Consider the first-order reaction described by the equation?

Answer 1

since its a first order reaction ….so half life = 0.693/rate constant = 0.693/5.18 X 10^-4 = 1337.838 s

http://chemwiki.ucdavis.edu/Physical_Chemistry/Kin…

second question —

ln [A] = -kt + ln [A]o

[A]o = initial conc = 0.0086 M
t = 2.4 hours = 2.4 X 60 X 60 = 8640 s
k = 5.18 X 10^-4 s^-1
[A] = conc. after 2.4 hours

putting the values…

ln [A] = -5.18 X 10^-4 X 8640 + ln 0.0086

ln [A] = -4.476 – 4.756

ln [A] = -9.232

converting ln to log …multiplying by 2.303

2.303 X log [A] = -9.232

log [A] = -9.232/2.303 = -4.009

taking antilog ….

[A] = 10^-4.009 = 9.795 X 10^-5 M

http://chemwiki.ucdavis.edu/Physical_Chemistry/Kin…

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