Problem 78 - Chemistry and Chemical Reactivity

Question

The substitution of \(\mathrm{CO}\) in \(\mathrm{Ni}(\mathrm{CO})_{4}\) by another molecule L [where \(\left.L \text { is an electron-pair donor such as } P\left(C H_{3}\right)_{3}\right]\) 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 \(\quad \mathrm{Ni}(\mathrm{CO})_{4} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{CO}\) Fast \(\quad \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{L} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3} \mathrm{L}\) (a) What is the molecularity of each of the elementary reactions? (b) Doubling the concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) 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. Does this agree with the mechanism described? (c) The experimental rate constant for the reaction, when \(\mathrm{L}=\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) is \(9.3 \times 10^{-3} \mathrm{s}^{-1}\) at \(20^{\circ} \mathrm{C}\) If the initial concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(0.025 \mathrm{M}\) what is the concentration of the product after 5.0 minutes?

Step-by-Step Solution

Answer

(a) Unimolecular, bimolecular. (b) Rate = k[Ni(CO)_4]; yes. (c) 0.0231 M product.

Step 1: Determine Molecularity of Reactions

The molecularity of a reaction refers to the number of molecules involved in a single step of the reaction. - For the reaction \( \mathrm{Ni}({\mathrm{CO}})_{4} \rightarrow \mathrm{Ni}({\mathrm{CO}})_{3}+ \mathrm{CO} \), only one molecule of \( \mathrm{Ni}({\mathrm{CO}})_{4} \) is involved, making it unimolecular.- For the reaction \( \mathrm{Ni}({\mathrm{CO}})_{3}+ \mathrm{L} \rightarrow \mathrm{Ni}({\mathrm{CO}})_{3} \mathrm{L} \), two molecules, \( \mathrm{Ni}({\mathrm{CO}})_{3} \) and \( \mathrm{L} \), are involved, making it bimolecular.

Step 2: Analyze Rate Data for Rate Law

Since doubling the concentration of \( \mathrm{Ni}({\mathrm{CO}})_{4} \) doubles the rate, the reaction rate is first-order with respect to \( \mathrm{Ni}({\mathrm{CO}})_{4} \). The fact that the concentration of \( \mathrm{L} \) does not influence the rate indicates a zero-order dependency on \( \mathrm{L} \). The rate law can be written as: \[ \text{Rate} = k[\mathrm{Ni}({\mathrm{CO}})_{4}] \]. This agrees with the mechanism described, as the slow step is the rate-determining step.

Step 3: Calculate Product Concentration

For a first-order reaction, the concentration as a function of time is given by the formula:\[ [A]_t = [A]_0 e^{-kt} \]where \([A]_t\) is the concentration at time \( t \), \([A]_0\) is the initial concentration, \( k \) is the rate constant, and \( t \) is the time. Plugging in the given values:- \( [\mathrm{Ni}({\mathrm{CO}})_{4}]_0 = 0.025 \mathrm{M} \)- \( k = 9.3 \times 10^{-3} \mathrm{s}^{-1} \)- \( t = 5.0 \times 60 \, \mathrm{s} = 300 \mathrm{s} \)Calculate:\[ [\mathrm{Ni}({\mathrm{CO}})_{4}]_t = 0.025 e^{-9.3 \times 10^{-3} \times 300} \approx 0.025 e^{-2.79} \approx 0.00186 \mathrm{M} \].Thus, \( [\mathrm{Ni}({\mathrm{CO}})_{3}\mathrm{L}] \approx 0.025 - 0.00186 = 0.02314 \mathrm{M} \).

Key Concepts

Molecularity

Understanding the concept of molecularity helps us determine how many molecules participate in a single step of a chemical reaction. This is a fundamental aspect when analyzing reaction mechanisms.
For instance, in the slow step of the exercise with the reaction \ \( \mathrm{Ni}(\mathrm{CO})_{4} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3} + \mathrm{CO} \), one molecule of \( \mathrm{Ni}(\mathrm{CO})_{4} \) is involved. This means it's a unimolecular reaction, meaning only one molecule undergoes transformation in that particular step.
On the other hand, the fast step involves two molecules: \( \mathrm{Ni}(\mathrm{CO})_{3} \) and \( \mathrm{L} \). Hence, this step is considered bimolecular, as it involves the collision and interaction between two molecules.

• **Unimolecular reactions**: Involve only one reacting molecule.
• **Bimolecular reactions**: Involve two reacting molecules coming together.

Understanding molecularity helps infer the mechanics of chemical reactions, especially when predicting which steps might be slower or faster.

Rate Law

The rate law of a reaction relates the rate of a chemical reaction to the concentration of its reactants. It's a mathematical expression that forms the foundation to predict reaction behaviour.
Based on experimental data from the exercise, where doubling the concentration of \( \mathrm{Ni}(\mathrm{CO})_{4} \) led to a doubling in reaction rate, we conclude that the reaction is first-order concerning \( \mathrm{Ni}(\mathrm{CO})_{4} \).
Conversely, changing the concentration of \( \mathrm{L} \) had no effect on the rate. This indicates a zero-order dependency on \( \mathrm{L} \).
In this case, the rate law can be written as:

• \( \text{Rate} = k[\mathrm{Ni}(\mathrm{CO})_{4}] \)

Zero-order means \( \mathrm{L} \) does not appear in the rate law, affirming its non-significant impact on rate. Hence, understanding rate laws is crucial for predicting how fast a reaction will proceed based on different conditions.

First-Order Reaction

A first-order reaction is characterized by the rate being directly proportional to the concentration of a single reactant. The equation for a first-order reaction is crucial for calculating the concentration changes over time.
The general formula to calculate the concentration of a reactant \ \( A \) at a given time in a first-order reaction is:

• \( [A]_t = [A]_0 e^{-kt} \)

where:

• \( [A]_t \) is the concentration at time \( t \).
• \( [A]_0 \) is the initial concentration.
• \( k \) is the rate constant.
• \( t \) is the time elapsed.

In the exercise's context, with the given rate constant and initial concentration, the calculations for the concentration after 5 minutes followed this model. Subsequently, we determined the remaining concentration of \( \mathrm{Ni}(\mathrm{CO})_{4} \) was approximately \( 0.00186 \mathrm{M} \), indicating the progress of the reaction and the concentration of formed products.