H 14 The substitution of CO in Ni(CO)4 by another e L [where L is an to an under

Question

H 14 The substitution of CO in Ni(CO)4 by another e L [where L is an to an understanding of some of the general principles that govern the chemistry of compounds having and led 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 O bonds. (See J. P. Day Slow Ni(CO). ? Ni(CO)s+ CO Fast Ni(CO)s+L? Ni(CO)3L Molecularity of the slow step- b. Doubling the concentration of Ni(CO), 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. c. The experimental rate constant for the reaction, when L = P(C6Hs), is 9.3 x 10-3 Ni(CO)4 is 0.028 M, what is the concentration of the product after 9.9 minutes? 1 pt 1 at 20?If the initial concentration of 5 item attempts remaining 12

Explanation / Answer

a. Molecularity of the slow step = 1

Molecularity of the fast step = 2

Reason: The no. of reactant molecules involved in a reaction is called its molecularity.

b. Rate = k * [Ni(CO)4]1 * [L]0

c. The first-order rate constant can be written as

k = 1/t * ln{[Ni(CO)4]0/[Ni(CO)4]}

i.e. 9.3*10-3 s-1 = 1/9.9*60 s * ln{0.028 M/[Ni(CO)4]}

i.e. [Ni(CO)4] = 1.117*10-4 M

Therefore, the concentration of product after 9.9 min = (0.028 - 1.117*10-4) M = 0.027888 M

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