1. 40 points Consider the reaction A +2B>C, with an overall rate law: a. d[C] -=

Question

1. 40 points Consider the reaction A +2B>C, with an overall rate law: a. d[C] -= Kobs The following experimental data are obtained for this reaction: (i) For [AJo-1 M and [BJo-5 x 103 M, Time, (sec) 0 [B], (M) 5x 10 4x 103 3x 103 2x 10 x 103 (ii) For [Ap-1 M and [B»-2 M, the initial rate dig at time-0 sec is 0.5 x 103 M see". 1.0 2.0 3.0 4.0 5.0 dt For [Ap = 2 M and [B]0-2M, the initial rate at time = 0 sec is 2 x 10" M sec-I For the rate law above, find the orders and with respect to reactants A and B, respectively, the overall order Ytot, and the effective rate constant coefficient kobs. b. For the reaction above, the following reaction mechanism is proposed: A+A-211 11 + B 12 12 + B C ki k2 k3 Assume that the first step in the mechanism is the rate limiting step, i.e. ki

Explanation / Answer

1.

For the given reaction,

a. from the given data

from (ii) when [B] = constant and [A] was doubled, the rate of reaction quadrupled,

so the order with respect to [A] alpha = 2

from (i) when [A] was kept constant ans [B] was varied

the rate constant for the reaction was a constant

rate constant = (5 x 10^-3 - 4 x 10^-3)/1 = 1 x 10^-3 M/s

rate constant = (4 x 10^-3 - 3 x 10^-3)/1 = 1 x 10^-3 M/s                     

So, order with respect to [B] beta = 0

rate law becomes,

rate = k[A]^2

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b. For the given reaction mechanism

with first step be the slow step,

rate = k1[A]^2

for [C]

d[C]/dt = k3[I2][B] ------(1)

[I2] is an intermediate

applying steady state for [I2]

rate of formation - rate of consumption

k2[I1][B] = k3[I2][B]

[I2] = (k2/k3)[I1] ----------(2)

[I1] is another intermediate, applying steady state rule

2k1[A]^2 = k2[I1][B]

[I1] = (2k1/k2)[A]^2/[B] -----(3)

feeding (3) in (2)

[I2] = (2k1/k3)[A]^2/[B]

feed in rate equation,

d[C]/dt = (2k1)[A]^2 ....actual rate law

So the mechanism gives matches with the rate law derrived in part a. above and so the values of alpha and beta also matches as obtained in part a.

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