1) use the integrated form of the rate law for a second order reaction to determ
ID: 1025344 • Letter: 1
Question
1) use the integrated form of the rate law for a second order reaction to determine the concentration at the 6 minute mark of a reactant that started at 0.900M and proceeded for 6 minutes with a rate constant of 0.0195M^-1 second^-1
2) plot ln(k), the natural log of k, versus 1/T (using kelvins) for the 4 data pairs in the box on the left for a decomposition reaction of HI. use the slope of the graph to determind the value of the activation energy, Ea.
3) graphically determine the order of the reaction in [Br-] to the right.
a] explain how the graph(s) supports your order answer
b] obtain the value of the rate constant, k, including units
c] state the reactions half life, show how
Use the integrated form of the rate law for a second-order reaction to determine the concentration at the 6.0-minute mark of a reactant that started at 0.900M and proceeded for 6.0 minutes with a rate constant of 0.0195 M1 second Plot In(k), the natural log of k, versus I/T (using Kelvins) for the four data pairs 2. in the box on the left for a decomposition reaction of HI. Use the slope of the graph to determine the value of the activation energy, Ea. (See Example 13.7 on page 539 of TRO. Attach a spreadsheet graph or use precisely ruled graph paper. Temperature, C k, Lmols 2.69x103 6.21x103 1.40x102 3.93x10 time (s) 440 460 480 500 10 15 20 25 30 35 40 45 50 (Br] 0.900 0.784 0.683 0.595 0.518 0.451 0.393 0.343 0.298 0.260 0.226 0.197 0.172 Graphically determine the order of the reaction in Br] to the right. It may help to refer to lab manual Appendix 5.6 (X-Y graphing). Attach a spreadsheet or precisely ruled graph paper. a) Explain how the graph (or graphs) supports your "order" answer. b) Obtain the value of the rate constant, k, including units, 60 65 State the reaction's half-life, showing how you got it.70 0.150 c) 0.130Explanation / Answer
Answer for ques.1
The integrated form of the second order rate equation is
1/[A] = 1/[Ao] + kt
= (1 / 0.9) + 0.0195 * 360
= 8.13
[A] = 0.12298M reactant remains after 6 minutes of reaction
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