Question

The value of $\Delta S$ for spontaneous process is

Answer 1

A spontaneous reaction is a reaction that favors the formation of products at the condition under which the reaction is occurring. A roaring bonfire is an example of spontaneous process. According to the second law of Thermodynamics, All spontaneous changes cause an increase in the entropy in the universe.

Complete answer:
In thermodynamic models the system and surrounding comprise of everything, that is, the universe, and so the following is true
$:\Delta {S_{univ}} = \Delta {S_{sys}} + \Delta {S_{surr}} - - - - - (1)$
The process of heat flow between two objects, one identified as the system and the other as the surrounding. There are three possibilities for such a process:
$1.$ The objects are at different temperatures, and heat flows from the hotter to the cooler object. This is always observed to occur spontaneously. Hotter object as the system and invoking the definition of entropy yields the following
$:\Delta {S_{sys}} = \dfrac{{ - {q_{rev}}}}{{{T_{sys}}}}$ and $\Delta {S_{surr}} = \dfrac{{{q_{rev}}}}{{{T_{surr}}}} - - - - - (2)$
The sign ${q_{rev}}$ denotes the loss of heat by the system and the gain of heat by the surroundings. Since ${T_{sys}} > {T_{surr}}$ In this scenario, the magnitude of the entropy change for the surroundings will be greater than that for the system, and so the sum of $\Delta {S_{sys}}$ and $\Delta {S_{surr}}$ will yield a positive value for $\Delta {S_{univ}}$. This process involves the increase in the entropy of the universe.
$2.$ The objects are at different temperatures, and heat flows from the cooler to the hotter object. This is never observed to occur spontaneously. Again the hotter object as the system and invoking the definition of entropy yields the following
$:\Delta {S_{sys}} = \dfrac{{{q_{rev}}}}{{{T_{sys}}}}$ and $\Delta {S_{surr}} = \dfrac{{ - {q_{rev}}}}{{{T_{surr}}}} - - - - - - (3)$
The sign of ${q_{rev}}$ denotes the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than for the system, but in the case, the signs of the heat changes will yields a negative value for $\Delta {S_{univ}}$ .this process involves a decrease in the entropy of the universe.
$3.$ The temperature difference between the objects is infinitesimally small, ${T_{sys}} \approx {T_{surr}}$ , and so the heat flow is thermodynamically reversible. The system and surroundings experience entropy changes that are equal in magnitude and therefore sum to yield a value of zero for $\Delta {S_{univ}}$ . This process involves no change in the entropy of the universe.
This result leads to a relation between entropy and spontaneity known as the second law of thermodynamics. The summary of these relation says,
$\Delta {S_{univ}} > 0$ , for spontaneous process
$\Delta {S_{univ}} < 0$ , nonspontaneous (spontaneous in opposite direction)
$\Delta {S_{univ}} = 0$ , reversible (system is at equilibrium)

Note:
Processes that involve an increase in entropy of the system $:$ are very often spontaneous.
Examples of spontaneous process $:$ A fire is exothermic, which means a decrease in the energy of the system as energy is released to the surrounding as heat. The products of a fire are composed mostly of gases such as carbon dioxide and water vapors, so the entropy of the system increases during most combustion reactions. This combination of a decrease in energy and an increase in entropy means that combustion reactions occur spontaneously.