The Total Entropy Change

If we are now to remove this entropy from the system - i.e. if we want Sg to equal - then during the spontaneous process an appropriate amount of heat must be transferred to the surroundings. This, of course, will result in an increase of the entropy of the surroundings. On the other hand, it is also apparent from our discussion that when the process A to B is reversible, there will be no entropy change for the system or the surroundings. 

We conclude, therefore, that the total entropy change AS,g, - that of the system (AS ) plus that of the surroundings (AS ) - must be positive for any natural (spontaneous) process  

1 represents the formal statement of the second law, the equal sign referring to the case of a reversible process, and provides the desired criterion that establishes  

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This expression gives the average entropy change per chain to get the average for the sample, we multiply by the number v of subchains in the sample. The total entropy change is... 

Because the engine operates in cycles, it experiences no change in its own properties therefore the total entropy change of the engine and its associated heat reservoirs is given by equation 8 ... 

Notice that the second law refers to the total entropy change, involving both system and surroundings. For many spontaneous processes, the entropy change for the system is a negative quantity. Consider, for example, the rusting of iron, a spontaneous process ... 

The system ot interest and its surroundings constitute the isolated system to which the second law refers (Fig. 7.15). Only if the total entropy change,... 

By considering the total entropy change, we can draw some far-reaching conclusions about processes going on in the universe. For instance, we saw in Section 6.3 that maximum work is achieved if expansion takes place reversibly, by matching the... 

EXAMPLE 7.12 Calculating the total entropy change for the expansion of an ideal gas... 

This equation shows how to calculate the total entropy change from information about the system alone. The limitation is that the equation is valid only at constant temperature and pressure. 

What is the total entropy change when 2.00 mol of water (A =6.01 kJ/mol) freezes at 0.0 °C in a freezer compartment whose temperature is held at -15 °C ... 

The problem asks for the total entropy change, which includes A S for the water and A S for the freezer compartment. When water freezes, heat flows from the water to its surroundings, the freezer compartment (see Eigure 14-61. Thus, q is negative for the water, whose entropy decreases. At the same time, q is positive for the freezer compartment, whose entropy increases. 

Equation gives the total entropy change A — A jS ystem + A Fjmygmjgjggs We can define the HFC as the system and the contents of the refrigerator as the immediate surroundings ... 

C14-0005. Calculate the total entropy change when 175 g of water at 0 °C freezes, transferring heat that causes HFC-134a to boil at -27 °C. 

In words, in any process that occurs at constant T and P, the free energy change for the system is negative whenever the total entropy change is positive that is, whenever the overall process is spontaneous. Defining a new function and imposing some restrictions provides a way to use properties of a system to determine whether a process is... 

If the configurational entropy A>Sm is assumed to represent the total entropy change LSm on mixing, the free energy of mixing is simply obtained by combining Eqs. (10) and (20). That is,... 

As a first approximation, assume that the total entropy change, ASy, is made up solely of configurational entropy. This can be determined by using the Boltzmann equation for the entropy of a disordered system ... 

If ASsurroundings is sufficiently positive to make the total entropy change positive, the process will be spontaneous. 

The second law of thermodynamics defines the conditions for a feasible reaction. It states that for a reaction to be feasible, the total entropy change for a reaction system and its surroundings must be positive, that is ... 

Similar complicated possibilities exist in interpreting 6S data, for one must consider the total entropy change for the reaction, including reactants, products, and solvent. 

The total pair-wise entanglement contribution is simply the sum of dS for strand pairs with all possible sets of internal coordinates. In a network of N strands there will be N(N — l)/2 such pairs. However, classification is only significant for pairs that are relatively close. Pairs separated by more than a few radii of gyration will belong to unentangled class exclusively. This assures that the total entropy change will be proportional only to N. 

A consideration of these relationships reveals8 that because E° is a thermodynamic parameter and represents an energy difference between two oxidation states and in many cases the spectroscopic or other parameter refers to only one half of the couple, then some special conditions must exist in order for these relationships to work. The special conditions under which these relationships work are that (a) steric effects are either unimportant or approximately the same in both halves of the redox couple and (b) changes in E° and the spectroscopic or other parameters arise mainly through electronic effects. The existence of many examples of these relationships for series of closely related complexes is perhaps not too unexpected as it is likely that, for such a series, the solvational contribution to the enthalpy change, and the total entropy change, for the redox reaction will remain constant, thus giving rise to the above necessary conditions. 

It may be necessary to transfer a quantity of heat, dQ, from the reservoir into the system in order to maintain constant temperature in the system the total entropy change of the system plus reservoir, dS, will then be... 

We can see from Table 7.2 that at 0°C the molar entropy of liquid water is 22.0 J-K -mol 1 higher than that of ice at the same temperature. This difference makes sense, because the molecules in liquid water are more disordered than in ice. It follows that when water freezes at 0°C, its entropy decreases by 22.0 J-K -mol-1. Entropy changes do not vary much with temperature so just below 0°C, we can expect almost the same decrease. Yet we know from everyday experience that water freezes spontaneously below 0°C. Clearly, the surroundings must be playing a deciding role if we can show that their entropy increases by more than 22.0 J-K -mol 1 when water freezes, then the total entropy change will be positive and freezing will be spontaneous. 

EXAMPLE 7.10 Determining the total entropy change for a process... 

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