CH8 - Energetics Flashcards

(11 cards)

1
Q

Why is it easier to measure enthalpy changes rather than energy changes?

A

As it is easier to consider constant pressure rather than constant volume when dealing with solids. At constant pressure, heat input is equal to enthalpy (thus U).

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2
Q

Why was entropy introduced?

A

In order to define the relative stability of minerals and predict the direction of a mineral reaction

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3
Q

Why do transformations not commonly occur at the thermodynamically defined equilibrium temperature?

A

As the free energy of the phases is the same. This means a reduction in G is needed to drive transformation i.e. undercooling
If activation energy is low, undercooling will be low

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4
Q

Why is superheating lower than supercooling?

A

As heating increases atomic mobility

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5
Q

Why is diamond considered to be in a metastable state?

A

As at low temperatures structural changes are difficult, thus its transformation to graphite at room temperature is insignificant.
If a complex mineral fails to transform to its equilibrium state, it may undergo an easier thermodynamic transformation (e.g. clinopyroxene to orthopyroxene)

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6
Q

What assumptions can be used to simplify the system?

A

Born Oppenheimer approximation, Harmonic approximation and phonon approximation

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7
Q

Born Oppenheimer approximation

A

The electronic and nuclear problems are considered separately - calculations related to electronic properties are solved at fixed nuclear coordinates

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8
Q

Harmonic approximation

A

initial calculation assumes vibrations behave harmonically. Underestimates density of states at high temperatures close to the dissociation point of bonds.

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9
Q

Phonon dispersion approximation

A

use the lattice rather than the motif to consider phonon displacements of the structure. Also limits the vibrations considered to a reasonable spatial frame

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10
Q

Undercooling

A

undercooled growth occurs when the liquid temperature is below the interface temperature. This occurs by cooling the liquid below its equilibrium transformation temp / by compositional segregation during solidification process

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11
Q

Hess’ law

A

Enthalpy change is the same regardless of it the reaction took place in one or several steps, provided reactants and products are the same

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