Gibbs Free Energy Or Free Energy
In the previous section, we have seen that entropy can be used as a criterion for the spontaneity of a process. In an isolated system, a process will be spontaneous if ΔSis positive during this process.
But natural processes seldom occur in isolated systems. In other systems, a process will be spontaneous if
⇒ \(\Delta S_{\text {universe }}\left(=\Delta S_{s y s}+\Delta S_{s u r t}\right)\) is positive. Many are quite inconvenient.
It is, thus, very useful to reformulate the spontaneity criterion in such a way that only the system is to be considered. For our purpose, J. ‘Willard Gibbs introduced a new thermodynamic function called Gibbs free energy” or free energy, denoted by‘ G immerse processes in which determination of ΔSt.
At constant temperature and pressure, the spontaneity ofa process can be determined from the value ofthe change in Gibbs free energy of the system.
As most of the physical or chemical changes occur at constant pressure, it is convenient to use the concept of free energy to determine the spontaneity of that change.
Definition Of Gibbis free energy:
It is the thermodynamic property of a system, whose decrease in a spontaneous process at constant temperature and pressure, measures the maximum useful energy obtainable in the form of work from the process.
In any spontaneous process occurring at constant temperature and pressure, the decrease in Gibbs free energy (-ΔG) = maximum useful or network performed by the system on the surroundings.
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Mathematical form of Gibbs free energy:
The total energy of a system, one part is free for doing useful work, and another part is unavailable, and cannot be converted into work. If the value of entropy of a system is S at T K, then the amount of unavailable energy of the system is TxS.
Therefore, the total energy of the system = G (free energy) + TS (unavailable energy) Generally enthalpy (H) is considered as the total energy of the system.
Thus, H = G+TS or, G = H-TS…………….(1)
Where G, H, and S are Gibbs free energy, enthalpy, and entropy ofthe system respectively. T is the temperature of the system in the Kelvin scale. Equation [1] is the mathematical form of Gibbs free energy or free energy.
Gibbs free energy is a state function:
Gibbs free energy (G), enthalpy (H), and entropy (S) of a system are related by the equation, G = H-TS. As H. S and T are state functions, Gibbs free energy (G) is also a state function.
Thus, the value of Gibbs free energy (G) of any system depends only on the present state of the system and not on how the system has reached its present state. Hence, the change in Gibbs free energy (AG) of any process doesn’t depend upon the nature of the process but depends only on the initial and final states ofthe process.
Gibbs free energy is an extensive property:
Enthalpy (H) and the product of entropy and absolute temperature (fxS) of the system depend on the amount of the substance present in the system.
With increasing amounts of the substance, values of these quantities are increased. Hence Gibbs free energy (G = H- TS) depends on the amount of the substance present in the system. Therefore, Gibbs free energy (G) is an extensive property of the system.
Change in Gibbs free energy in a process occurring at constant temperature and pressure: The change in Gibbs free energy in a process occurring at a constant temperature and pressure,
ΔG = Δ(H-TS) = ΔH-Δ(TS)
∴ ΔG = (ΔH- TAS) [Δ(TS) = TΔS as T = constant]
The above equation represents the relation between the entropy change (AS), enthalpy change (AH), and absolute temperature (T) for a physical or chemical change occurring at a particular temperature and pressure. Using this equation, it is possible to predict the spontaneity of a process at constant temperature and pressure.
Change in Gibbs free energy for a physical or chemical change and process
At a particular temperature (T) and pressure (P), if the change in enthalpy and change in entropy for a physical or chemical process are AH and AS, respectively, then the change in Gibbs free energy,
ΔG = ΔH -TAS……………….(1)
If ΔSsys and ΔSsurr are the changes in entropies of the system and its surroundings, respectively, then the total change in entropy for the process
⇒ \(\Delta S_{\text {total }}=\Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}\) The heat gained by the system at fixed pressure and temperature, qsys= change in enthalpy (ΔH). So sys=ΔH
Therefore, the heat lost by the surroundings at a fixed pressure and temperature \(\left(q_{\text {surr }}\right)=-q_{\text {sys }}=-\Delta H.\)
So, the change in entropy ofthe surroundings for a physical or chemical change at a particular temperature and pressure,
⇒ \(\Delta S_{\text {surr }}=\frac{q_{\text {surr }}}{T}=-\frac{\Delta H}{T}\)
And the total change in entropy at a particular temperature and pressure,
⇒ \(\Delta S_{\text {total }}=\Delta S_{\text {sys }}+\Delta S_{\text {surt }}=\Delta S-\frac{\Delta H}{T} \quad\left[\text { where } \Delta S=\Delta S_{\text {sys }}\right]\)
Or, \(T \Delta S_{\text {total }}=T \Delta S-\Delta H \text { or, } \Delta H^{\prime}-T \Delta S=-T \Delta S_{\text {total }}\) ……………………….(2)
Comparing equations (1) and (2) \(\Delta G=-T \Delta S_{\text {total }}\)…………………….(3)
Equation (3) represents the relation between changes in Gibbs free energy (AG) for a physical or chemical change at a fixed temperature and pressure and the total change in entropy of the system and surroundings \(\left(\Delta S_{\text {total }}\right)\) for that process.
We know that a process will be spontaneous if the total change in entropy of the system and surrounding in the process is positive; i.e.,
⇒ \(\Delta S_{\text {total }}\left(=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}\right)>0.\) According to
equation [3], \(\text { if } \Delta S_{\text {total }}>0 \text {, then } \Delta G<0 \text {. }\)
Thus, a physical or chemical change at a fixed temperature and pressure will be spontaneous ifthe change in Gibbs free energy (AG) is negative i.e., ΔG < 0.
If in a process, the total change in entropy ofthe system and its surroundings is -ve, i.e.,
⇒ \(\Delta S_{\text {total }}\left(=\Delta S_{s v s}+\Delta S_{s u r r}\right)<0,\) then the process will be non-spontaneous, but the reverse the process will be spontaneous. From equation [3], if
⇒ \(\Delta S_{\text {total }}<0 \text {, then } \Delta G>0\text {. }\)
Thus, for a physical or chemical change at a fixed pressure and temperature, if the Gibbs free change is positive (ΔG > 0), then the process will be nonspontaneous but the reverse one will be spontaneous.
The total change in entropy of the system and surroundings will be zero, i.e.,
⇒ \(\Delta S_{\text {total }}\left(=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}\right)=0\) for a process at equilibrium. According to equation [3], if
⇒ \(\Delta S_{\text {total }}=0 \text { then } \Delta G=0\).
Therefore, the Gibbs free energy change will be zero, i.e., ΔG = 0 for a physical change or chemical reaction at equilibrium under the condition ofa fixed pressure and temperature.
Reaction; A→ B (at constant and pressure)
If &G(=GB- GA) <0 at a fixed temperature and pressure; then the transformation of A to B will be spontaneous.
If ΔG(= GB- Ga) > 0 at a fixed temperature and pressure; then the transformation of A to B will be non-spontaneous. But the transformation of B to A will be spontaneous as GB > GA and for the reverse process the value of ΔG(= GA- GB) is negative.
If ΔG(=Gb- Ga) = 0, then A and B will be in equilibrium. In this condition, the rate of transformation of B into A or A into B will be the same. So no net change will occur either in the forward or in the reverse direction.
Effect of ΔH and ΔS on the value of ΔG for any physical change or chemical reaction:
Effect of temperature on the change in Gibbs free energy and the spontaneity of a process
We know that ΔG = ΔH- TΔS. In this relation, the values of both ΔH and ΔS may be either positive or negative, but the temperature (in the Kelvin scale) is always positive. Again TΔS is a temperature-dependent quantity.
With the increase or decrease in temperature, the magnitude of TΔS increases or decreases, and the value of ΔH almost remains unchanged. So, ΔG depends on temperature. From the signs of ΔH and ΔS, we can predict the effect of temperature on ΔG.
The different possibilities have been discussed in the following table:
Numerical Examples
Question 1. \(\mathrm{Br}_2(l)+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{BrCl}(\mathrm{g})\); Whether the reaction is spontaneous or not at a certain pressure & 298 K ? [Aff=29.3 kJ.mol-1, ΔS=104.1 J. K-1.mol-1]
Answer:
We know, ΔG = ΔH- ΔTS
∴ ΔG =29.3 ×10³ -(298× 104.1)
=-1721.8 J- mol-1
As ΔG is negative, this process is spontaneous.
Question 2. At a certain pressure and 27°C, the values of ΔG and ΔH ofa the process are- 400 kj and 50 kj respectively. Is the process exothermic? Is it spontaneous? Determine the entropy change of the process.
Answer:
As per given data ΔH = 50 kj. As the value of ΔH is positive, it is not an exothermic process.
For the process ΔG =-400 kj at a certain pressure and 27°C. As ΔG is negative, it is a spontaneous process.
ΔG = ΔH- TΔS
∴ -400 = 50-300 × ΔS
or, ΔS =1.5kJ.K-1
=1500 kJ.k-1
Question 3. Values of ΔH to ΔS for the given reaction are -95.4kJ and -198.3 J.K-1 respectively: \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightarrow 2 \mathrm{NH}_3(g)\) State whether the reaction will be spontaneous at 500 K or not. Consider AH and AS are independent of temperature.
Answer:
We know, ΔG = ΔH- TΔS
Given, ΔH = -95.4 kj , ΔS = -198.3 J.K-1 and T = 500 K.
∴ ΔG= [-95.4 ×10³- 500 × (-198.3)]J
= 3750 J
= 3.75 kJ
As the value of ΔG is positive at constant pressure and 500 K temperature, it is not a spontaneous process.
Question 4. In and the ASreaction,= + 35 JA(s). K-1+. State B(g)-C(g)whether+ D(g)the reaction, ΔH =31 will k-1 be spontaneous at 100°C and 1100°C or not? Consider ΔH and ΔS are independent of temperature.
Answer:
We know, ΔG = ΔH- TΔS. Now at 100°C,
ΔG = ΔH-TΔS = 31 × 10³- (273 + 100) × 35
=+ 17945J
∴ At 1100°C, = 31 × 10³-(273 + 1100) × 35
=-17055 J
At 100°C, AG for the given reaction is positive, so the reaction will be non-spontaneous. On the contrary, at 1100°C, AG for the given reaction is negative, so it will be spontaneous.
Question 5. Is the vaporization of water at 50°C and 1 atm spontaneous? Given: For vaporization of water at that temperature and pressure, ΔH = 40.67 kj.mol-1 and ΔS = 108.79 J.K-1.mol-1.
Answer:
We know, ΔG = ΔH- TΔS
∴ ΔG=ΔH- TΔS = 40.67 × 10³- (323 × 108.79)
= + 5530.83 J.mol-1
As ΔG = +ve, so vaporization will be non-spontaneous.
Question 6. At 25°C and 1 atm, the heat of formation of 1 mol of water is -285.8 kj. mol-1. State whether the formation reaction will be spontaneous at that temperature and pressure or not. Given: The molar entropies of H2(g) ,O2 (g) & H2O(Z) at 25°C and 1 atm are 130.5, 205.0 and 69.9 J.K-1.mol-1 respectively.
Answer:
The equation for the formation reaction of water:
⇒ \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H=-285.8 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)
The change in entropy for this reaction, \(\Delta S=S_{\mathrm{H}_2 \mathrm{O}(l)}-\left(S_{\mathrm{H}_2(g)}+\frac{1}{2} S_{\mathrm{O}_2(g)}\right)\)
= \(\left[69.9-130.5-\left(\frac{1}{2} \times 205\right)\right] \mathrm{J} \cdot \mathrm{K}^{-1}=-163.1 \mathrm{~J} \cdot \mathrm{K}^{-1}\)
Therefore, the change in Gibbs free energy at 25°C and 1 atm for the formation of1 mol of H2O(Z) from 1 mol of H2(g) and
⇒ \(\text { and } \frac{1}{2} \mathrm{~mol} \text { of } \mathrm{O}_2(\mathrm{~g}), \Delta G=\Delta H-\mathrm{T} \Delta S\)
= -285.8¹ 103- 298 × (-163.1) =-237.196 kJ
As ΔG is negative at 25°C and 1 atm, so the formation of water at this temperature and pressure will be spontaneous.
Determination of temperature at which equilibrium is established in a physical or chemical change
We know that at a given temperature and pressure the change in free energy (AG) for a reaction is zero when the reaction is at equilibrium. Therefore, at equilibrium,
⇒ \(\Delta G=\Delta H-T \Delta S=0 \text { or, } \mathbf{T}=\frac{\Delta \boldsymbol{H}}{\Delta S}\)……………………………..(1)
Applying equation no.(1), we can determine the temperature at which equilibrium is established in a physical or chemical change.
Example: The values of the enthalpy and entropy changes ofthe system are + 40.7 kj- mol-1 and 109.1 J. K-1mol-1respectively for the process, H2O(Z) H2O(g) at 1 atm pressure. At which temperature equilibrium will be established between water and water vapor?
Answer:
At equilibrium ΔG = 0 and T \(=\frac{\Delta H}{\Delta S} .\)
Given: ΔH = 40.7 kj.mol-1 = 40.7 ×103 J.mol-1 and AS = 109.11 -K-1- mol-1
∴ \(T=\frac{40.7 \times 10^3 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{109.1 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}}=373 \mathrm{~K}\)
373 K (100°C) is the normal boiling point of water. So, at 1 atm and 100°C, water and its vapor will remain in equilibrium.
The temperature at which a physical or chemical change becomes spontaneous We have seen. that in any physical or chemical change, if the signs of AH and AS are the same (either + or – ), then the sign of AG as well as the spontaneity of that change depends on temperature.
Suppose, for a physical or chemical change at a particular temperature and pressure, ΔH > Δ and ΔS > Δ. According to the relation, ΔG = ΔH- TΔS the process is spontaneous because when TΔS > ΔH, ΔG<Δ This means that
ΔH- TΔS < 0 or TΔS > ΔH thus,
⇒ \(T>\frac{\Delta H}{\Delta S}\)
Hence, the process will be spontaneous when \(T>\frac{\Delta H}{\Delta S}\), if \(T<\frac{\Delta H}{\Delta S}\)
ΔS the process will be non-spontaneous, but the reverse process will be spontaneous. For a physical or chemical change if ΔH < 0 and ΔS < 0, then it can be shown that the process will be spontaneous when
⇒\(T<\frac{\Delta H}{\Delta S}. \text { If } T>\frac{\Delta H}{\Delta S}\)
The process will be non-spontaneous, but the reverse process will be spontaneous.
Question 1. In the reaction, A-B+ C, ΔH = 25 kj. mol-1and ΔS = 62.5 J.K-1. At which temperature the reaction will occur spontaneously at constant Pressure?
Answer:
The condition for the spontaneity of a reaction at a given temperature & pressure is ΔG < 0.
We know ΔG = ΔH- TΔS For a spontaneous reaction,
ΔH- TΔS < 0 or, \(\Delta H<T \Delta S \quad \text { or, } T \Delta S>\Delta H \quad \text { or, } T>\frac{\Delta H}{\Delta S}\)
Given: ΔH = 25 × 103 J and ΔS = 62.5 J . K-1
Therefore \(T>\frac{25 \times 10^3 \mathrm{~J}}{62.5 \mathrm{~J} \cdot \mathrm{K}^{-1}} \quad \text { or, } T>400 \mathrm{~K} \text {. }\)
Question 2. H2O(g) H2O(l) ; Δh = -40.4 kj.mol-1 ΔS = -108.3 J.K-1.mol-1. At which temperature the process will be spontaneous at a constant 1 atm?
Answer:
In this process ΔH<0 and ΔS <0. For such type of process, the temperature at which the reaction will occur spontaneously \(T<\frac{\Delta H}{\Delta S}\)
Given: ΔH=-40.4 × 10³ J mol-1, ΔS=-108.3 J.K-1.mol-1
∴ \(T<\frac{\Delta H}{\Delta S} \quad \text { or, } T<\frac{40.4 \times 10^3}{108.3} \quad \text { or, } T<373 \mathrm{~K}\)
∴ The process will be spontaneous below 373 K (100°C).
Question 3. H2(g)+Br(Z)-+2HBr(g); ΔH=-72.8kJ (1 atm, 25°C) If molar enterpoies of H2(g), Br2(l),HBr(g) are 130.5, 152.3 and 198.3j.k-1.mol-1 respectively then at which temperature the reaction will be spontaneous?
Answer:
Change in entropy for the given reaction,
⇒ \(\Delta S=2 S_{\mathrm{HBr}(g)}-\left[S_{\mathrm{H}_2(g)}+S_{\mathrm{Br}_2(l)}\right] \)
=\([2 \times 198.3-(130.5+152.3)] \mathrm{J} \cdot \mathrm{K}^{-1}=113.8 \mathrm{~J} \cdot \mathrm{K}^{-1}\)
As ΔH < 0 and ΔS > 0 for the given reaction, the reaction will be spontaneous at any temperature.
Question 4. For the reaction A(g) + B(g)→ C(s) +D(l); ΔH =-233.5 kj and ΔS = -466.1 J. K-1 At what temperature, equilibrium will be established? In which directions the reaction will proceed above and below that temperature?
Answer:
At constant temperature and pressure, the equilibrium temperature ofa reaction \(T=\frac{\Delta H}{\Delta S}.\)
Given:ΔH = -233.5 × 10³ J and ΔS = -466.1 J .K-1
Therefore \(T=\frac{233.5 \times 10^3}{466.1} \mathrm{~K}=500.9 \mathrm{~K}=227.9^{\circ} \mathrm{C}\)
∴ At 227.9°C, the reaction will attain equilibrium.
When T > 500.9 K, the magnitude of TΔS is greater than that of ΔH. Then according to the equation, ΔG = ΔH- TΔS, the value of AG will be positive. So the reaction will be non-spontaneous above 500.9 K.
When T < 500.9 K, the magnitude of TΔS is less than that of ΔH. According to the equation, ΔG = ΔH- TΔS, ΔG will be negative. Therefore, the reaction will be spontaneous below 500.9 K
The standard free energy of formation of a substance and the standard free energy change in a chemical reaction
Standard free energy of formation:
The standard free energy of the formation of a compound is denoted by \(\Delta G_f^0\) and its unit is kj.mol-1 (or kcal-mol-1).
Definition:
The free energy change associated with the formation of 1 mol of a pure compound from its constituent elements present at standard state is termed as the standard free energy of formation of that compound.
The value of standard free energy of formation of any element at 25°C and 1 atm pressure is taken as zero. For the elements having different allotropic forms, the standard free energy of formation of the most stable allotrope at 25°C and 1 atm pressure is taken as zero.
For example,\(\left(\Delta G_f^0\right)\) [C (graphite)] = 0 but \(\left(\Delta G_f^0\right)\) [C dimond] ≠0.
Standard free energy of formation (Gf°) of some elements and compounds at 25°C:
Standard free energy change in a chemical reaction:
In any reaction, the change in free energy (ΔG) depends on temperature, pressure, and concentration. Thus to compare the ΔG of different reactions, the standard free energy change is calculated.
Definition:
It is defined as the change in free energy when the specified number of moles of reactions (indicated by the balanced chemical equation] in their standard states are completely converted to the products in their standard states.
It is expressed by ΔG°. At a particular temperature, the standard free energy change in a reaction is calculated from the standard free energy of formation ofthe reactants and products at that temperature.
The change in standard free energy in a reaction, ΔG° =
The sum of the standard ‘free energy of formation of the products – the sum of the standard free energy of formation of the reactants \(\)
Or,\(\Delta G^0=\sum n_1 \Delta G_{f, 1}^0-\sum n_j \Delta G_{f, j}^0\)
Where n1 and n2 to are the number of moles of i -th product and j-th reactant In the balanced chemical equation whereas ΔG °f,t, and ΔG °f,j are the standard free energy of formation of the j -th product and j-th reactantrespectively
Example: in the case of the reaction aS+bB-cC+dD:
⇒ \(\Delta G^0=\left[c \times \Delta G_f^0(C)+d \times \Delta G_f^0(D)\right]\)
⇒ –\(\left[a \times \Delta G_f^0(A)+b \times \Delta G_f^0(B)\right]\)
The standard free energy change (ΔG°) in a reaction can also be determined from the values of the standard change enthalpy (ΔH0) and standard change in entropy (ΔS0), using the following equation,
⇒ \(\Delta G^0=\Delta H^0-T \Delta S^0\)
ΔG° =ΔH° TΔS° ……………………………(1)
ΔG° and the spontaneity of a physical or chemical change:
If ΔG° < 0 for any physical or chemical change, then the process will be spontaneous under standard conditions.
If the value of ΔG° > 0 for any physical or chemical change, then the process will be nonspontaneous under standard conditions. However, the reverse process will be spontaneous under standard conditions.
Free energy change in a chemical reaction, reaction equilibrium, and equilibrium constant
ΔG° in a reaction can be determined either from the values of the standard free energies of formation of the participating reactants and products or from the equation,
ΔG°= ΔH°- TΔS°. But if a reaction occurs in a condition other than standard condition then the free energy change of the reaction can be calculated using the following relation.
⇒ \(\Delta G=\Delta G^0+R T \ln Q\)…………………………..(1)
Where Q = reaction quotient [a detailed discussion on reaction quotient has been made, T = temperature in Kelvin scale, R = the universal gas constant, ΔGs the free energy change in a reaction at constant pressure and a constant temperature TK, ΔG°= the standard free energy change in the reaction at TK
Relation between standard free energy change (AG°) and equilibrium constant (K The relation between ΔG° and K can be derived from the above equation [1] For a reaction at equilibrium, ΔG = 0 at constant temperature and pressure. Also at equilibrium, the reaction quotient (Q) = equilibrium constant (JC). Therefore, according to the equation [1], at equilibrium
∴ 0 = ΔG° +RT in K°
or, ΔG-RT in k………………………..(2)
Or, ΔG° = -2.03 RT Log K………………………..(3)
Equations (2) and (3) show the relationship between the standard free energy change in a reaction (ΔG0) and the equilibrium constant (K) of the reaction at a particular temperature (T).
Hence, using these equations [2 and 3], he value of the equilibrium constant (K) of a reaction at a particular temperature (F) can be determined from the value of the standard free energy change of the reaction (ΔG°) at the same temperature (T).
Alternatively, the value of ΔG° can be determined from the value of K by using equations [2] and [3]. According to equation [2] (or [3]);
- If ΔG° is negative, then In K (or logic) will be positive. Thus K > 1
- If ΔG° is positive, then In K (or logic) will be negative Thus K < 1.
- If ΔG° is equal to zero, then In K (or logic) will be equal to zero. Thus K = 1.
Numerical Examples
Question 1. In the given reaction, calculate the standard free energy change at 25°C: \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightarrow 2 \mathrm{NH}_3\) [Given that, ΔH° = -91.8 kJ and ΔS° = -198 J. K-1]
Answer:
We know, ΔG° = ΔH°- TΔS°
ΔH° = -91.8 kJ , ΔS°=-198 J.K-1 , F
= (273 +25) = 298K
ΔG° =-91.8 × 10³- 298 × (-198)
=-32796J
=-32.796kJ
Question 2. In the given reaction, calculate standard free energy change at 25°C: 2NO(g) + 0,(g) 2NO2(g). Is the reaction spontaneous under standard conditions?\(\left[\right. Given: At 25^{\circ} \mathrm{C}, \Delta G_f^0[\mathrm{NO}(\mathrm{g})]=86.57 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} and \Delta G_f^0\left[\mathrm{NO}_2(g)\right]=51.30 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}.\)
Answer:
⇒ \(\left.\Delta G^0=\sum \Delta G_f^0 \text { (products }\right)-\sum \Delta G_f^0(\text { reactants })\)
In the given case,
⇒ \(\Delta G^0=2 \Delta G_f^0\left[\mathrm{NO}_2(g)\right]-2 \Delta G_f^0[\mathrm{NO}(g)]-\Delta G_f^0\left[\mathrm{O}_2(g)\right]\)
⇒ \(\text { Given: } \left.\Delta G_f^0[\mathrm{NO}(g)]=86.57 \mathrm{k}\right] \cdot \mathrm{mol}^{-1}\)
⇒ \(\Delta G_f^0\left[\mathrm{O}_2(\mathrm{~g})\right]=0 \text { and } \Delta G_f^0\left[\mathrm{NO}_2(\mathrm{~g})\right]=51.30 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)
[∴ Standard free energy of formation of an element =0]
Since the value of ΔG° is negative, so the reaction will be spontaneous under standard conditions.
Question 3. At 25°C, the standard free energy change for a reaction is 5.4 kJ. Calculate the value of the equilibrium constant of the reaction at that temperature.
Answer:
We know, ΔG° = -RTlnK.
∴ We have, ΔG° = 5.4 kJ
= 5.4 × 10³ J and T
= 298 K
Or, log k = 0.95
∴ K= 0.113
Therefore, at 25°C the value of the equilibrium constant for the given reaction will be 0.113.
Question 4. The equilibrium constant for a reaction is 1.6 × 10-6at 298K. Calculate the standard free energy change (ΔG°) and standard entropy change (ΔS°) of the reaction at that temperature. Given, at 298 K, ΔH° = 25.34 kJ
Answer:
We know. ΔG° = -RTinK
∴ ΔG° =-8.314 × 298 ln(1.6 × 10-6)
= 33064J = 33.064 kJ
∴ The standard free energy change at 298 K = + 33.064 kJ
Again we know, ΔG°=ΔH°- TΔS°
Here, ΔH°=25.34 kJ
∴ +33.064 = 25.34- 298 × ΔS°
So, the standard entropy change for the reaction = 25.9j. k-1
Question 5. At 298 K, the standard free energy of formation of H2O(l) = 23.7 kJ.mol-1 calculates the value of equilibrium constant temperature for the following reaction:
⇒\(2 \mathrm{H}_2 \mathrm{O}(t) \rightarrow 2 \mathrm{H}_2(g)+\mathrm{O}_2(g)\)
Answer:
Standard free energy change for the given reaction,
⇒ \(\Delta G^0=2 \Delta G^0\left[\mathrm{H}_2(g)\right]+\Delta G^9\left[\mathrm{O}_2(g)\right]-2 \Delta G^9\left\{\left[\mathrm{H}_2 \mathrm{O}(l)\right]\right.\)
The standard free energy of the formation of an element Is
taken as zero, so
⇒ \(\Delta G_f^0\left[\mathrm{O}_2(g)\right]=0 \text { and } \Delta G_j^9\left[\mathrm{H}_2(g)\right]=0 \text {. }\)
Therefore \(\left.\left.\Delta G^0=[0+0-2 \times(-237.13)] \mathrm{k}\right]=+474.25 \mathrm{k}\right]\)
∴ +474.26 × 10³ =-8.314 × 298 in k or k
= -191.42
∴ K= 7.36 × 10-84
So, the equilibrium constant for the reaction = 7.36 × 10-84
The Third Law Of Thermodynamics
Statement of the third law of thermodynamics:
The entropy of a pure and perfectly crystalline substance at 0 K temperature is equal to zero.
Explanation:
In a perfect crystalline solid, all the constituent particles are perfectly arranged in a well-ordered manner. Defects like point defects, line defects, etc., that can generally be observed in crystal lattices, are found to be absent in a perfect crystal.
The constituent particles become motionless and attain the lowest energy in such as substance at absolute zero. As a result, the randomness of those constituent particles becomes zero. For this reason, the value of entropy of a pure and perfectly crystalline substance is zero.
Explanation of thermodynamic probability:
A system can achieve a particular thermodynamic state in various ways or configurations. The number of ways in which a particular state can be achieved is called thermodynamic probability and is designated by the symbol ‘W’.
With increasing thermodynamic probability the randomness as well as the entropy of the system increases. So we can expect that there exists a relationship between the two quantities S and W. Lauding Boltzmann introduced a relation between S and W. The relation is S = k InW; where k- Boltzmann constant.
At absolute zero, all the constituent particles would occupy the minimum energy state, and hence there is only one way of arranging the constituent particles in different energy levels i.e., W = 1. This gives S = 0. So, for a pure and perfect crystalline substance, the value of entropy is zero at zero kelvin. This is the statement of the third law of thermodynamics.
Application of the third law of thermodynamics: The absolute value of entropy can be determined by making use of this law. If the increase in entropy of a substance is ΔS due to an increase in temperature from OK to TK then ΔS = ST- SQ; where, ST and SQ are the entropies of that substance at TK, and 0 K, respectively. According to the third law of thermodynamics, SQ = 0. Therefore, S = ST. So by measuring ΔS, it is possible to determine the absolute value of entropy of a substance at T K.
Heat or enthalpy of neutralization
Heat or enthalpy of neutralization Definition:
The change in enthalpy that occurs when 1 gram equivalent of an acid is completely neutralized by 1 gram equivalent of a base or vice-versa in a dilute solution at a particular temperature is called the enthalpy (or heat) of neutralization.
The change in enthalpy that occurs when 1 mol of H+ ions reacts completely with mol of OH– ions in a dilute solution to form 1 mol water at a particular temperature is known as the Enthalpy (or heat) of neutralization.
The enthalpy of neutralization is denoted as ΔHN, where subscript TV ‘indicates ‘neutralization’.
Neutralization of strong acid and strong base:
If both the acid and base are strong, then the value of heat of neutralization constant is found to be almost and this value is ~57.3 kj.