Confusion About Gibbs Free Energy & Enthalpy

[dmission]

Hi all -- two quick questions.
1) When is Gibbs Free Energy be equal to 0 when a reaction is at equilibrium (is boiling point a good example?)
2) Is there a good way to qualitatively understand enthalpy? I know it's H = deltaU + PdeltaV, but beyond that I'm a bit at a loss.

Thanks.

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[geeyouknit]

Hi all -- two quick questions.
1) When is Gibbs Free Energy be equal to 0 when a reaction is at equilibrium (is boiling point a good example?)
2) Is there a good way to qualitatively understand enthalpy? I know it's H = deltaU + PdeltaV, but beyond that I'm a bit at a loss.

Thanks.

Not sure what the first question means? Wording is kind of confusing.
Let me take a shot at the second one.
Enthalpy describes the amount of energy in the system at a certain state, A. To get to another state, B, the system requires energy or expels excess energy.
Enthalpy is also just a relative term. People called some things to have 0 enthalpy ( I guess the state of being an element, like O2 or Sulfur) and when that state changed to SO2, the measured the amount of energy (in heat) that was transferred was called to be the change in enthalpy, Delta H.

Hope that helps a little. Anyone, please feel free to correct me if I'm wrong.

 

[MD Odyssey]

Hi all -- two quick questions.
1) When is Gibbs Free Energy be equal to 0 when a reaction is at equilibrium (is boiling point a good example?)

Not sure what you are asking here. If the change in the Gibbs energy is zero, this implies that the enthalpy is equal to the product of the change in entropy and the temperature.

2) Is there a good way to qualitatively understand enthalpy? I know it's H = deltaU + PdeltaV, but beyond that I'm a bit at a loss.

Thanks.

The change in enthalpy of a reaction is really the change in internal energy of the system during the reaction, when the work done by the system on the environment is negligible. For most applications, it seems that chemists tend to equate heat and enthalpy, even though they aren't precisely the same thing. For most practical applications, the change in enthalpy of a reaction is more or less considered to be the same thing as the heat evolved during the reaction.

Of course, there is more to the story - determining whether a particular reaction is spontaneous or not requires a knowledge of the change in entropy of the reaction and the temperature that the reaction occurs at.

 

[MT Headed]

From EK Chemistry section 3.11:
"Enthalpy is a man-made property"
"Enthalpy is not a measure of some intuitive property"
"Enthalpy of the universe does not remain constant"
"Enthalpy cannot be intuited. Just memorize the equation."

It's kind of like heat but not really. It's kind of like energy but it counts some energies twice. For the purposes of the MCAT, memorize the equation, answer the enthalpy question quickly, and move on. Enthalpy is confusion, by design.

 

[dmission]

Thanks for the replies.

And, sorry, what I meant by the first question was this:
When exactly would gibbs free energy be equal to 0? Would it be when a reaction is at equilibrium (and if so, is a reaction at equilibrium when a substance is at its boiling point?)

 

[muhali3]

If the free energy change of a reaction is zero, then the equilibrium constant equals 1.

Gibbs free energy is zero when the products and reactants are of equal energy. The reaction is not spontaneous in either direction. Neither the products or reactants would be favored.

Equilibrium is when the forward reaction rate equals the reverse reaction rate.

At the boiling point of a liquid, the vapor pressure equals the atmospheric pressure. Molecules are vaporizing into the gas phase at the same rate molecules are condensing back into the liquid phase.

But this is a phase change, not a chemical reaction, so the concept of equilibrium doesn't apply. (afaik, anyone please feel free to correct me)

However, at the boiling point, the free energy change equals zero. deltaG = deltaH - T x deltaS

Boiling is an endothermic process (+H), but has a positive entropy change (+S), so by increasing the temperature of the liquid, delta G would begin to approach zero, and start boiling when deltaG = 0.

 

[dmission]

Couldn't boiling be viewed as a phase change reaction, though? For example H20(l)->H20(g)?

 

[muhali3]

Well, if you want to look at it that way, then yes, at the boiling point, the phase change reaction is at equilibrium.

But afaik, equilibrium is thought of in terms of a chemical reaction.

Where Keq = ([C]^c[D]^d )/ ([A]^a^b)

 

[indianjatt]

You are absolutely correct. Delta G = 0 is at an equilibrium state. There is no tendency for the reaction to proceed forward or backward.

 

[dmission]

Thanks. So can delta G or delta H then determine direction of a reaction?

 

[MD Odyssey]

Thanks. So can delta G or delta H then determine direction of a reaction?

In general, there are two things which determine the direction of a reaction: the change in the energy of the system and the entropy gained or lost by the system. The Gibbs energy takes both of these factors into account:

The enthalpy term is what shows the change in internal energy of the system plus any mechanical work done on the system. The latter term includes the change in entropy of the system. Both the entropy and enthalpy terms can be either positive or negative, thus it's possible for a reaction to be energetically favorable at one temperature while being unfavorable at another. Note that this equation requires T to be in Kelvin rather than degrees Celsius or Fahrenheit.

Permit me a subtle example. For short-chained, saturated hydrocarbons, the most likely conformation is one with all hydrogen atoms gauche to each other - the staggered configuration we see in a Newman projection. However, if I have a very long, saturated hydrocarbon, then the most likely conformation becomes something more complicated. This is due to the fact that, for a large hydrocarbon, the entropic term actually dominates and forces the change in Gibbs energy to be negative.
​

 

[dmission]

In general, there are two things which determine the direction of a reaction: the change in the energy of the system and the entropy gained or lost by the system. The Gibbs energy takes both of these factors into account:

The enthalpy term is what shows the change in internal energy of the system plus any mechanical work done on the system. The latter term includes the change in entropy of the system. Both the entropy and enthalpy terms can be either positive or negative, thus it's possible for a reaction to be energetically favorable at one temperature while being unfavorable at another. Note that this equation requires T to be in Kelvin rather than degrees Celsius or Fahrenheit.

Permit me a subtle example. For short-chained, saturated hydrocarbons, the most likely conformation is one with all hydrogen atoms gauche to each other - the staggered configuration we see in a Newman projection. However, if I have a very long, saturated hydrocarbon, then the most likely conformation becomes something more complicated. This is due to the fact that, for a large hydrocarbon, the entropic term actually dominates and forces the change in Gibbs energy to be negative.
​

So you can just use deltaG to determine the direction? How would you do that exactly?
Thanks.

 

[MD Odyssey]

So you can just use deltaG to determine the direction? How would you do that exactly?
Thanks.

Well, given the change in entropy and enthalpy for the reaction, you compute the change in the Gibbs energy for the reaction a given temperature. Those particular values are tabulated and can typically be looked up in the CRC.

If the change in the Gibbs energy is negative, then the reaction is obviously exergonic. The converse obviously holds if the change in the Gibbs energy turns out to be positive.