The Gibbs free energy (commonly called G) is a thermodynamic potential defined as the difference of the enthalpy H, minus the product of the temperature T, by the entropy S of the system:
G = H - T S
Gibbs free energy is measured in Joules (according to the International System), in ergs (for the Cegesimal System of Units), in calories or in electron volts (for electro Volts).
In processes that occur at constant pressure and temperature, the variation of Gibbs free energy is ΔG = ΔH - T ΔS. In such processes, (G) represents the energy available in the system that can be converted into work.
For example, in exothermic chemical reactions, enthalpy decreases while entropy increases. In the Gibbs function these two factors are counteracted, but only when the Gibbs energy decreases does the reaction occur spontaneously.
So if the variation of G is negative, the process is spontaneous. When the Gibbs function reaches its minimum, the system reaches a stable state of equilibrium. In synthesis, in a process for which the pressure and the temperature remain constant, we can affirm:
- If the process is spontaneous, then ΔG < 0
- When the system is in equilibrium: ΔG = 0
- In a non-spontaneous process G increases: ΔG> 0.
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The Gibbs free energy (G) is calculated using the definition given at the beginning:
G = H - T⋅S
In turn, the enthalpy H is a thermodynamic potential defined as:
H = U + P V
Next, a step-by-step analysis will be made, to know the independent variables of which the Gibbs energy is a function:
1- From the first law of thermodynamics we have that the internal energy U is related to the entropy S of the system and its volume V for reversible processes through the differential relationship:
dU = dQ - dW = TdS - PdV
From this equation it follows that the internal energy U is a function of the variables S and V:
U = U (S, V)
2- Starting from the definition of H and taking the differential, we obtain:
dH = dU + d (P V) = dU + VdP + PdV
3- Substituting the expression for dU obtained in (1) we have:
dH = TdS - PdV + VdP + PdV = TdS + VdP
From this it is concluded that the enthalpy H depends on the entropy S and the pressure P, that is:
H = H (S, P)
4- Now the total differential of Gibbs free energy is calculated obtaining:
dG = dH -TdS -SdT = TdS + VdP -TdS -SdT
Where dH has been replaced by the expression found in (3).
5- Finally, when simplifying, we obtain: dG = VdP - SdT, being clear that the free energy G depends on the pressure and the temperature T as:
G = G (P, T)
From the analysis in the previous section, it follows that the internal energy of a system is a function of entropy and volume:
U = U (S, V)
Then the differential of OR will be:
dU = ∂SU |V dS + ∂VU |S dV = TdS - PdV
From this partial derivative expression, the so-called Maxwell thermodynamic relations can be derived. Partial derivatives apply when a function depends on more than one variable and are easily calculated by applying the theorem in the next section.
∂VT |S = -∂SP |V
To arrive at this relationship, the Clairaut-Schwarz theorem on partial derivatives, which states the following:
"The mixed derivatives of second order with the interchanged variables are equal, as long as the functions to be derived are continuous and differentiable".
Based on what is shown in point 3 of the previous section:
H = H (S, P) and dH = TdS + VdP
It can be obtained:
∂PT |S = ∂SV |P
We proceed in a similar way with Gibbs free energy G = G (P, T) and with Helmholtz free energy F = F (T, V) to obtain the other two Maxwell thermodynamic relationships.
1- Associated with the internal energy U: ∂VT |S = -∂SP |V
2- The one obtained from the enthalpy H: ∂PT |S = ∂SV |P
3- Related to the Helmholtz energy F: ∂TP |V = ∂VS |T
4- Linked to the Gibbs free energy G: ∂TV |P = -∂PS |T
Calculate the variation of Gibbs free energy for 2 moles of ideal gas at a temperature of 300K during an isothermal expansion that takes the system from an initial volume of 20 liters to a final volume of 40 liters.
Recalling the definition of Gibbs free energy we have:
G = H - T S
Then a finite variation of F will be:
ΔG = ΔH - T ΔS, since ΔT = 0
In ideal gases, the enthalpy only depends on its absolute temperature, but since it is an isothermal process, then ΔH = 0 and ΔG = - T ΔS.
For ideal gases the entropy change of an isothermal process is:
ΔS = nR ln (Vtwo/ V1)
What applied to the case of this exercise we have:
ΔS = 2 moles x 8.314 J / (K mol) x ln (40L / 20L) = 11.53 J / K
Then we can get the change in Helmholtz energy:
ΔG = - 300K x 11.53 J / K = -3457.70 J.
Taking into account that Gibbs free energy is a function of temperature and pressure G = G (T, P); determine the variation of G during a process in which the temperature does not change (isothermal) for n moles of a monatomic ideal gas.
As shown above, the change in Gibbs energy only depends on the change in temperature T and volume V, so an infinitesimal variation of it is calculated according to:
dG = -SdT + VdP
But if it is a process in which the temperature is constant then dF = + VdP, so a finite pressure variation ΔP leads to a change in the Gibbs energy given by:
ΔG = + ∫ VdP = + ∫ (n R T) dP / P = + n R T ln (ΔP)
Using the ideal gas equation:
P V = n R T
During an isothermal process it happens that:
d (P V) = P dV + V dP = 0
That is:
dP / P = - dV / V
So the above result can be written as a function of the volume variation ΔV:
ΔG = + ∫ VdP = + ∫ (n R T) dP / P = - ∫ (n R T) dV / V = -n R T ln (ΔV)
Considering the following chemical reaction:
Ntwo0 (g) + (3/2) Otwo (g) ↔️ 2NOtwo (g) at temperature T = 298 K
Find the variation of the Gibbs free energy and, using the result obtained, indicate whether or not it is a spontaneous process.
Here are the steps:
- Step One: Enthalpies of Reaction
ΔHr = 2 * ΔH (NOtwo (g)) - ΔH (Ntwo0 (g)) = 2 * 33.2-81.6 = -15.2kJ / mol
- Second step: the reaction entropy variation
ΔSr = 2 * S (NOtwo (g)) - S (Ntwo0 (g)) - (3/2) S (Otwo (g)) = 2 * 240.1 - 220.1 - 1.5 * 205.2 = -47.7 J / (mol * K).
- Third step: variation in the Gibbs function
This value will determine the balance between the decreasing energy and the increasing entropy to know if finally the reaction is spontaneous or not.
ΔGr = ΔHr - T ΔSr = -15.2 -298 * (- 47.7) = -985.4 J / mol
As it is a negative variation of the Gibbs energy, it can be concluded that it is a spontaneous reaction at the temperature of 298 K = 25 ºC.
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