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: Slow Ni(CO)_4 rightarrow Ni(CO)_3 + CO Fast Ni(CO)_3+L righrarrow Ni(CO)_3L What is the molecularity of each of the elementary reactions? Molecularity of the slow step = Molecularity of the fast step = 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. (Use k for the rate constant.) Rate = The expermiental rate constant for the reaction, when L = P(C_6H_5)_3, is 9.3 Times 10^-3 s^-1 at 20 degree C. If the initial concentration of Ni(CO)_4 is 0.029 M. what is the concentration of the product after 7.4 minutes? Concentration =

Explanation / Answer

a) Molecularity refers to number of molecules participating in the reaction. Accordingly

Molecularity for reaction -1 = 1 and for reaction-2 = 2

b) let the rate be represented as r =K[ NI(Co)4]m [ L]

Doubling [NI(Co)4 doubled the rate so m= 1 (first order reactionand doubling L does not have any effect on the rate so n=0

so r= K [Ni(Co)4]

for a first order reaction ln [A]= ln[A]0- kt

where [A] and [A]0 represent concentration of [A] at any time t and [A]0 is concentration at t=0 ( initial concentration)

given [A]0= 0.029M k=9.3*10-3sec-1 and t= 7.4 mintes= 7.4*60= 444sec

ln [A]= ln(0.029)- 9.3*10-3*444, [A] =0.000467M

for a first oder

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