Two isomers (A and B) of a given compound dimerize as follows. Both processes ar

Question

Two isomers (A and B) of a given compound dimerize as follows.

Both processes are known to be second order in the reactant, and k1 is known to be 0.250 L mol?1 s?1 at 25°C. In a particular experiment, A and B were placed in separate containers at 25°C, where [A]0 = 1.00???10?2 M and [B]0 = 2.50???10?2 M. After each reaction had progressed for 3.89 min, [A] = 3.00[B]. In this case the rate laws are defined as the following.

Rate = ? d[A] dt = k1[A]2

Rate = ? d[B] dt = k2[B]2

(a) Calculate the concentration of A2 after 3.89 min.

(b) Calculate the value of k2.

(c) Calculate the half-life for the experiment involving A.

Explanation / Answer

Part a

For second order reaction

1/[A] = ( k1t + 1/[A]0 )

= ( 0.250 L/mol·s x 3.89 min x 60s/min) +

(1/1.0×10-²mol/L)

1/[A] = 158.35 L/mol

[A] = 0.00632 mol/L
From the stoichiometry of the reaction

[A2] = - ?[A]/2 = ([A]0 - [A]) / 2

= (1.0×10-²mol/L - 0.00632mol/L) / 2

= 0.00184 mol/L

Part b

Given that

[A] = 3.00[B]

[B] = (1/3) [A] = (1/3) x 0.00632 mol/L

= 0.0021 mol/L

Again for second order reaction for the second reaction

1/[B] = ( k2t + 1/[B]0 )

k2t = (1/[B] - 1/[B]0)

k2 x 3.89min x 60s/min = ( 1/0.0021mol/L) - (1/2.5×10-²mol/L )

k2 = 1.868 L/mol·s

Part C

For half life

t = t1/2

[A] = [B]0/2

From the second order reaction

2/[A]0 = k1 x t1/2 + 1/[A]0

t1/2 = 1/ ( k1 x [A]0 )

= 1/ ( 0.250 L/mol·s x 1.0×10-²mol/L )
= 400 s

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