Kinetic Pre-Lab Questions PDF

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Kinetic Pre-Lab Questions 1. Why is it important that we know the initial concentration of NaOH very accurately and yet for crystal violet, we do not need to know its initial concentration? (We aren't using volumetric glassware when we add 3 drops of crystal violet) Since the concentration of the hydroxide ion is 1000 times larger than the concentration of crystal violet, and it does not change obviously during the reaction, so the concentration of the hydroxide ion is considered to be constant over time. As well, the rate of the reaction is depend on the concentration of crystal violet and it can be determined by absorbance. As a result, the initial concentration of the hydroxide ion is important to be known accurately to calculate the rate of the reaction, but the initial concentration of crystal violet is not needed to be known. 2. Write out the procedure in your own words in point form. You do not need to include procedural elements related to the use of the Vernier interface. You may write it out by hand or type it and print it. NaOH solution preparation 1. label the Erlenmeyer flasks as A and B, and the volumetric flask as C. 2. Use A flask to obtain 80 mL of the 0.100 M NaOH 3. pipet 20.00 mL of A into the 150.00 mL B flask 4. pipet 20.00 mL of water into B flask 5. swirl to mix 6. pipet 10.00 mL of A into the 50 mL C flask 7. fill the flask with water 8. inverting the flask to mix for 10 times

9. pipet 10.00 mL of A into a test tube 10. label the tube 11. repeat this step 12. place three tubes in a beaker with 35°C water 13. place the thermometer in A and record the temperature Rate of reaction measurement 1. filling an empty cuvette 75% with A 2. seal it with the lid. 3. calibrate the colorimeter 4. prepare the Vernier interface to collect data 5. adding three drops of the crystal violet solution 6. seal with a lid and mix for 2 seconds 7. wipe off its outside 8. ensure the run is proceeded 9. save the run 10. repeat for B and C Rate of reaction measurement at a higher temperature 1. record the temperature as it is stable 2. repeat from last section 3. send a copy of the raw data Clean up 1. dispose of the waste properly 2. rinse the glassware with water

3. A similar reaction, the reaction between phenolphthalein and NaOH, is first order with respect to NaOH. If the rate constant is calculated as 1.40 x 10-2 L mol-1 s-1 at 20.95 °C and 5.72 x 10-2 L mol-1 s-1 at 47.05 °C, what is the activation energy of the reaction? ln(kT2/ kT1)=−Ea/R(1/T2−1/T1) where T1 is the first temperature, T2 is the second temperature, Ea is energy of activation for the reaction, R is 8.314 J K-1 mol-1 , kT1 is the pseudo rate constant at T1 and kT2 is pseudo rate constant at T2 so In(5.72 x 10-2 L mol-1 s-1/1.40 x 10-2 L mol-1 s-1) =−Ea/8.314 J K-1 mol1

(1/320.2K−1/294.1K)

Ea= 4.22 × 10^4 J /mol As a result, the activation energy of the reaction is 4.22

× 10^4 J /mol....