Q1.) The mechanism of a reaction A B is investigated. At the beginning of the re

Question

Q1.) The mechanism of a reaction A B is investigated. At the beginning of the reaction, the concentration of A is 1.86103M. After 77.6 s, the concentration of A is 4.80105M. Using the integrated rate laws, calculate the half-life of the reaction at 77.6 s for the following conditions.

(a) The reaction is first order in A.
________s

(b) The reaction is second order in A.
________s

Q2.) When dissolved in water, ammonium cyanate, NH4CNO, isomerizes to urea, H2NCONH2, in the following reaction.

NH4CNO(aq) H2NCONH2(aq)

In this second-order reaction, the rate constant, k, is equal to 0.0597 L/mol·min. From an initial concentration of 0.197 M ammonium cyanate, calculate the concentration of ammonium cyanate after the reaction has progressed for 49.6 minutes.

Explanation / Answer

a)

use integrated rate law for 1st order reaction

ln[A] = ln[A]o - k*t

ln(4.80*10^-5) = ln(1.86*10^-3) - k*77.6

-9.94 = -6.29 - k*77.6

k*77.6 = 3.66

k = 4.713*10^-2 s-1

Given:

rate constant, k = 4.713*10^-2 s-1

use relation between rate constant and half life of 1st order reaction

t1/2 = (ln 2) / k

= 0.693/(k)

= 0.693/(4.713*10^-2)

= 14.7 s

Answer: 14.7 s

b)

use integrated rate law for 2nd order reaction

1/[A] = 1/[A]o + k*t

1/(4.80*10^-5) = 1/(1.86*10^-3) + k*77.6

20833.33 = 537.63 +k*77.6

k*77.6 = 20295.7

k = 2.615*10^2 s-1

Given:

rate constant, k = 2.615*10^2 M-1.s-1

use relation between rate constant and half life of 2nd order reaction

t1/2 = 1/([A]o*k)

= 1/(0.00186*2.615*10^2)

= 2.06 s

Answer: 2.06 s

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