At given temperature the given hydrocarbons are to be predicted in order of increasing vapour pressure and have to be explained. Concept introduction: Vapour pressure: In a closed system, vapour pressure is the pressure produced by a vapour in thermodynamic equilibrium with its solid or liquid phase at a given temperature. Intermolecular forces are Van der Waals forces. They are weak and have two types viz., London dispersion forces, dipole-dipole forces and hydrogen bonding. Hydrogen bonding is relatively the strongest one. Intermolecular forces are the forces acting between molecules whereas Intramolecular forces are the forces that operate within a molecule. Hydrogen bonding is a special type of Dipole-dipole forces but stronger than the former. London dispersion forces exist in non-polar covalent compounds whereas dipole-dipole forces exist in polar covalent compounds but both are weak. Larger the molecular size, stronger the London dispersion force. Arrangement of major types of intermolecular forces in increasing order of strength: London dispersion forces < Dipole-dipole forces < Hydrogen bonding
At given temperature the given hydrocarbons are to be predicted in order of increasing vapour pressure and have to be explained. Concept introduction: Vapour pressure: In a closed system, vapour pressure is the pressure produced by a vapour in thermodynamic equilibrium with its solid or liquid phase at a given temperature. Intermolecular forces are Van der Waals forces. They are weak and have two types viz., London dispersion forces, dipole-dipole forces and hydrogen bonding. Hydrogen bonding is relatively the strongest one. Intermolecular forces are the forces acting between molecules whereas Intramolecular forces are the forces that operate within a molecule. Hydrogen bonding is a special type of Dipole-dipole forces but stronger than the former. London dispersion forces exist in non-polar covalent compounds whereas dipole-dipole forces exist in polar covalent compounds but both are weak. Larger the molecular size, stronger the London dispersion force. Arrangement of major types of intermolecular forces in increasing order of strength: London dispersion forces < Dipole-dipole forces < Hydrogen bonding
Science that deals with the amount of energy transferred from one equilibrium state to another equilibrium state.
Chapter 11, Problem 11.70QP
Interpretation Introduction
Interpretation:
At given temperature the given hydrocarbons are to be predicted in order of increasing vapour pressure and have to be explained.
Concept introduction:
Vapour pressure: In a closed system, vapour pressure is the pressure produced by a vapour in thermodynamic equilibrium with its solid or liquid phase at a given temperature.
Intermolecular forces are Van der Waals forces. They are weak and have two types viz., London dispersion forces, dipole-dipole forces and hydrogen bonding. Hydrogen bonding is relatively the strongest one.
Intermolecular forces are the forces acting between molecules whereas Intramolecular forces are the forces that operate within a molecule.
Hydrogen bonding is a special type of Dipole-dipole forces but stronger than the former.
London dispersion forces exist in non-polar covalent compounds whereas dipole-dipole forces exist in polar covalent compounds but both are weak.
Larger the molecular size, stronger the London dispersion force.
Arrangement of major types of intermolecular forces in increasing order of strength:
Part I.
a)
Draw reaction mechanism for the transformations of benzophenone to benzopinacol to benzopinaco lone
b) Pinacol (2,3-dimethyl, 1-3-butanediol) on treatment w/ acid gives a mixture of pina colone
(3,3-dimethyl-2-butanone) and 2, 3-dimethyl - 1,3-butadiene. Give reasonable mechanism
the formation of
the products
For
3. The explosive decomposition of 2 mole of TNT (2,4,6-trinitrotoluene) is shown below:
Assume the C(s) is soot-basically atomic carbon (although it isn't actually atomic carbon in real life).
2
CH3
H
NO2
NO2
3N2 (g)+7CO (g) + 5H₂O (g) + 7C (s)
H
a. Use bond dissociation energies to calculate how much AU is for this reaction in kJ/mol.
Chapter 11 Solutions
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