
For chemical reactions, the calculation of change in entropy is normally done
A. At constant pressure
B. At constant temperature
C. At constant temperature and pressure both
D. At constant volume
Answer
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Hint: To calculate the entropy change in a chemical reaction, we have to make use of the standard entropies of the reactants involved and products formed in the reaction. Knowing the definition of the standard entropy of a substance will help us answer this question.
Complete Step by Step Solution:
Entropy is one of the most confusing concepts in thermodynamics. Although multiple descriptions of it exist, for our purposes it can be described as a measure of the degree to which energy (such as thermal energy) becomes spread out or dispersed within a system. If the energy of a particular system is spread out or dispersed among its constituent particles (atoms or molecules) to a greater extent, its entropy is high.
In a chemical reaction, molecules of different species react with each other to form molecules of entirely new species. A system of molecules of a particular species can disperse energy through it to a certain extent at a given temperature and pressure i.e., it has a certain, fixed value of entropy at a given temperature and pressure. The entropy of a system will change if the temperature and pressure on the system changes.
The entropy of one mole of a pure substance measured at 1 atm pressure and 25\[^\circ C\]is called the standard entropy of the substance. It is denoted as\[S^\circ \] .
Consider a chemical reaction \[aA + bB + cC + ... \to pP + qQ + rR + ...\] with each reactant and product being in its pure state. The change in entropy of this reaction will be given as:
\[\Delta S^\circ = [pS{^\circ _P} + qS{^\circ _Q} + rS{^\circ _R} + ...] - [aS{^\circ _A} + bS{^\circ _B} + cS{^\circ _C} + ...]\]
\[ \Rightarrow \Delta S^\circ = \sum {S{^\circ _{products}} - \sum {S{^\circ _{reac\tan ts}}} } \] where a,b,c,p,q,… are the stoichiometric coefficients.
Since the standard entropies of substances are determined at a constant temperature and pressure, the change in entropy of a reaction is also calculated at constant temperatures and pressures.
Thus, option C is correct.
Note: When the temperature of the system increases, the thermal energy of the system increases as well. This excites the particles of the system, and their random motion is amplified. This enables a greater degree of dispersion of the thermal energy. Hence, the entropy increases. When the pressure on the system increases, it compresses the particles of the system into a smaller space which reduces its random motion and, hence, its ability to disperse energy. Thus, entropy decreases.
Complete Step by Step Solution:
Entropy is one of the most confusing concepts in thermodynamics. Although multiple descriptions of it exist, for our purposes it can be described as a measure of the degree to which energy (such as thermal energy) becomes spread out or dispersed within a system. If the energy of a particular system is spread out or dispersed among its constituent particles (atoms or molecules) to a greater extent, its entropy is high.
In a chemical reaction, molecules of different species react with each other to form molecules of entirely new species. A system of molecules of a particular species can disperse energy through it to a certain extent at a given temperature and pressure i.e., it has a certain, fixed value of entropy at a given temperature and pressure. The entropy of a system will change if the temperature and pressure on the system changes.
The entropy of one mole of a pure substance measured at 1 atm pressure and 25\[^\circ C\]is called the standard entropy of the substance. It is denoted as\[S^\circ \] .
Consider a chemical reaction \[aA + bB + cC + ... \to pP + qQ + rR + ...\] with each reactant and product being in its pure state. The change in entropy of this reaction will be given as:
\[\Delta S^\circ = [pS{^\circ _P} + qS{^\circ _Q} + rS{^\circ _R} + ...] - [aS{^\circ _A} + bS{^\circ _B} + cS{^\circ _C} + ...]\]
\[ \Rightarrow \Delta S^\circ = \sum {S{^\circ _{products}} - \sum {S{^\circ _{reac\tan ts}}} } \] where a,b,c,p,q,… are the stoichiometric coefficients.
Since the standard entropies of substances are determined at a constant temperature and pressure, the change in entropy of a reaction is also calculated at constant temperatures and pressures.
Thus, option C is correct.
Note: When the temperature of the system increases, the thermal energy of the system increases as well. This excites the particles of the system, and their random motion is amplified. This enables a greater degree of dispersion of the thermal energy. Hence, the entropy increases. When the pressure on the system increases, it compresses the particles of the system into a smaller space which reduces its random motion and, hence, its ability to disperse energy. Thus, entropy decreases.
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