For instance, is your half-life is 50 seconds, the rate constant is .0139 second
ID: 910623 • Letter: F
Question
For instance, is your half-life is 50 seconds, the rate constant is .0139 seconds. Questions: 1. Show that if a first order reaction runs for 10 half-lives, only 0.098% of the original starting material remains. 2. What type of reaction is the hydrolysis of t-butylchloride (see Chapter 9 in Brown et al)? 3. Derive the expression that relates rate constant to the half-life. Hint: the concentration at this time (t1/2) is half the concentration with which you started. Substitute this information into the equation for a first-order reaction. Do the units you use to measure the Volume of base effect you calculated value of k? Why or why not? Referring to the equations for the treatment of your data. if your were to plot ln(Vinfinite/ Vinfinite-Vs) vs. t. what would the slope equal? Grading Scheme Kinetics Reproducibility of plots 20 15 10 5 0 Calculation of k 20 15 10 5 0Explanation / Answer
After one half-life, we have ½ of the original material remaining.
After two half-lives, we have ¼ of the original material. At this rate, after any number (n) of half-lives, we have (1/2)n of the original material.
Here n = 10
(1/2)10 = 1/1024 = 0.00097656 of the original material
percent form = 0.00097656*100/1 = 0.097656% ~ 0.098% (round up).
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