At a certain temperature, the half-life of the first order decomposition of phen

Question

At a certain temperature, the half-life of the first order decomposition of phenol (shown below) is 1.94 hr. phenol rightarrow cyclopentadiene + carbon monoxide Answer the following questions about the decomposition of phenol and report all answers to three significant figures. If the initial concentration of phenol is 4.89 times 10^-4 M, calculate the time (in hr) required for the concentration of phenol to decrease to 27.5 % of the initial concentration. 0.896 hr If the initial concentration of phenol is 4.89 times 10^-4 M, calculate the concentration (in M) after 1.48 hr. 2.82*10^- 5 M

Explanation / Answer

half life = 1.94 hour

use the relation between rate constant and half life for 1st order reaction,

rate constant, k = 0.693/half life

= 0.693/1.94 hour

= 0.357 hour-1

1)

[A]o = 4.89*10^-4 M

[A] = 27.5 % of [A]o = 0.275*4.89*10^-4 M = 1.345*10^-4 M

use integrated rate law for 1st order reaction:

ln [A] = –kt + ln [A]o

ln(1.345*10^-4) = - (0.357)*t + ln (4.89*10^-4)

(0.357)*t = 1.291

t = 3.62 hour

Answer: 3.62 hour

2)

ln [A] = –kt + ln [A]o

ln[A] = - (0.357)*1.48 + ln (4.89*10^-4)

ln [A] = -8.152

[A] = 2.88*10^-4 M

Answer: 2.88*10^-4 M

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