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erma4kov [3.2K]
3 years ago
6

In general, which attractions are stronger: intermolecular or intramolecular? Research and explain a quantitative comparison bet

ween the strength of a chemical bond and the strength of intermolecular forces of attraction.
- I understand that intramolecular forces are stronger but I don’t know what to put for the quantitative data
Chemistry
1 answer:
murzikaleks [220]3 years ago
8 0

Answer is: intramolecular attractions are stronger.

Intramolecular attractions are the forces between atoms in molecule.

There are several types of intramolecular forces: covalent bonds, ionic bonds.

Intermolecular forces are the forces between molecules. The stronger are intermolecular forces, the higher is boiling point of compound, because more energy is needed to break interaction between molecules.

There are several types of intermolecular forces: hydrogen bonding, ion-induced dipole forces, ion-dipole forces andvan der Waals forces.

Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.

Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.

20 kJ/350 kJ = 0.057 = 5.7 %.


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At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ / mol. When 1.697 g of compound
melisa1 [442]

Answer:

13.85 kJ/°C

-14.89 kJ/g

Explanation:

<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>

<em />

The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

1.697g.\frac{1mol}{101.67g} .\frac{(-3039.0kJ)}{mol} =-50.72kJ

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qcomb + Qcal = 0

Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ

The heat capacity (C) of the calorimeter can be calculated using the following expression.

Qcal = C . ΔT

where,

ΔT is the change in the temperature

Qcal = C . ΔT

50.72 kJ = C . 3.661 °C

C = 13.85 kJ/°C

<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>

Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ

The heat of combustion per gram of B is:

\frac{-56.09 kJ}{3.767g} =-14.89 kJ/g

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