Determine the rate constant for each of the following first-order reactions: A r

Question

Determine the rate constant for each of the following first-order reactions: A rightarrow B + C, given that the concentration of A decreases to one-fourth its initial value in 125 min; 2 A rightarrow D + E, given that [A]_0 = 0.0421 mol middot L^-1 and that after 63 s the concentration of D increases to 0.00132 mol middot L^-1; 3 A rightarrow F + G, given that [A]_0 = 0.080 mol middot L^-1 and that after 11.4 min the concentration of F rises to 0.015 mol middot L^-1. In each case, write the rate law for the rate of loss of A.

Explanation / Answer

The rate equation for a first order reaction is:

ln(CAo/CA) = kt

part (a)

A = B + C

rate law = RA = kCA

t = 125 mins

CA = CAo/4

put values to get

ln(4) = k*125

k = 0.01109 min-1

part (b)

2A = D +E

rate law = RA = kCA

CA/2 = D/1=E/1

CAo = 0.0421

to find CA

2 mols of A gives 1 mole of D

D = 0.00132. Hence, A reacted = 0.00132*2

CA = 0.0421-0.00264

   = 0.03946

substitute values in main equation to get

ln(0.0421/0.03946) = k*63

k = 0.001028 s-1

part (c)

3A = F + G

rate law = RA = kCA

CAo = 0.08

t = 11.4 min

F = 0.015

3 moles of A gives 1 mole of F, hence A reacted = 0.015*3

A left = CA = 0.08-(0.015*3) = 0.035

substituting in the formula

ln(0.08/0.035)= k*11.4

k = 0.0725 min-1

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