1. 17.4: Heat Transfer & Changes in the Entropy of Surroundings
Cameron Cantine

2. The Second Law
The second law states that for a process to be considered spontaneous, there must be a net increase in the entropy of the universe.
The equation used to determine the change in entropy of the universe is ΔSuniv = ΔSsystem + ΔSsurroundings
This means that in order for a process to be spontaneous, the ΔSsurroundings must be greater than the negative ΔSsystem

3. For Example...
Knowing that the ΔSsystem when water freezes is negative, for the process to be spontaneous, must the absolute value of ΔSsurroundings be greater or less than the ΔSsystem?
Greater! For example if the negative number was -5, only an x value of more than 5 will make -5 + x > 0

4. Explanation:
It would be greater! Now let’s consider how the entropy of the surroundings can be increased. Using the last example, the exothermic process of water freezing, heat is released into the surroundings. Entropy can be defined as the randomness or disorder of a substance and, as heat increases the speed of particles in a system, when heat is lost, the entropy of that system will decrease.

5. ΔSsurr is also dependant on temperature:
Entropy is a measure of joules/kelvins meaning that it is a measure of energy over temperature. This means that the higher the temperature, the lesser the effect of the energy change. In the case of the freezing of water, spontaneity is lost above 0°C. This is due to the fact that the change in the entropy of the surroundings is not high enough to counteract the change in the system when put into the formula ΔSuniv = ΔSsystem + ΔSsurroundings. If the absolute value of the change in entropy of the surroundings is lower than the absolute value of the change in the system, like in the example above, they will not add to be greater than 0, so they will not be spontaneous.

6. Quantifying Entropy Changes in the Surroundings
When a process has -qsys, it is emitting heat into the surroundings so there’s positive change in entropy of the surroundings and vice versa. Because the absolute values of the changes are the same, we can say that ΔSsurr is proportional to the -qsys. It is also known that ΔSsurr is inversely proportional to the temperature.

7. so...?
These two trends can be combined to create the following equation:
ΔSsurr = (-qsys)/(T) (T being in kelvins)
Also, because qsys = ΔHsys at a constant pressure, this equation is made true: (Where “H” expresses enthalpy)
ΔSsurr = (-ΔHsys)/(T) (at constant pressures)

8. 17.5 Gibbs Free Energy
At a constant temperature and pressure, we can use our first formula, ΔSuniv = ΔSsys + ΔSsurr, with this most recent one, (-ΔHsys)/(T) = ΔSsurr, to create the following formula:
ΔSuniv = ΔSsys - (ΔHsys/T)
This formula can be switched around to look like this: -T(ΔSuniv) = ΔHsys -T(ΔSsys) The right side of this equation represents the change in Gibbs Free Energy which is shown by the equation G = H - TS, so at a constant temperature, we can equate ΔG to ΔH - TΔS creating the following equation: ΔG = ΔH - TΔS
-T(ΔSuniv) = ΔHsys - T(ΔSsys) can be written simply as -T(ΔSuniv) = ΔH - TΔS

9. Summary:
Using the two formulas, -T(ΔSuniv) = ΔH - TΔS and ΔG = ΔH - TΔS, because both of the left sides equate to the same right side we end up with this formula to express the significance of the change in Gibbs Free Energy:
ΔG = -T(ΔSuniv) This means that ΔG is negatively proportional to ΔSuniv, and because ΔSuniv determines if something is spontaneous, ΔG will now also, just with the opposite sign.
So, when there is a negative change in Gibbs Free Energy, the process is spontaneous.

10. continued:
Because we can now find ΔG using just the ΔS and ΔH of the system we can find the spontaneity of chemical reactions, because both of these change values are calculable.

11. This diagram shows the reason Gibbs Free Energy is also called chemical potential. It is showing that the direction of spontaneous change is controlled by it.

12. Now let’s apply this:
Determine if the following reaction is spontaneous:
302(g) ➝ 203(g) ΔH°rxn = kJ
This reaction is endothermic as we can see ΔH is positive and because the number of moles of gas decreases from the left side to the right side of the equation the change in entropy (ΔS) is negative. When you subtract a negative from a positive you are always left with a positive, so ΔG must be positive rendering this reaction nonspontaneous at all temperatures! If ΔS is positive and ΔH is negative, the reaction will be spontaneous at all temperatures.

13. Temperature has influence when ΔS and ΔH have the same sign
When both ΔS and ΔH are positive values, meaning the reaction is endothermic and faces an increase in entropy, a low T value will make the process spontaneous, but once the reaction reaches a certain higher temperature, it becomes nonspontaneous.
When both are negative, meaning the reaction is exothermic and something makes the reaction lose entropy, the reaction will be spontaneous at a low temperature, but high T values will make it nonspontaneous.
10 - (2)4 = 2, but if T raises from 2 to 30, meaning it warms up, 10 - (30)4 = Remember, a negative ΔG means that the process is spontaneous.

14. How to find when ΔG changes sign:
When the signs of ΔS and ΔH are the same, this does not matter because the reaction will either be positive or negative at all temperatures. However, when the signs are different, you can find the value of T that will change the spontaneity of the reaction. This can be done by making ΔG = 0 and then by solving the equation ΔG = ΔH - TΔS for T Consider a reaction where ΔH = kJ and ΔS = J/K. Does ΔG become more negative or more positive as temperature increases?

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