Change In Entropy And Spontaneity Of A Process
We have seen that there are many spontaneous processes (like the melting of ice above 0°C and 1 atm) in which the entropy ofthe system increases (ASurr > 0). Again, there are some spontaneous processes (such as the transformation of water into ice below 0°C and 1 atm) in which the entropy of the system decreases (Δ < 0).
Hence, the change in entropy of the system alone cannot predict the spontaneity of a process; instead, we must consider the change in entropy of the system as well as that of the surroundings. In a process, if the change in entropy of the system and its surroundings are AS and ASsurr respectively, then the total change in entropy in the process
⇒ \(\Delta S_{\text {total }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}\)
The system and its surroundings constitute the universe. So,
⇒ \(\Delta S_{\text {total }}=\Delta S_{\text {univ }}=\Delta S_{s y s}+\Delta S_{\text {surr }}\)
All spontaneous processes occur irreversibly, and in any irreversible process, the entropy of the universe increases. So, for a spontaneous process,
⇒ \(\Delta S_{\text {univ }}=\left(\Delta S_{\text {sys }}+\Delta S_{\text {surr }}\right)>0\)
In a reversible process, any change in the entropy of the system is exactly balanced by the entropy change in the surroundings. Therefore, in a reversible process.
⇒ \(\Delta S_{\text {sys }}=-\Delta S_{\text {surr }} \text { or, }-\Delta S_{\text {sys }}=\Delta S_{\text {surr }}\)
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So, for a reversible process,
⇒ \(\Delta S_{u n i v}=\Delta S_{s y s}+\Delta S_{s u r r}=0\) When the spontaneous process reaches equilibrium, the value of
⇒ \(\Delta S_{u n i v}\left(=\Delta S_{s y s}+\Delta S_{\text {surr }}\right)=0.\).
For any spontaneous process,
⇒ \(\Delta S_{u n i v}\left(=\Delta S_{s y s}+\Delta S_{s u r r}\right)>0\)
For any reversible process, \(\Delta S_{u n i v}\left(=\Delta S_{s y s}+\Delta S_{s u r r}\right)>0\)
At equilibrium ofa process, \(\Delta S_{u n i v}\left(=\Delta S_{s y s}+\Delta S_{s u r r}\right)=0\)
At normal atmospheric pressure, water spontaneously converts into ice below o°C although the entropy change of the system (ΔSsys) in v.n is negative:
H2O molecules in ice are held orderly at fixed positions, which makes them unable to move about within ice. On the other hand, H2O molecules in water are not held at fixed positions as in ice and are capable of moving throughout the water.
Therefore, the molecular randomness in water is quite greater than that in ice. Thus entropy ofthe system decreases when water transforms into ice. So, in this process, Δ< 0.
As this transformation is an exothermic process, the heat released by the system is absorbed by the surroundings. As a result, the randomness as well as the entropy ofthe surroundings increases. So, ΔSys> 0.
However, in this process, the increase in entropy of the surroundings is greater than the decrease in entropy of the system. Consequently, the total change in entropy or the entropy ofthe universe (ASuntv) becomes positive. This favors the spontaneous conversion of water into ice below 0°C temperature and at normal atmospheric pressure.
Change in entropy and condition of spontaneity of a process in an isolated system
As an isolated system does not interact with its surroundings the total energy of such a system always remains constant is any process in it.
Therefore, the driving force for any spontaneous process in an isolated system is the change in entropy of the system. In such type of system, since surroundings remain unchanged, \(\Delta S_{\text {surr }}=0\) hence for any spontaneous process occurring in an isolated system,
⇒ \(\Delta S_{s y s}+\Delta S_{s u r r}>0 \text { or }, \Delta S_{s y s}\)
With the progress of a spontaneous process occurring in an isolated system, the entropy of the system gradually increases. When the process reaches equilibrium, the entropy
ofthe system gets maximized, and no further change in its value takes place.
Hence, at equilibrium, of a process occurring in an isolated system,
⇒ \(S_{\text {sys }}=\text { constant or, } d S_{\text {sys }}=0 \text { or, } \Delta S_{\text {sys }}=0 \text {. }\).
Change in entropy and spontaneity of exothermic and endothermic reactions
Change in entropy and spontaneity of an exothermic reaction:
In an exothermic reaction, heat is released by the reaction system. The released heat is absorbed by the surroundings, causing the randomness as well as the entropy of the surroundings to increase. Thus, ASsurr is always positive for exothermic reactions. However, the entropy of the system may decrease or increase for such type of reactions.
If the total entropy ofthe products is greater than the total entropy of the reactants in an exothermic reaction, then \(\Delta S_{s y s}>0\) In this case the value \(\Delta S_{u n i v}\) is greater or less than \(\Delta S_{\text {surr }}\). As a result, the reaction occurs sponataneouslty.
In an exothermic reaction if the total entropy of the reactants is greater than the total entropy of the products then
⇒ \(\Delta S_{s y s}<0\). Such type of reactions will be spontaneous if the numerical value of \(\Delta S_{\text {surr }}\) is greater than \(\Delta S_{s y s}\) because only in this condition \(\Delta S_{\text {univ }}>0\).
Change in entropy and spontaneity of an endothermic reaction: In an endothermic reaction, the heat is absorbed by the system from its surroundings. Causing the randomness as well as the entropy of the surroundings to decrease. So, in an endothermic reaction,
ΔSsurr < 0. Hence, an endothermic reaction will be spontaneous only when \(\Delta S_{s y s}\) is +ve and its magnitude is greater than that of \(\Delta S_{\text {surr }}\)
Entropy and the second law of thermodynamic
The second law of thermodynamics is expressed in various ways. One statement of this law is—”All spontaneous processes occur irreversibly and proceed in a definite direction We know that, for any spontaneous or natural changes, the total change in entropy of the system and the surroundings is positive, i.e., for any spontaneous change
⇒ \(\Delta S_{s y s}+\Delta S_{s u r r}>0.\).
Therefore, in all natural or spontaneous processes, the entropy of the universe continuously increases In other words all natural or spontaneous processes move in that direction which leads to the increase in entropy of the universe. Conversely, any process, which does not increase the entropy ofthe universe, will not occur spontaneously.
Second law of thermodynamics gives entropy For all the natural processes, the entropy of the universe is gradually increasing and approaching a maximum.
Processes occurring in nature are spontaneous. In all these processes, energy may be transformed into different forms although the total energy of the universe remains constant the entropy of the universe does not remain constant It is always increasing due to natural processes.
Entropy and Spontaneity in Law of Thermodynamics Numerical Examples
Question 1. The latent heat of fusion of ice at 0°C is 6025.24 J-mol-1 Calculate molar entropy of the process at 0°C.
Answer:
The change in entropy due to melting of lmol of ice
= \(\frac{\text { Molar latent heat (or enthalpy) of fusion of ice }}{\text { Melting point of ice }}\)
⇒ \(=\frac{6025.24 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{(273+0) \mathrm{K}}=22.07 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)
Question 2. The enthalpy change for the transformation of water into vapor at the standard boiling point is 40.8 kl. mol-1. Calculate the entropy change for the process.
Answer:
Change in entropy for the process
⇒ \(\frac{\text { Molar enthalpy of vaporisation of water }}{\text { Boiling point of water }}\)
= \(\frac{40.8 \times 10^3 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{(273+100) \mathrm{K}}\)
⇒ \(109.38 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)
Question 3. The enthalpy of vaporization of benzene at 80°C (boiling point) is 31-mol-1. What will be the change in entropy for the transformation of 31.2 g of benzene vapor into liquid benzene at 80°C?
Answer:
⇒ \(31.2 \mathrm{~g} \text { benzene }=\frac{31.2}{78}=0.4 \mathrm{~mol}[\text { Molar mass }=78] .\)
Enthalpy of condensation for lmol benzene vapor =(-)x enthalpy of vaporization of lmol liquid benzene =-31 kj. Therefore, the enthalpy of condensation of 0.4 mol of benzene =-31 × 0.4
=-12.4 kj.
∴ The change in entropy for the transformation of0.4 mol of benzene vapor into liquid benzene at 80°C,
⇒ \(\Delta S=-\frac{12.4 \times 10^3 \mathrm{~J}}{(273+80) \mathrm{K}}=-35.127 \mathrm{~J} \cdot \mathrm{K}^{-1}\)
Question 4. 1 mol or on Ideal gas If. expanded from lu Initial: volume of II, lo (lie Hind volume of 1(H) t, ul 25C. What will be changed In enthalpy for this process?
Answer:
⇒ \(\Delta S=2.303 n R \log \frac{V_2}{V_1}\)
⇒ \(=2.303 \times 8.314 \log \frac{100}{1}=38.29 \mathrm{~J} \cdot \mathrm{K}^{-1}\)
So, the change in entropy for this process = +398.29j
Question 5. The pressure of 1 mol of an Ideal gas confined In a cylinder fitted with a piston is 50 atm. The gas is expanded reversibly when the cylinder Is kept in contact with a thermostat at 25°C. During expansion, the pressure of the gas is decreased from 90 to 9 atm. Calculate the change in entropy in (Ids process, ff the heat absorbed by the gas during expansion he 5705 f, then calculate the change in entropy of the surroundings?
Answer:
We know \(\Delta S=2.303 n R \log \frac{p_1}{p_2}\)
Given, Px = 50 atm, P2 – 5 atm and n = 1
∴ The change in entropy of the system (i.e., gas)
⇒ \(\Delta S=2.303 \times 8.314 \log \frac{50}{5}=19.15 \mathrm{~J} \cdot \mathrm{K}^{-1}\)
Here surroundings are at a fixed temperature (25°C). During expansion, the heat absorbed by the gas from the surroundings = 5705 J.
Therefore, at 25’C, the heat released by the surroundings =-5705 j.
∴ Change in entropy of the surroundings,
⇒ \(\Delta S_{\text {surr }}=-\frac{5705}{(273+25)} \mathrm{J} \cdot \mathrm{K}^{-1}=-19.14 \mathrm{~J} \cdot \mathrm{K}^{-1}\)
Question 6. At 1 atm and 298 K, entropy change of the reaction, \(4 \mathrm{Fe}(s)+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3(s) \text { is }-549.4 \mathrm{~J} \cdot \mathrm{K}^{-1} \text {. }\) In this reaction, if A// = -1648 kj, then predict whether the reaction is spontaneous or not.
Answer:
As the reaction enthalpy is negative the reaction is exothermic. Therefore, the heat released by the given reaction will be equal to the heat absorbed by the surroundings Consequently the entropy of the surroundings will increase. Heat absorbed by the surroundings =(-) x heat released by the system
=-ΔH = -(-1648)kJ
= +1648 kj.
∴ Change in entropy of the surroundings
⇒ \(\Delta S_{\text {surr }}=-\frac{\Delta H}{T}=\frac{1648 \times 10^3}{298} \mathrm{~J} \cdot \mathrm{K}^{-1}=5530.2 \mathrm{~J} \cdot \mathrm{K}^{-1}\)
Thus, in this reaction \(\Delta S_{\text {untv }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}\)
= (-549.4 + 5530.2)J.K-1= +4980.8 J.K-1
Since \(\Delta S_{u n I v}>0\) the reaction will occur spontaneously.
Question 7. At 1 atm and 298 K, ΔH° value for the reaction \(2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) \text { is }-572 \mathrm{~kJ}\) . Calculate the change in entropy ofthe system and surroundings for this reaction. Is this reaction spontaneous at that temperature and pressure? Given: Standard molar entropies of H2(g), O2(g) & H2O(f) at 298K are 130.6, 205.0, and 69.90 J. K-1. mol-1 respectively
Answer:
The change in entropy ofthe given reaction
Δ = \(\left[2 S^0\left(\mathrm{H}_2 \mathrm{O}, l\right)\right]-\left[2 \times S^0\left(\mathrm{H}_2, g\right)+S^0\left(\mathrm{O}_2, g\right)\right]\)
=\( 2 \times 69.9-(2 \times 130.6+205)] \mathrm{J} \cdot \mathrm{K}^{-1}\)
= \(-326.4 \mathrm{~J} \cdot \mathrm{K}^{-1}\)
Change in entropy ofthe surroundings in the reaction
⇒ \(\Delta S_{\text {surr }}\)=\(-\frac{\Delta H^0}{298}\)
= \(\frac{572 \times 10^3}{298} \mathrm{~J} \cdot \mathrm{K}^{-1}\)
=\(+1919.4 \mathrm{~J} \cdot \mathrm{K}^{-1}\)
∴ Total change in entropy,
⇒ \(\Delta S_{\text {univ }}\)= \(\Delta S_{\text {sys }}+\Delta S_{\text {surr }}\)
= -326.4+1919.4
= \(+1593 \mathrm{~J} \cdot \mathrm{K}^{-1}\)
Since \(\Delta S_{u n i v}>0,\) then the reaction will occur spontaneously at 298 K and 1 atm.
Question 8. The molar enthalpy of fusion and the molar entropy of fusion for ice at 0°C and 1 atm are 6.01 kj- mol-1 and 22.0 J K-1.mol-1 respectively. Assuming ΔH and ΔS are independent of temperature, show that the inciting of ice at 1 atm is not spontaneous, while the reverse process is spontaneous.
Answer:
A process is spontaneous when the change in entropy of the universe \(\left(\Delta S_{u n i v}=\Delta S_{s y s}+\Delta S_{s u r r}\right)\) is positive. The transformation of the office into water involves the process.
⇒ \(\mathrm{H}_2 \mathrm{O}(s)\rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
⇒ \(\Delta H =6.01 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} ; \Delta S=22.0 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)
Now , \(\Delta S_{\text {sys }}=22.0 \mathrm{~J} \cdot \mathrm{mol}^{-1}\)
⇒ \(\Delta S_{\text {surr }}=\frac{\Delta H}{T}=\frac{-6010}{271}\)
=\(-22.17 \mathrm{~J} \cdot \mathrm{mol}^{-1}\)
Thus, \(\Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}=22.0+(-22.17) \mathrm{J} \cdot \mathrm{mol}^{-1}\)
= \(-0.17 \mathrm{~J} \cdot \mathrm{mol}^{-1}\)
Hence, the total entropy change in the process is negative at 271 K. Therefore, the transformation of ice into water at -2°C is not spontaneous. The reverse process i.e., the conversion of water to ice at -2°C is spontaneous. This is because at -2°C and atm pressure the overall entropy (ΔSuniv) is +0.17 J . mol-1