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    cyclopentadiene + carbon monoxide



Answer the following questions about the decomposition of phenol and report all answers to three significant figures.



1. If the initial concentration of phenol is 4.89×10-4 M, calculate the time (in hr) required for the concentration of phenol to decrease to 27.5 % of the initial concentration.

2. If the initial concentration of phenol is 4.89×10-4 M, calculate the concentration (in M) after 1.48 hr.

Explanation / Answer

Ans 1 -

t1/2 = 0.693 / K

N = N0 e -kt

t = time

N = concentration of reactant at any time

N0 = initial concentration

initial concentration = 4.89×10-4 M

if the concentration is decreased by 27.5 % tha 100-27.5 = 72.5 % is left

so,

K = 0.693 / t1/2

t half = 1.94 hr

1.94 hr = 6984 sec

K = .693 / 6984

K = 9.92 x 10-5

concentration after time is = 72.5 % x 4.89×10-4 M

= .0725 x 4.89×10-4 M

= 3.545 x 10-5

N = 3.545 x 10-5 M

putting in eq -

N = N0 e -kt

3.545 x 10-4 = 4.89 x 10-4 e-9.92 x 10-5 t

ln [3.545 x 10-4 / 4.89 x 10-4 ] = - K t / 2.303

-.32 = - 9.92 x 10-5 t

0.32/ 9.92 x 10-5 = t

t = 3225.8 sec

t = .896 hr

approx = .9 hr

Ans 2 - N = N0 e -kt

K = 0.693 / t 1/2

K =.693 / 1.94 = 0.357

N = unknown

N = 4.89 x10-4 e -0.357 t

N = 4.89 x10-4 e -0.357 x 1.48

N = 4.89 x10-4 x .589

N =2.88 x10-4 M

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