Spontaneous Processes
Processes can be spontaneous and nonspontaneous. Spontaneous process is a process that occurs without any ongoing outside intervention while nonspontaneous process needs outside intervention to occur. Below are some examples of spontaneous processes:- A waterfall runs downhill, but never up, spontaneously.
- A lump of sugar spontaneously dissolves in a cup of coffee.
- Water freezes spontaneously below 0oC, and ice melts spontaneously above 0oC.
- Heat flows from a hotter object to a colder one.
- A shiny nail left outdoors will eventually rust.
An spontaneous process has a difinite direction in which it occurs. The spontaneity of the process can depend on temperature. For example, the endothermic process of melting ice under atmospheric pressure. When T > 0oC, ice melts spontaneously; the reverse process, liquid water turning ice at this temperature is not spontaneous. However when T < 0oC, liquid water converts into ice spontaneously and at this temperature converting ice into water is not spontaneous. When T = 0oC, the normal boiling point of water, the solid and liquid phases are in equilibrium.
Example:
Predict whether the following processes are spontaneous, nonspontaneous and at equilibrium.
a. When a piece of metal heated to 150oC is added to water at 40oC, the water gets hotter.
b. Water at room temperature decomposes into H2(g) and O2(g).
Answer:
a. The process is spontaneous, heat is always transferred from a hotter object to a colder one.
b. The process is nonspontaneous, at room temperature the reverse occurs.
Reversible and Irreversible Processes
A reversible process is a process in which the state of a system can change. The change in the system can be restored to its original state by exactly reversing the change. Example is the interconversion of ice and water, at T = 0oC the process is at equilibrium, now decreasing the temperature, liquid water will turn into ice and increasing the temperature will melt the ice into water.
An irreversible process is one that cannot be simply reversed to restore the system and its
surroundings to their original state. When a system changes by an irreversible process, it must take a different path ( with a different value of q and w) to get back to its original state.
There are two very important concepts regarding reversible and irreversible processes:
1. Whenever a chemical system is at equilibrium, reactants and products can inter-convert reversibly.
2. In any spontaneous process, the path between reactants and products is irreversible
An irreversible process is one that cannot be simply reversed to restore the system and its
surroundings to their original state. When a system changes by an irreversible process, it must take a different path ( with a different value of q and w) to get back to its original state.
There are two very important concepts regarding reversible and irreversible processes:
1. Whenever a chemical system is at equilibrium, reactants and products can inter-convert reversibly.
2. In any spontaneous process, the path between reactants and products is irreversible
Entropy
Entropy is open described as the measure of disorder or randomness of the system. The more disorder or random the system, the larger its entropy. Like internal energy and enthalpy, entropy is also a state function. The change in the entropy of the system, ∆S = Sfinal - Sinitial, depends only on the initial and final states of the system and not on the particular pathway by which the system changes. A positive value of ∆S indicates that the final state is more disordered than the initial state. Example when gas expands to a large volume , its entropy increases. A negative value of ∆S indicates that the final state is more disordered or less random than initial state.
Example 1
Predict whether the entropy increases or decreases for the following processes:
a. freezing ethanol
b. evaporating a beaker of liquid bromine at room temperature
c. dissolving glucose in water
d. cooling a nitrogen gas from 80oC to 20oC.
Answer:
a. Entropy decreases,the ethanol molecules are held together rigidly.
b. Entropy increases, bromine gas molecules occupy more positions in nearly empty space.
c. Entropy increases, glucose is dissolved in water and so the more disorder the molecules.
d. Entropy decreases, decreasing temperature decreases the molecular motion of molecules.
Example 2
By considering the disorder of the products, predict whether the ∆S is positive or negative for the following processes:
a. H2O(l) → H2O(g)
b. Ag+(aq) + Cl-(aq) → AgCl(s)
c. 4Fe(s) + 3O2(g) → 2Fe2O3(s)
d. CO2(s) → CO2(g)
Answer:
a. Entropy is positive there is a large increase in volume.
b. Entropy is negative, ions are confined in more ordered positions.
c. Entropy is negative, molecules are converted to solid, solid is more ordered than gas.
d. Entropy is positive molecules becomes more disordered in gas form.
∆Suniv = ∆Ssys + ∆Ssurr
In terms of ∆Suniv, the second law can be expressed as follows: In any reversible process, ∆Suniv = 0, whereas in any irreversible (spontaneous) process, ∆Suniv > 0.
Reversible process (Equilibrium) : ∆Suniv = ∆Ssys + ∆Ssurr = 0
Irreversible process (Spontaneous) : ∆Suniv = ∆Ssys + ∆Ssurr > 0
For a spontaneous process, the second law says that ∆Suniv must be greater than zero, but it does not restrict ∆Ssys and ∆Ssurr. Thus, it is possible that ∆Ssys and ∆Ssurr to be positive or negative provided the sum of two quantities must be greater than zero. For an equilibrium process, the ∆Suniv must be equal to zero, this means that the value of ∆Ssys and ∆Ssurr must have the same magnitude but opposite in sign.
A special case of the second law concerns the entropy change in an isolated system, one that doesn't exchange energy or matter with its surroundings. In any process that occurs in an isolated system leaves the surroundings completely unchanged. Therefore, ∆Ssurr = 0 for such process. Thus for a special case of an isolated system, second law becomes:
Reversible process (Equilibrium) : ∆Suniv = 0
Irreversible process (Spontaneous) ; ∆Suniv > 0
Example 1
Predict whether the entropy increases or decreases for the following processes:
a. freezing ethanol
b. evaporating a beaker of liquid bromine at room temperature
c. dissolving glucose in water
d. cooling a nitrogen gas from 80oC to 20oC.
Answer:
a. Entropy decreases,the ethanol molecules are held together rigidly.
b. Entropy increases, bromine gas molecules occupy more positions in nearly empty space.
c. Entropy increases, glucose is dissolved in water and so the more disorder the molecules.
d. Entropy decreases, decreasing temperature decreases the molecular motion of molecules.
Example 2
By considering the disorder of the products, predict whether the ∆S is positive or negative for the following processes:
a. H2O(l) → H2O(g)
b. Ag+(aq) + Cl-(aq) → AgCl(s)
c. 4Fe(s) + 3O2(g) → 2Fe2O3(s)
d. CO2(s) → CO2(g)
Answer:
a. Entropy is positive there is a large increase in volume.
b. Entropy is negative, ions are confined in more ordered positions.
c. Entropy is negative, molecules are converted to solid, solid is more ordered than gas.
d. Entropy is positive molecules becomes more disordered in gas form.
Second Law of Thermodynamics
Second law of thermodynamics relates the spontaneity of reaction to entropy. The entropy of the universe increases in a spontaneous process and remain unchanged in an equilibrium process. We must both consider the entropy of the system and the surroundings. The total change in the entropy, called the change in the entropy of the universe and denoted ∆Suniv, is the sum of the changes in the entropy of the system, ∆Ssys, and of the surrounding, ∆Ssurr.∆Suniv = ∆Ssys + ∆Ssurr
In terms of ∆Suniv, the second law can be expressed as follows: In any reversible process, ∆Suniv = 0, whereas in any irreversible (spontaneous) process, ∆Suniv > 0.
Reversible process (Equilibrium) : ∆Suniv = ∆Ssys + ∆Ssurr = 0
Irreversible process (Spontaneous) : ∆Suniv = ∆Ssys + ∆Ssurr > 0
For a spontaneous process, the second law says that ∆Suniv must be greater than zero, but it does not restrict ∆Ssys and ∆Ssurr. Thus, it is possible that ∆Ssys and ∆Ssurr to be positive or negative provided the sum of two quantities must be greater than zero. For an equilibrium process, the ∆Suniv must be equal to zero, this means that the value of ∆Ssys and ∆Ssurr must have the same magnitude but opposite in sign.
A special case of the second law concerns the entropy change in an isolated system, one that doesn't exchange energy or matter with its surroundings. In any process that occurs in an isolated system leaves the surroundings completely unchanged. Therefore, ∆Ssurr = 0 for such process. Thus for a special case of an isolated system, second law becomes:
Reversible process (Equilibrium) : ∆Suniv = 0
Irreversible process (Spontaneous) ; ∆Suniv > 0
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