At the beginning of this chapter, we introduced the concept of a spontaneous process: a process that occurs naturally without the need for continuous external intervention.
Consider the following two processes:
\[ NaOH(s) \rightarrow Na^+(aq)+OH^-(aq) \qquad \Delta H < 0 \]
\[ NaHCO_3(s) \rightarrow Na^+(aq)+HCO_3^-(aq) \qquad \Delta H > 0 \]
Both processes occur spontaneously, yet one is exothermic and the other is endothermic. Clearly, enthalpy alone cannot determine whether a process is spontaneous.
We therefore require a reliable criterion of spontaneity, which \( \Delta H \) does not provide.
The entropy change of a process can be divided into two contributions:
Together, these contributions determine the entropy change of the universe:
\[ \boxed{ \Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{sys}} + \Delta S_{\mathrm{surr}} } \]
This relationship allows us to examine not only what happens within a system, but also the effect of that process on everything else around it.
The Carnot cycle suggested that entropy behaves as a state function. Clausius took this idea one step further and proposed that the Second Law of Thermodynamics could be expressed in terms of entropy.
The result is remarkably simple:
\[ \boxed{ \Delta S_{\mathrm{univ}} > 0 } \]
for any spontaneous process.
Likewise,
\[ \boxed{ \Delta S_{\mathrm{univ}} = 0 } \]
for a reversible process.
And
\[ \boxed{ \Delta S_{\mathrm{univ}} < 0 } \]
corresponds to a process that cannot occur spontaneously.
The entropy criterion is more powerful than simply examining heat flow or enthalpy changes.
A process may increase the entropy of the system while decreasing the entropy of the surroundings, or vice versa. The determining factor is the net effect on the universe:
\[ \Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{sys}} + \Delta S_{\mathrm{surr}} \]
If the total is positive, the process is spontaneous.
If the total is negative, the process is not spontaneous.
Thus, entropy provides the criterion for spontaneity that enthalpy alone cannot provide.
One of the most profound consequences of the Second Law is that it gives a preferred direction to natural processes.
Ice melts in a warm room, but liquid water does not spontaneously assemble itself into ice at room temperature. Heat flows naturally from hot objects to cold objects, but not in the reverse direction.
These processes occur in the direction that increases the entropy of the universe.
In this sense, the Second Law provides a thermodynamic explanation for the arrow of time: the distinction between past and future arises because spontaneous processes proceed in the direction of increasing entropy.
Big picture: Entropy is not merely another thermodynamic state function. The Second Law elevates entropy to a special role by making it the criterion for spontaneity. A process is spontaneous if it increases the entropy of the universe, and this principle governs the natural direction of change throughout the physical world.
Consider \(10.0\ \mathrm{g}\) of ice at \(0^\circ\mathrm{C}\) melting to form liquid water at \(0^\circ\mathrm{C}\) in a room at \(25^\circ\mathrm{C}\). Calculate \(\Delta S_{\mathrm{H_2O}}\), \(\Delta S_{\mathrm{surr}}\), and \(\Delta S_{\mathrm{univ}}\).
Use
\[ \Delta H_{\mathrm{fus}} = 6.01\ \mathrm{kJ\,mol^{-1}} \]
First calculate the number of moles of water:
\[ n = \frac{10.0\ \mathrm{g}} {18.02\ \mathrm{g\,mol^{-1}}} = 0.555\ \mathrm{mol} \]
The heat absorbed by the ice as it melts is
\[ q_{\mathrm{H_2O}} = n\Delta H_{\mathrm{fus}} = (0.555\ \mathrm{mol}) (6.01\ \mathrm{kJ\,mol^{-1}}) = 3.34\ \mathrm{kJ} \]
The phase change occurs at the melting temperature of water:
\[ T_{\mathrm{melt}} = 273.15\ \mathrm{K} \]
Therefore,
\[ \Delta S_{\mathrm{H_2O}} = \frac{q_{\mathrm{H_2O}}} {T_{\mathrm{melt}}} = \frac{3.34\times10^3\ \mathrm{J}} {273.15\ \mathrm{K}} \]
\[ \Delta S_{\mathrm{H_2O}} = +12.2\ \mathrm{J\,K^{-1}} \]
The surroundings lose the same amount of heat that the ice absorbs:
\[ q_{\mathrm{surr}} = -3.34\times10^3\ \mathrm{J} \]
The room acts as a large thermal reservoir at \(25^\circ\mathrm{C}=298.15\ \mathrm{K}\), so
\[ \Delta S_{\mathrm{surr}} = \frac{q_{\mathrm{surr}}} {T_{\mathrm{surr}}} = \frac{-3.34\times10^3\ \mathrm{J}} {298.15\ \mathrm{K}} \]
\[ \Delta S_{\mathrm{surr}} = -11.2\ \mathrm{J\,K^{-1}} \]
\[ \Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{H_2O}} + \Delta S_{\mathrm{surr}} \]
\[ \Delta S_{\mathrm{univ}} = 12.2\ \mathrm{J\,K^{-1}} - 11.2\ \mathrm{J\,K^{-1}} \]
\[ \boxed{ \Delta S_{\mathrm{univ}} = +1.0\ \mathrm{J\,K^{-1}} } \]
Interpretation: The entropy of the water increases because liquid water has more accessible molecular arrangements than ice. The entropy of the surroundings decreases because the room gives up heat. However, the increase in the entropy of the water is larger than the decrease in the entropy of the surroundings, so \(\Delta S_{\mathrm{univ}}>0\). Therefore, the melting process is spontaneous.
Consider \(10.0\ \mathrm{g}\) of liquid water at \(0^\circ\mathrm{C}\) warming to \(25^\circ\mathrm{C}\) in a room at \(25^\circ\mathrm{C}\). Calculate \(\Delta S_{\mathrm{H_2O}}\), \(\Delta S_{\mathrm{surr}}\), and \(\Delta S_{\mathrm{univ}}\).
Use
\[ C_p = 75.38\ \mathrm{J\,mol^{-1}\,K^{-1}} \]
First calculate the number of moles of water:
\[ n = \frac{10.0\ \mathrm{g}}{18.02\ \mathrm{g\,mol^{-1}}} = 0.555\ \mathrm{mol} \]
The water warms from \(T_1=273.15\ \mathrm{K}\) to \(T_2=298.15\ \mathrm{K}\), so
\[ \Delta S_{\mathrm{H_2O}} = nC_p\ln\left(\frac{T_2}{T_1}\right) \]
\[ \Delta S_{\mathrm{H_2O}} = (0.555\ \mathrm{mol}) (75.38\ \mathrm{J\,mol^{-1}\,K^{-1}}) \ln\left(\frac{298.15}{273.15}\right) \]
\[ \Delta S_{\mathrm{H_2O}} = +3.67\ \mathrm{J\,K^{-1}} \]
The heat absorbed by the water is
\[ q_{\mathrm{H_2O}} = nC_p\Delta T = (0.555\ \mathrm{mol}) (75.38\ \mathrm{J\,mol^{-1}\,K^{-1}}) (25.00\ \mathrm{K}) \]
\[ q_{\mathrm{H_2O}} = 1.05\times10^3\ \mathrm{J} \]
Therefore, the room loses this heat:
\[ q_{\mathrm{surr}} = -1.05\times10^3\ \mathrm{J} \]
Since the room is a large thermal reservoir at \(298.15\ \mathrm{K}\),
\[ \Delta S_{\mathrm{surr}} = \frac{q_{\mathrm{surr}}}{T_{\mathrm{surr}}} = \frac{-1.05\times10^3\ \mathrm{J}}{298.15\ \mathrm{K}} \]
\[ \Delta S_{\mathrm{surr}} = -3.51\ \mathrm{J\,K^{-1}} \]
\[ \Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{H_2O}} + \Delta S_{\mathrm{surr}} \]
\[ \Delta S_{\mathrm{univ}} = 3.67\ \mathrm{J\,K^{-1}} - 3.51\ \mathrm{J\,K^{-1}} \]
\[ \boxed{ \Delta S_{\mathrm{univ}} = +0.16\ \mathrm{J\,K^{-1}} } \]
Interpretation: The water gains entropy as it warms because its thermal energy becomes more dispersed. The surroundings lose entropy because the room transfers heat to the colder water. Since the entropy gained by the water is slightly larger than the entropy lost by the surroundings, \(\Delta S_{\mathrm{univ}}>0\), so the warming process is spontaneous.
Consider \(1.00\ \mathrm{mol}\) of a monatomic ideal gas \(\left(C_V=\frac{3}{2}R\right)\) initially at \(10.0\ \mathrm{L}\) and \(298\ \mathrm{K}\). The gas expands isothermally against a constant external pressure of \(1.00\ \mathrm{atm}\) until its volume reaches \(24.4\ \mathrm{L}\).
Calculate \(\Delta S_{\mathrm{sys}}\), \(\Delta S_{\mathrm{surr}}\), and \(\Delta S_{\mathrm{univ}}\).
Entropy is a state function, so we may calculate \(\Delta S_{\mathrm{sys}}\) using any convenient reversible path connecting the same initial and final states.
For an ideal gas undergoing an isothermal change:
\[ \Delta S_{\mathrm{sys}} = nR \ln\left( \frac{V_2}{V_1} \right) \]
\[ \Delta S_{\mathrm{sys}} = (1.00)(8.314) \ln\left( \frac{24.4}{10.0} \right) \]
\[ \Delta S_{\mathrm{sys}} = +7.42\ \mathrm{J\,K^{-1}} \]
The work performed by the gas is
\[ w = -P_{\mathrm{ext}}\Delta V \]
\[ w = -(1.00\ \mathrm{atm}) (24.4-10.0\ \mathrm{L}) \]
\[ w = -14.4\ \mathrm{L\,atm} \]
\[ w = -1.46\times10^3\ \mathrm{J} \]
Since the process is isothermal for an ideal gas,
\[ \Delta U = 0 \]
and therefore
\[ q_{\mathrm{sys}} = -w = +1.46\times10^3\ \mathrm{J} \]
The surroundings lose the same amount of heat:
\[ q_{\mathrm{surr}} = -1.46\times10^3\ \mathrm{J} \]
Assuming the surroundings remain at \(298\ \mathrm{K}\),
\[ \Delta S_{\mathrm{surr}} = \frac{q_{\mathrm{surr}}} {T} \]
\[ \Delta S_{\mathrm{surr}} = \frac{-1.46\times10^3} {298} \]
\[ \Delta S_{\mathrm{surr}} = -4.90\ \mathrm{J\,K^{-1}} \]
\[ \Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{sys}} + \Delta S_{\mathrm{surr}} \]
\[ \Delta S_{\mathrm{univ}} = 7.42 - 4.90 \]
\[ \boxed{ \Delta S_{\mathrm{univ}} = +2.52\ \mathrm{J\,K^{-1}} } \]
Interpretation: The gas expands spontaneously because its entropy increases substantially. Although the surroundings lose entropy as they supply heat to the gas, the entropy gained by the gas is larger than the entropy lost by the surroundings. The result is \(\Delta S_{\mathrm{univ}}>0\), which is the criterion for a spontaneous process.
A problem will be chosen at random. Select the correct values for \(\Delta S_{\mathrm{sys}}\), \(\Delta S_{\mathrm{surr}}\), and \(\Delta S_{\mathrm{univ}}\).
| Quantity | Your answer |
|---|---|
| \(\Delta S_{\mathrm{sys}}\) | |
| \(\Delta S_{\mathrm{surr}}\) | |
| \(\Delta S_{\mathrm{univ}}\) |
Big picture: Entropy provides the criterion for spontaneity that enthalpy alone cannot provide. A process is spontaneous if it increases the entropy of the universe. This principle explains why ice melts in a warm room, why heat flows from hot objects to cold objects, and ultimately why natural processes exhibit a preferred direction in time.