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Chapter 14: Problem 112

In a hydrocarbon solution, the gold compound $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\( decomposes into ethane \)\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ and a different gold compound, \(\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3} .\) The following mechanism has been proposed for the decomposition of $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}:$ Step 1: $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \underset{k-1}{\stackrel{k_{1}}{\rightleftharpoons}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3} \quad$ (fast) Step 2: $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{\mathrm{k}_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad$ (slow) Step 3: $\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{k_{3}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3} \quad$ (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary steps? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (f) What would be the effect on the reaction rate of adding \(\mathrm{PH}_{3}\) to the solution of $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} ?$

Short Answer

Expert verified
The overall reaction is \((CH_3)_3AuPH_3 \longrightarrow C_2H_6+(CH_3)AuPH_3\). The intermediates are \((CH_3)_3Au\) and \((CH_3)Au\). The molecularity of steps 1, 2, and 3 are bi-molecular, uni-molecular, and bi-molecular, respectively. The rate-determining step is step 2, and the predicted rate law is \(Rate= k_2\left(\frac{K_1[(CH_3)_3AuPH_3]}{[PH_3]}\right)\). Adding PH3 to the solution will decrease the reaction rate.

Step by step solution

01

Overall Reaction

To find the overall reaction, we need to add up all the elementary steps in the mechanism and cancel out any species which appear on both the reactants and products side. Step 1: \((CH_3)_3AuPH_3\rightleftharpoons (CH_3)_3Au + PH_3\) (fast) Step 2: \((CH_3)_3Au\longrightarrow C_2H_6 + (CH_3)Au\) (slow) Step 3: \((CH_3)Au + PH_3\longrightarrow (CH_3)AuPH_3\) (fast) Summing up the steps, we have: \((CH_3)_3AuPH_3 \longrightarrow C_2H_6+(CH_3)AuPH_3\) Upon canceling out common species, we get the overall reaction: \((CH_3)_3AuPH_3 \longrightarrow C_2H_6+(CH_3)AuPH_3\)
02

Intermediates

Intermediates are the species that are formed during reaction steps and then consumed in subsequent steps. We can identify the intermediates by examining each step in the mechanism. In this case, the intermediates are \((CH_3)_3Au\) and \((CH_3)Au\).
03

Molecularity of Elementary Steps

Molecularity is the number of molecules or ions that participate in a reaction step. In this case, we have three elementary steps: 1. Step 1: 1 reactant molecule and 2 product molecules (bi-molecular) 2. Step 2: 1 reactant molecule and 2 product molecules (uni-molecular) 3. Step 3: 2 reactant molecules and 1 product molecule (bi-molecular)
04

Rate-Determining Step

The rate-determining step is the slowest step in the mechanism, which determines the overall reaction rate. In this case, step 2 is the slowest step: \((CH_3)_3Au\longrightarrow C_2H_6 + (CH_3)Au\). Therefore, the rate-determining step is step 2.
05

Rate Law Prediction

Since the rate-determining step is step 2, the rate law can be written using the concentrations of species involved in step 2: Rate Law: \(Rate = k_2[(CH_3)_3Au]\) However, the concentration of an intermediate is not directly measurable. So, we need to express the concentration of intermediate \((CH_3)_3Au\) in terms of other species using the equilibrium constant \(K_1\) from step 1: \(K_1 = \frac{[(CH_3)_3Au][PH_3]}{([(CH_3)_3AuPH_3])}\) Solving for the intermediate concentration: \([(CH_3)_3Au] = \frac{K_1[(CH_3)_3AuPH_3]}{[PH_3]}\) Substitute into the rate law: \(Rate= k_2\left(\frac{K_1[(CH_3)_3AuPH_3]}{[PH_3]}\right)\)
06

Effect of Adding PH3

To understand the effect of adding PH3 on the reaction rate, analyze the rate law expression: \(Rate= k_2\left(\frac{K_1[(CH_3)_3AuPH_3]}{[PH_3]}\right)\) As we can see from the rate law, the reaction rate is inversely proportional to the concentration of PH3. If we add more PH3 to the solution, the reaction rate will decrease.

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Most popular questions from this chapter

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at $30{ }^{\circ} \mathrm{C}\( is \)4.0 \times 10^{-2} M^{-1} \mathrm{~s}^{-1}$. If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is $2.5 \times 10^{-4} \mathrm{M}\( and that of \)\mathrm{Cl}_{2}\( is \)2.0 \times 10^{-2} \mathrm{M},$ what is the rate of formation of \(\mathrm{H}^{+}\) ?

The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form \(\mathrm{NOF}\) and \(\mathrm{F}\) has an activation energy of $E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\(. and a frequency factor of \)A=6.0 \times 10^{8} M^{-1} \mathrm{~s}^{-1}$. The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g) $$ (a) Calculate the rate constant at \(100^{\circ} \mathrm{C}\). (b) Draw the Lewis structures for the \(\mathrm{NO}\) and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the shape for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

Consider two reactions. Reaction (1) has a half-life that gets longer as the reaction proceeds. Reaction (2) has a half-life that gets shorter as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right) .(\mathbf{a})\) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest. (a) $E_{a}=75 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-20 \mathrm{~kJ} / \mathrm{mol}$ (b) $E_{a}=100 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=+30 \mathrm{~kJ} / \mathrm{mol}$ (c) $E_{a}=85 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-50 \mathrm{~kJ} / \mathrm{mol}$
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