reactant Concentration Average Rate (1/T avg ) Calculated Reaction order (X) Show work! Average Reaction order Reaction Order (integer) H 2 O 2 0.4 mL .044 (.831 + .924 -.544)/3 = .404 n = 0 .831 H 2 O 2 0.3 mL .032 ©2016 2 Carolina Biological Supply Company
Reactant Concentration Average Rate (1/T avg ) Calculated Reaction order (X) Show work! Average Reaction order Reaction Order (integer) .924 H 2 O 2 0.2 mL .022 -.544 H 2 O 2 0.1 mL .035 You must show your full calculation work for credit. Reactant Concentration Average Rate (1/T avg ) Calculated Reaction order (X) Show work! Average Reaction order Reaction Order (integer) HCl (H + ) 100% .044 (.08 + .305 - .144)/3 = .08 p = 0 .08 HCl (H + ) 75% .043 .305 HCl (H + ) 50% .038 -.144 HCl (H + ) 25% .042 ©2016 2 Carolina Biological Supply Company
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Rate Law R = k [I–]^1 [H 2 O 2 ]^0 [H ]^0 Overall Order of the Reaction 1 + 0 + 0 = 1 Question 4: Use the rate-law expression you determined for Data Table 5 to answer the following questions. Answer using doubled, tripled, quadrupled, halved, quartered, no change, etc. A) If the concentration of I was doubled, how would that affect the reaction rate? Double. B) If the concentration of H 2 O 2 was halved, how would that affect the reaction rate? No change. C) If the concentrations of I and H 2 O 2 were both doubled, how would that affect the reaction rate? It would double due to the concentration of I-, but the reaction rate would not be impacted by the concentration of H2O2. D) If the concentration of H + (HCl) was doubled, how would that affect the reaction rate? No change. ©2016 2 Carolina Biological Supply Company