C Homework 10 Adaptve Follow-Up c) Item 8 ± Using Integrated Rates Laws - Part A

Question

C Homework 10 Adaptve Follow-Up c) Item 8 ± Using Integrated Rates Laws - Part A Constants I Berodic Table The reactant concentration in a zero-order eaction was 0.100 mol Lter 100 s and 400M102 mol L- ar 365 s. What is the rate constant for this reaction? Express your answer with the appropriate The integrated rate laws for zero,rat, and second-onder reaction may be arranged such hat they resamble the equation for a straight Iney mz + b Order Integrated Rate Law Graph Slope View Available Hints) w.tk RValue Units X Incorrect: Try Again: 4 attempts remaining Part What was the initial reactant ooncentration for the reaction desorbed in Part A? Express your answer with the appropriate unitis View Availablie Hine AValue Units

Explanation / Answer

part-A
[A]0 = 0.1mole/L

[A]t   = 4*10^-2 mole/L
t     = 100sec
[A]t   = -Kt+ [A]0
4*10^-2 = -K*100+0.1
0.04-0.1 = -K*100
-0.06     = -K*100
K        = 0.0006mole/L-sec
part-B initial concentration [A]0 = 0.1M

part-C
        [A]0 = 5.7*10^-2mole/L
        [A]t = 1.1*10^-3 mole/L
         t    = 70sec
    ln[A]t = -Kt+ln[A]0
     ln1.1*10^-3 = -k*70 + ln5.7*10^-2
     -6.8124        = -K*70-2.8647
    -6.8124+2.8647   = -K*70
     -3.9477         = -K*70
       K              = 3.9477/70   = 0.0564sce^-1
part_D
        1/[A]t - 1/[A]0   = Kt
        1/0.033 - 1/0.31   = K*835
        30.30-3.2258       = K*835
          K = 0.03242L- mole^-1 sec^-1
       
      

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.