The substitution of CO in Ni(CO)4 by another molecule L [where L is an electron-
ID: 483924 • Letter: T
Question
The substitution of CO in Ni(CO)4 by another molecule L [where L is an electron-pair donor such as P(CH3)3was 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:
1. What is the molecularity of each of the elementary reactions?
a.Molecularity of the slow step =
b.Molecularity of the fast step =
2. Doubling the concentration of Ni(CO)4 increased the reaction rate by a factor of 2. Doubling the concentration of had no effect on the reaction rate. Based on this information, write the rate equation for the reaction.
Rate= _____________
(Use k for the rate constant.)
3. The experimental rate constant for the reaction, when L= P(C6H5)3 is 9.3x10-3s-1 , is at 20 °C. If the initial concentration of Ni(CO)4 is 0.029 M, what is the concentration of the product after 3.8 minutes?
Concentration = _________________ M
*****THE OTHER CHEGG USER DID #3 WRONG WHO ANSWERED THIS QUESTION PREVIOUSLY, SO PLEASE DON NOT COPY HIS/HERS. THANKS*******
Slow Ni(CO)4 ----> Ni(CO)3 + CO Fast Ni(CO)3 + L ----> Ni(CO)3LExplanation / Answer
For the given substitution reaction,
1. slow step involves only one reactant so,
a. molecularity of slow step = 1
fast step involves two substrates so,
b. molecularity of the fast step = 2
2. When concentration of Ni(CO)4 is doubled, rate doubles so order with respect o Ni(CO)4 is first order.
the rate equation thus would be,
Rate = k[NiCO)4]
3. with,
k = 9.3 x 10^-3 s-1
[Ni(CO)4]o = 0.029 M
t = 3.8 x 60 = 228 s
concentration after 3.8 min [Ni(CO)4] would be,
ln[Ni(CO)4] = ln[Ni(CO)4]o - kt
ln[Ni(CO)4] = ln(0.029) - 9.3 x 10^-3 x 228
[Ni(CO)4] = 0.0035 M
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