|Lecture Notes::Lec 5_5 June||
So what determines whether a process is spontaneous? Two components contribute under common laboratory and biological conditions:
More generally we can say that the entropy of the Universe increases in a spontaneous process. This is enshrined as the second law of thermodynamics.
Note that while the First Law says the Energy of the universe is constant, the entropy is not. Entropy has sometimes been referred to as "times arrow" because it is what gives the Universe a direction - we can only go forward in time.
So what is entropy? Entropy is often described as a measure of disorder (randomness).
Examples: Opening a small vail of gas in a room allows the gas to disperse throughout the room. The gas has now become more disordered - it is much harder to localize the positions of the gas particles in the larger space. The gas in the large space has more entropy. It is much easier to describe the positions of the ions in a small crystal than when they are dissolved into a large volume of water. The ions dissolved in water have more entropy.
Aside: What about processes which lead to more ordered systems? For example babies eat some pretty disordered looking stuff (pureed food), but grow into highly organized beings. Is the Second Law being violated? No. Although the material incorporated into the baby is more ordered than its food, most of the food was transformed into carbon dioxide and water which dispersed over the world's atmosphere - it has become MUCH more disorganized. As a result the Universe has increased in entropy. Similarly, though even less obviously, plants organize small molecules at the expense, ultimately, of the Sun's dispersal over the Universe.
Another way of describing entropy is to describe the number of arrangements (microstates) available to a system. (Example on p787 of text.)
If the entropy increase of the Universe is the ultimate driving force for reactions, what's going on? There must be a relationship between deltaH in the system and the entropy of the Universe. Apparently the transfer of heat affects entropy.
Let's look at how this might occur. When heat is transferred to the surroundings the result is an increase in the temperature, T, of the surroundings. As a result the KE and thus motion of the particles is increased, which in turn means their distribution is harder to describe - it has become more random. Entropy in the surroundings has increased! So heat and entropy are indeed connected when we look at the entire system.
We must consider the system and its surroundings to determine the overall change in entropy and thus the spontaneity of a given process. In most cases we need only analyze the local surroundings to understand what is happening, though ultimately it is the entire Universe which determines spontaneity.
If we look at an exothermic reaction as an example, the reaction mixture loses bond energy as enthalpy, while the surroundings gain heat which increases its entropy. The entropy increase from the enthalpy of the reaction results in a net increase in the entropy of the Universe and thus favors the reaction. Because of this enthalpic entropy increase most exothermic reactions are favored, however in some reactions the entropy of the reaction decreases sufficiently to make them non-spontaneous.
To make things even more complex, the net entropy change, and therefore spontaneity of a reaction depends also on the temperature. To understand the temperature dependence, consider the vaporization of water as an example:
Why should such a change occur with temperature? At either side of 100°C deltaH will be about the same as will deltaS, within the system.
Apparently the change must be due to changes in the surroundings. Let's look at an exothermic process because its a bit easier to follow. Essentially, the fractional change in entropy in the surroundings due to enthalpy will be greater at low temperatures. Generally we can therefore say:
Thus deltaSsur depends on heat/temperature (with temperature in Kelvins):
At constant pressure we use deltaH for heat exchange. We also generally look at the enthalpy and entropy of the system, so an exothermic process, for which the deltaH of the surroundings is +, the deltaH of the system is -, and for endothermic deltaH is +. Thus
Note that the (-) sign occurs because we are looking at the surroundings, but deltaH is defined for the system:
-deltaHsys = +deltaHsur
We can now begin to see why the vaporization of water changes in spontaneity with T:
Note that this represents a continuum - that is the % liquid vs. vapor changes continuously as the vapor pressure increases, until at 100°C the vapor pressure = atmospheric pressure and all of the water evaporates.
To predict whether a process is spontaneous then, we need to determine the deltaS of the Universe. This is not always convenient however, so another expression is more often used in chemistry and the life sciences, the "free energy" = G, defined as:
The free energy is particularly useful because it tells us the energy from a process available to do useful work at constant pressure - often an important consideration in life!
For a process occurring at constant pressure then,
Thus a negative deltaG must correspond to to an increase in the entropy of the Universe. This can be seen if we divide both sides of the free energy expression by T:
but we showed earlier that:
Thus a process with a negative deltaG is spontaneous!
Discussion of deltaG vs. negative and positive values of deltaH and deltaS.
First let's review some important aspects of free energy.
As an example of picking conditions to solve a problem let's look at a variation of an example in your text.
Find the entropy of vaporization of Br2 given the enthalpy of vaporization, deltaH, and a table of vapor pressure versus T. deltaH = 3.1 x 104J/mol. So how do we approach this problem?
- Write the equation:
deltaG = deltaH -TdeltaS
-TdeltaS = deltaG - deltaH
TdeltaS = deltaH - deltaG
deltaS = (deltaH - deltaG)/T
We know deltaH and T, but what about deltaG? Let's think about the process. Remember that deltaG tells us about spontaneity. Is there a condition where the system has no net tendency to either vaporize or condense? If the system doesn't tend either direction then deltaG must be 0! This occurs in any system at equilibrium. An obvious equilibrium position is at the boiling point where the vapor pressure = atmospheric pressure. So all we need to do is look up the boiling point and set our calculations to the appropriate temperature! T = 333K.
- We can now place values into our equation:
deltaS = (deltaH - deltaG)/T = (3.1 x 104J/mol - 0)/333K
deltaS = 93 JK-1mol-1
© R A Paselk
Last modified 5 June 2006