**Spontaneity, Enthalpy, Entropy, and Free Energy**

By Hao Zhang

**Laws of Thermodynamics**

- 1
^{st}-energy is constant in the universe; energy is neither created nor
eliminated.
- 2
^{nd} - entropy is always increasing in the universe.
- 3
^{rd} – for a perfect crystal at zero kelvin, the entropy is zero.

Spontaneity- The likelihood of a reaction to proceed without outside intervention. A
spontaneous reaction can be fast or slow. Spontaneous reactions will happen by themselves.

The equation for spontaneity is

D G = D H - TD
S

- D G is the free energy
- D H is the enthalpy
- D S is the entropy
- T is the temperature in kelvin

If D G is **negative**, the reaction is **spontaneous**.

If D G is **positive**, the reaction is not **spontaneous**.

The standard free energy: D G° =
D H° - TD
S°

This equation assumes that the reactant and product are both in their standard states.

Here is what it breaks down to:

Entropy; D S |
Enthalpy; D H |
Spontaneity |

+ D S |
+ D H |
At all temperature |

+ D S |
- D H |
At high temperature |

- D S |
- D H |
At low temperature |

- D S |
+ D H |
Not spontaneous |

D G is dependent on pressure. The equation that links them
together is:

D G = D G°
+ RTln(P)

- D G is the free energy
- D G° is the standard free energy,
for a gas the standard state is always 1 atm.
- R= 8.3145 J
- T is the temperature in kelvin.
- P is the partial pressure.

Free energy is also related to the equilibrium constant k and Q.

D G = D G°
+ RTln(Q)

At equilibrium, D G is 0 because at the product and reactant
have equal free energy.

0 = D G° + RTln(Q)

D G° = -RTln(k).

Free energy is the max work obtained from the reaction.

Work_{max}= D G

**Enthalpy**

Aaah, yes, enthalpy. If you've just heard of enthalpy, you will have no idea what it
is. When you're done with enthalpy, you'll still not have a clear good idea of what it is,
but you will know how to do problems with it, and that's what is important.

The enthalpy of a system, H, is simply defined as:

H = E + PV

Enthalpy is equal to the total energy of the system, plus the pressure of the system
times the volume of the system. It's sort of hard to grasp what enthalpy is, from that
definition. Instead of enthalpy (H) itself, you will usually deal with a *change* of
enthalpy (DH). And if pressure is constant, and the only work allowed to work on the
system is through volume, then:

D H = q

Yep, you should think of enthalpy as sort of like heat. It's not heat exactly, but if
those two conditions are met, then it is heat.

So if the change of enthalpy is increasing, that means it is gaining an increase of
energy, and therefore is endothermic. If the change of enthalpy is decreasing, that means
it is losing heat to the surroundings, or exothermic.

Final thought: +D H = Endo, -D H
= Exo.

How do you find the change of H? It's the enthalpy of the final products, minus the
enthalpy of the reactants. Or...

D H = D H_{products}
- D H_{reactants}

So if the products have less energy than reactants, then D H
will be negative, indicating energy was lost. And vice versa.

*
*

Entropy

Entropy, D S, is the degree of chaos in the universe.
For example, when a glass cup breaks on the floor, the glass breaks into little pieces.
The cup changes from a unified object to many smaller pieces.

D S is **positive** if things are going to **more
disorder**.

D S is **negative** if things are going to **less
disorder**.

The entropy of the universe is positive; the universe is becoming more disordered.

D S_{universe} = D S_{system}
+ D S_{surrounding}

The entropy for the three states of matter

Solid < Liquid < Gas

The molecules in the gas phase can be arranged in more ways than in the solid or liquid
state. The gas phase also has less intermolecular forces to hold the molecules together.

Heat flow changes D S_{surrounding. }If the system
is exothermic, heat is released; the heat increases the disorder of the surrounding. If
the system is endothermic, the surrounding loses heat. With less heat in the surrounding,
the molecules move slower and entropy decreases.

D S_{surrounding} = - D
H/T

- D H is the enthalpy of the system.
- T is the temperature in kelvin.

D S_{system} |
D S_{surrounding} |
D S_{universe} |
Spontaneity |

+ |
+ |
+ |
yes |

- |
- |
- |
no |

+ |
- |
? |
D S_{system >
}D S_{surrounding} |

- |
+ |
? |
D S_{system <
}D S_{surrounding} |

**Hess's Law**

This has to do with enthalpy again. Yay. Remember that reactions have a certain
enthalpy change with it: if it's positive, then the reaction will absorb energy; if
negative, the reaction will give off energy. We can write the equation, and then the
delta-H associated with it to the right of that. Like for example, to show the enthalpy
change for the boiling of one mole of water to water vapor, you can write like this:

H_{2}O_{ (l)} ---> H_{2}O_{ (g)} D H = 44 kJ

So before one mole of water evaporates, 44 kJ of energy must be given to it.

I had to look up the 44 kJ in a table or something; there's no way anyone can just look
at the equation and come up with it. But how about reactions that can't be found in a
table? There are tons of reactions; you just can't carry a 30 pound book with you and look
them all up. Hess's Law exists to make this easier for you. Technically, it means that the
enthalpy change between two states is not dependant on the pathway it takes to get there.
At first, this doesn't seem to help you one bit; but it means if you can add two or more
equations to get the desired equation, then you can add their respective enthalpy changes
to get the enthalpy change of this equation. There are two rules you will need to use in
these problems. We'll use the equation above to illustrate these two rules:

**
****If you have to multiply each side of a reaction by some number X, then multiply the
respective enthalpy change of that reaction by X also.**

D H° = S
n_{pD }H° _{f (products)
}- S n_{pD }H° _{f (reactants)}

D S° = S
n_{pD }S° _{f (products)
}- S n_{pD }S° _{f (reactants)}

D G° = S
n_{pD }G° _{f (products)
}- S n_{pD }G° _{f (reactants)}