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

Question

At a certain temperature, the half-life of the first order decomposition of cyclopentene (shown below) is 2.19 hr. cyclopentene rightarrow cyclopentadiene + dihydrogen Answer the following questions about the decomposition of cyclopentene and report all answers to three significant figures. If the initial concentration of cyclopentene is 7.43 times 10^-2 M, calculate the time (in hr) required for the concentration of cyclopentene to decrease to 44.5 % of the initial concentration 2.56 hr If the initial concentration of cyclopentene is 7.43 times 10^-2 M, calculate the concentration (in M) after 1.11 hr. 0.135 M

Explanation / Answer

1.Equation for calculating rate constant of a firsr order reaction is

k = 2.303/t{log a/a-x}

where k = rate constant

t = time

a = initial concentration

a-x = concentration at time t

equation for half-life of first order reaction t1/2 = 0.693/k

given  t1/2 = 2.19 hr

initial concentration (a) = 7.43 x 10-2 M

concentration at time 2.56 hr is 44.5% of initial concentration = {7.43 x 10-2 x 44.5} / 100

= 0.033 M

First calculate k from t1/2

k = 0.693/t1/2

= 0.693/2.19 hr

=0.316 hr-1

using this find the time required to derease the concentration by 44.5%

equation t =  2.303/k{log a/a-x}

= 2.303/0.316{log 0.0743/0.033}

= 2.56 hr

2. concentration after 1.11 hr

we know k =0.316 hr-1

t = 1.11hr

subtituting in the equation k = 2.303/t{log a/a-x}

we can calculate a-x = concentration at time t

so 0.316 = 2.303/1.11{log 0.0743/a-x}

{log 0.0743/a-x} = 0.316 x 1.11 /2.303

{log 0.0743/a-x} = 0.152

log 0.0743 - log(a-x) = 0.152

-1.129 -log(a-x) = 0.152

log (a-x) = -1.281

a-x = antilog of ( -1.281) = 0.0523 M

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