The substitution of CO in Ni(CO)_4 by another molecule L [where L is an electron

Question

The substitution of CO in Ni(CO)_4 by another molecule L [where L is an electron-pair donor such as P(CH_3)_3 ] was studied some years ago and led to an understanding of some of the general principles that govern the chemistry of compounds having metal-CO bonds. (See J. P. Day, F. Basolo, and R. G. Pearson: Journal of the American Chemical Society, Vol. 90, p. 6927, 1968.) A detailed study of the kinetics of the reaction led to the following mechanism: What is the monocularity of each of the elementary reactions? Doubling the concentration of Ni(CO)4 increased the reaction rate by a factor of 2. Doubling the concentration of L had no effect on the reaction rate. Based on this information, write the rate equation for the reaction. The experimental rate constant for the reaction, when L = P(C_6H_5)_3, is 9.3 x 10^3 s^-1 at 20 degree C. If the initial concentration of Ni(CO)_4 is 0.019 M, what is the concentration of the product after 7.6 minutes?

Explanation / Answer

a)
For slow step, number of molecules colliding is 1
So Molecularity of slow step = 1

For fast step, number of molecules colliding is 2
So Molecularity of fast step = 2

b)
Rate depends on Ni(CO)4 such that doubling concentration will double the rate. So order will be 1
Answer:
rate = K [Ni(CO)4]

c)
This is 1st order reaction
t = 7.6 min = 7.6*60 s = 456 s

Let us 1st find the final concentration of reactant using:
[A] = [Ao]*e^(-K*t)
       = 0.019 * e^(-9.3*10^-3 * 456)
       = 2.74*10^-4 M

use:
decrease in ceoncentration of reactant = increase in concentration of product
increase in concentration of product = 0.019 M - 2.74*10^-4 M
                                                                            = 0.0187 M
Answer: 0.0187 M

a)
For slow step, number of molecules colliding is 1
So Molecularity of slow step = 1

For fast step, number of molecules colliding is 2
So Molecularity of fast step = 2

b)
Rate depends on Ni(CO)4 such that doubling concentration will double the rate. So order will be 1
Answer:
rate = K [Ni(CO)4]

c)
This is 1st order reaction
t = 7.6 min = 7.6*60 s = 456 s

Let us 1st find the final concentration of reactant using:
[A] = [Ao]*e^(-K*t)
       = 0.019 * e^(-9.3*10^-3 * 456)
       = 2.74*10^-4 M

use:
decrease in ceoncentration of reactant = increase in concentration of product
increase in concentration of product = 0.019 M - 2.74*10^-4 M
                                                                            = 0.0187 M
Answer: 0.0187 M

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