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Flashcards in General Chemistry-Thermochemistry Deck (190)
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How can you derive the standard free energy change using Keq?

/\G degree rxn= -RT ln Keq


What do the letters represent in the equation /\G degree rxn= -RT ln Keq?

R: ideal gas constant
T: Temp in Kelvins
Keq: The equilibrium constant


The greater the value of Keq the ____________ the value of _________. In the /\G degree rxn= -RT ln Keq equation.

More positive the value of its natural logorithm. The more positive the natural logarithm, the more negative the standard free energy change. The more negative the standard free energy change, the more spontaneous the reaction.


What happens once a reaction begins, when calculating /\G degree rxn?

The standard state conditions (specifically 1 M solutions) no longer apply


How do you change the equation when calculating /\G degree rxn, once a reaction is in progress?

Relate the /\G rxn to the reaction quotient Q:
/\G rxn= /\G degree rxn + RT ln Q= RT ln Q/Keq


When Q/Keq is less than 1, what happens when using the /\G rxn equation?

Then the natural logarithm will be negative, and the free energy change will be negative, so the reaction will spontaneously proceed forward until equilibrium is reached.


What happens when the ratio Q/Keq is greater than 1?

Then the natural logarithm will be positive, and the free energy change will be positive. The reaction will spontaneously move in the reverse direction until equilibrium is reached.


What happens if Q/Keq is equal to 1?

The reaction quotient is equal to the equilibrium constant; the reaction is at equilibrium and the free energy change is zero.


How can reaction profiles be altered?

By the presence of a catalyst.


What happens when a catalyst is added?

The overall free energy change of the reaction is not altered, but the activation energy required to accomplish the reaction is reduced significantly.