We can see here that the catalytic cycle for the hydrogenation of alkenes using Wilkinson's catalyst involves several steps.
What are the steps involved?Here's an overview of the catalytic cycle, including the necessary details:
i. Chemical structure of the catalyst:
Wilkinson's catalyst, also known as chloridotris(triphenylphosphine)rhodium(I), has the following chemical structure: [RhCl(PPh3)3]
ii. Molecular geometry of the catalyst:
The Wilkinson's catalyst has a trigonal bipyramidal geometry around the rhodium center. The three triphenylphosphine (PPh3) ligands occupy equatorial positions, while the chloride (Cl) ligand occupies an axial position.
iii. Type of reactions involved:
The catalytic cycle involves several reactions, including:
Oxidative addition: The rhodium center undergoes oxidative addition, reacting with molecular hydrogen (H2) to form a dihydride intermediate.Alkene coordination: The alkene substrate coordinates to the rhodium center, forming a π-complex.Hydrogenation: The coordinated alkene undergoes hydrogenation, resulting in the addition of hydrogen atoms to the double bond and formation of a metal-alkyl intermediate.Reoxidation: The metal-alkyl intermediate reacts with a hydrogen molecule to regenerate the rhodium dihydride species.iv. Starting material, reagent, and solvent:
The starting material is an alkene, and the reagent is Wilkinson's catalyst ([RhCl(PPh3)3]). The reaction is typically carried out in a suitable solvent, such as dichloromethane (CH2Cl2) or tetrahydrofuran (THF).
v. Major and minor end-products:
The major end-product of the hydrogenation reaction is the fully saturated alkane, resulting from the addition of hydrogen across the double bond. The minor end-product may include cis- or trans-configured alkanes if the original alkene substrate possesses geometric isomers.
vi. Intermediates:
The intermediates in the catalytic cycle include:
Rhodium dihydride complex: [RhH2(PPh3)3]Alkene-Rhodium π-complex: [Rh(η2-alkene)(PPh3)3]Metal-alkyl intermediate: [Rh(alkyl)(PPh3)3]These intermediates play a crucial role in facilitating the hydrogenation reaction and enabling the catalytic cycle to proceed.
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A 300.0 mL quantity of hydrogen is collected over water at 19.5 C and a total atmospheric pressure of 750. mm Hg. The partial pressure of water at this temperature is 17.0 mm Hg
The partial pressure of hydrogen in the collected gas sample is 733.0 mm Hg (calculated by subtracting the partial pressure of water, 17.0 mm Hg, from the total atmospheric pressure, 750.0 mm Hg).
When a gas is collected over water, the presence of water vapor affects the total pressure observed. In this case, the total atmospheric pressure is given as 750.0 mm Hg, and the partial pressure of water vapor at 19.5°C is 17.0 mm Hg.
To determine the partial pressure of hydrogen, we need to subtract the partial pressure of water vapor from the total atmospheric pressure. Partial pressure refers to the pressure exerted by an individual gas component in a mixture. In this scenario, the collected gas is primarily hydrogen, with water vapor being the other component.
By subtracting the partial pressure of water vapor (17.0 mm Hg) from the total atmospheric pressure (750.0 mm Hg), we can find the partial pressure of hydrogen:
Partial pressure of hydrogen = Total atmospheric pressure - Partial pressure of water vapor
Partial pressure of hydrogen = 750.0 mm Hg - 17.0 mm Hg
Partial pressure of hydrogen = 733.0 mm Hg
Therefore, the partial pressure of hydrogen in the collected gas sample is 733.0 mm Hg.
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How many liters of dry H₂ are produced at STP from 8.5 g Al according to the following reaction?
2 Al (s) + 6 HCI (aq) →2 AICI3 (aq) + 3 H₂ (g)
14.1 L
7.06 L
10.6 L
21.2 L
The volume of H₂ gas produced at STP from 8.5 g of Al according to the following reaction is option (3) 10.6 L
We can use the stoichiometry method to determine the volume of hydrogen gas produced from a given amount of aluminium in the following reaction:2 Al (s) + 6 HCI (aq) →2 AICI3 (aq) + 3 H₂ (g)To use the stoichiometry method, the balanced equation of the reaction is required.
From the equation, we know that the mole ratio of Al to H₂ is 2:3. This means that 2 moles of Al will produce 3 moles of H₂ gas. Therefore, we can find the number of moles of H₂ gas produced by the number of moles of Al.
Using the given mass of Al, we can find the number of moles of Al:
mass of Al = 8.5 g
Molar mass of Al = 26.98 g/mol
Number of moles of Al = mass of Al/molar mass of Al= 8.5 g/26.98 g/mol= 0.315 mol
From the mole ratio in the balanced equation, we know that 2 moles of Al will produce 3 moles of H₂.
Therefore, the number of moles of H₂ produced is:
Number of moles of H₂ = 3/2 × number of moles of Al= 3/2 × 0.315 mol= 0.473 mol
To find the volume of hydrogen gas produced, we need to use the ideal gas law equation which is PV = nRT.
Since the volume is required at STP, we can use the molar volume of gas at STP, which is 22.4 L/mol.To find the volume of H₂ gas, we can use the following formula:
Volume of H₂ gas = number of moles of H₂ × molar volume at STP= 0.473 mol × 22.4 L/mol= 10.6 L
Therefore, the volume of H₂ gas produced at STP from 8.5 g of Al is 10.6 L. The correct option is option (3) 10.6 L.
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