Enthalpy change of solution 174-175 Flashcards

1
Q

Define the standard enthalpy change of solution.

A

The enthalpy change that takes place when one mole of a compound is completely dissolved in water under standard conditions.

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2
Q

What two processes take place when a solid dissolves?

A
  • breakdown of the ionic lattice into gaseous ions.
  • hydration of the ions.
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3
Q

Explain what is involved in the breakdown of an ionic lattice when a substance is dissolved in a liquid?

What is the breakdown of an ionic lattice opposite to?

A

It involves breaking down the crystal lattice and separating the ions. Overcoming the attractive forces between the oppositely charged ions requires energy.

The process is the exact opposite of that producing the lattice enthalpy. Enthalpy change involved in breaking ionic lattice=-HLE.

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4
Q

Explain what happens in the hydration part of the standard enthalpy change of solution.

A

In the process the gaseous ions bond with the water molecules. The positively charged ions will be attracted to the slightly negative oxygen atoms in water, and the negatively charged ions will be attracted to the slightly positive hydrogen atoms in water.

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5
Q

Define the standard enthalpy change of hydration.

A

The enthalpy change that takes place when one mole of isolated gaseous ions is dissolved in water forming one mole of aqueous ions under standard conditions.

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6
Q

Is the enthalpy of hydration an endothermic or exothermic process?

A

Energy is released when the ions form bonds with water molecues. So hydration is an exothermic process.

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7
Q

What do you need to know to be able to calculate the lattice enthalpy of an ionic solid?

A
  • The enthalpy changes of hydration of the constituent gaseous ions and the enthalpy change of solution of the ionic solid.
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8
Q

Knowing the following produce a born-Haber cycle and calculate the lattice enthalpy:

K*(g) + Cl-(g) → KCl(s) ΠLE(lattice enthalpy)

KCl(s) + aq → K*(aq) + Cl-(aq) Πs = +26 (enthalpy change of solution.)

K*(g) + aq → K*(aq) Πhyd= -322

Cl-(g) + aq → Cl-(aq) Πhyd= -363

A

Answer is on page 175 on book.

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