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The Second Law of Thermodynamics
Δ S univ >0 spontaneous
Δ S univ <0 nonspontaneous (spontaneous in opposite direction)
Δ S univ = 0 reversible (system is at equilibrium)

For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, q surr is a good approximation of q rev , and the second law may be stated as the following:

Δ S univ = Δ S sys + Δ S surr = Δ S sys + q surr T

We may use this equation to predict the spontaneity of a process as illustrated in [link] .

Will ice spontaneously melt?

The entropy change for the process

H 2 O ( s ) H 2 O ( l )

is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?

Solution

We can assess the spontaneity of the process by calculating the entropy change of the universe. If Δ S univ is positive, then the process is spontaneous. At both temperatures, Δ S sys = 22.1 J/K and q surr = −6.00 kJ.

At −10.00 °C (263.15 K), the following is true:

Δ S univ = Δ S sys + Δ S surr = Δ S sys + q surr T = 22.1 J/K + −6.00 × 10 3 J 263.15 K = −0.7 J/K

S univ <0, so melting is nonspontaneous ( not spontaneous) at −10.0 °C.

At 10.00 °C (283.15 K), the following is true:

Δ S univ = Δ S sys + q surr T = 22.1 J/K + −6.00 × 10 3 J 283.15 K = +0.9 J/K

S univ >0, so melting is spontaneous at 10.00 °C.

Check your learning

Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of S univ ?

Answer:

Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.

Got questions? Get instant answers now!

The third law of thermodynamics

The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal ( W = 1). According to the Boltzmann equation, the entropy of this system is zero.

S = k ln W = k ln ( 1 ) = 0

This limiting condition for a system’s entropy represents the third law of thermodynamics    : the entropy of a pure, perfect crystalline substance at 0 K is zero.

We can make careful calorimetric measurements to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies are given the label S 298 ° for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K. The standard entropy change (Δ S °)    for any process may be computed from the standard entropies of its reactant and product species like the following:

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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