Gibbs Energy Change and Equilibrium
Energy can take many forms, including kinetic energy produced by an object’s movement, potential energy produced by an object’s position, heat energy transferred from one object to another due to a temperature difference, radiant energy associated with sunlight, the electrical energy produced in galvanic cells, the chemical energy stored in chemical substances, and so on. All of these different types of energy may be transformed from one form to the other.
For example, as water in a dam reservoir falls, its potential energy is turned into kinetic energy, and if the falling water is utilized to power turbines, the kinetic energy of the water is converted into electrical energy. However, if the water collides with rocks near the dam’s base. The kinetic energy of the object is transformed into thermal energy.
As a result, the various kinds of energy are quantitatively tied to one another. Thermodynamics is the study of such quantitative relationships between various kinds of energy. Energy shifts occur as a result of physical and chemical processes. The study of energy transitions in these processes is the main focus of thermodynamics.
Thermodynamics is the branch of science that studies the many kinds of energy, their quantitative connections, and the energy changes that occur in physical and chemical processes.
Gibbs Energy
J.W. Gibbs was an American theoretician. He introduced a new thermodynamic function called Gibbs energy denoted as G. The second law of thermodynamics states that Î”S_{total} = (Î”S_{system} + Î”S_{Surrounding}) must be positive for all spontaneous processes. To assess the spontaneity of a process, two entropy changes, Î”S_{System} and Î”S_{Surrounding} must be determined. As a result, it is more straightforward to describe the criteria of spontaneity in terms of the system’s thermodynamic features alone, without respect for the surroundings and this problem was solved by Gibbs.
Gibbs energy (G) is defined as,
G = H – TS
where:
- S is the entropy of the system,
- H is Enthalpy, and
- T is the Temperature.
G is also a state function because, H, T, and S are state functions.
The change in Gibbs Energy (Î”G) on the initial and final state of the system and not on the path connecting the two states. The change in Gibbs energy at constant temperature and pressure is defined as:
Î”G = Î”H – T Î”S
where:
- Î”S is the change in entropy of the system and
- Î”H is the change in Enthalpy.
Gibbs Energy and Spontaneity
The total entropy change can be written as,
Î”S_{total} = Î”S_{system} + Î”S_{surrounding} = Î”S + Î”S_{surr}
According to the second law of thermodynamics, Î”S_{total} > 0 at constant temperature and pressure for the process to be spontaneous. If Î”H is the enthalpy change accompanying the reaction, that is the enthalpy change of the system then the change in enthalpy of the surroundings is (-Î”H)
Therefore,
Î”S_{surr} = -Î”H / T
Hence the total entropy is given by,
Î”S_{total} = Î”S – Î”H / T
This equation shows that Î”S_{total} is expressed in terms of the properties of the system only.
Rearranging the equation we get,
-T Î”S_{total} = -T Î”S + Î”H
or
-T Î”Stotal = Î”H -T Î”S
Combining the above two equations, we get,
Î”G = -T Î”S_{total}
This equation indicates that Î”G and Î”S_{total} have opposite signs because T is always positive. Thus, for a spontaneous process carried out at a constant temperature and pressure Î”S_{total }> 0 and hence Î”G < 0.
Gibbs energy of a system decreases in a spontaneous change that takes place at constant temperature and pressure. On contrary, for a non-spontaneous reaction Î”S_{total }and hence Î”G > 0.
Gibbs energy of a system increases in a non-spontaneous change that takes place at constant temperature and pressure. The end of the spontaneous process is an equilibrium that corresponds to a minimum in G. Hence the change in Gibbs energy is:
- Î”G < 0, the process is spontaneous.
- Î”G > 0, the process is non-spontaneous.
- Î”G = 0, the process is at equilibrium.
Factor affecting the Spontaneity
Consider the equation,
Î”G = Î”H – T Î”S
The value Î”G determines whether a physical or chemical change will occur spontaneously. The equations Î”H and Î”S correspond to the values of the system alone. The equation states that two elements influence the spontaneity of reactions
- Î”H is the amount of heat transmitted at constant pressure and temperature, and
- Î”S is the rise or reduction in molecular disorder.
The spontaneous process is favoured by a decrease in enthalpy (-Î”H) and increase in entropy (Î”S) On the other hand non-spontaneous reaction is favoured by an increase in enthalpy (+Î”H) and decrease in entropy (-Î”S). The term temperature in the equation is an important component in determining the relative relevance of the enthalpy and entropy contributions to Î”G. If Î”H and Î”S in the equation are both positive or both negative, the sign of Î”G and hence the spontaneity of the reaction, depends on temperature.
Î”H |
Î”S |
Î”G |
Spontaneity of reaction |
Negative (exothermic) |
Positive |
Negative |
Reactions are spontaneous at all temperatures. |
Negative (exothermic) |
Negative |
Negative or Positive |
Reactions become spontaneous at low temperatures when |T. Î”S| < |Î”H|. |
Positive (endothermic) |
Positive |
Negative or Positive |
Reactions become spontaneous at low temperatures when T.Î”S < Î”H. |
Positive (endothermic) |
Negative |
Positive |
Reactions are non-spontaneous at all temperatures. |
Temperature of Equilibrium
At equilibrium, i.e., Î”G = 0, the process is neither spontaneous nor non-spontaneous because it is balanced between spontaneous and non-spontaneous behavior. (+Î”H)
So,
Î”G = Î”H – T Î”S = 0
Hence,
Î”H = TÎ”S or T = Î”H / Î”S
T is the temperature at which the transition from spontaneous to non-spontaneous behavior happens. T is calculated on the assumption that Î”H and Î”S are temperature independent. In reality, Î”H and Î”S change with temperature. However, for modest temperature changes, the variance in them will not add considerable mistakes.
Î”G and Equilibrium constant
All of the substances (reactants and products) in a chemical reaction may not be in their normal forms. As a result of the connection, the change in Gibbs energy of a reaction is related to the change in standard Gibbs energy.
Î”G = Î”GÂ° + RT ln Q
where:
- Î”GÂ° is the standard Gibbs energy change (change in Gibbs energy when all the substances are in their standard state).
- Q is the reaction quotient.
The expression of the reaction quotient is similar to that of the equilibrium constant, but there is one single difference between them, i.e., Equilibrium concentrations or partial pressures of products and reactants are included in the equilibrium constant. Whereas Q is expressed in terms of reactant beginning concentration partial pressures and product final concentrations or pressures.
For Example, consider the below example:
aA +bB â‡¢ cC + dD
For the above reaction, the reaction Quotient is given by
or
When the values of concentration or partial pressure are other than equilibrium values. When the reaction reaches equilibrium, the concentrations and partial pressure reach their equilibrium values and at this stage, Q = K. At equilibrium, Î”G = 0 and Q = K, then the standard Gibbs energy equation becomes,
0 = Î”GÂ° + RT ln K
Hence,
Î”GÂ° = -RT ln K = -2.303RT log_{10}K
This equation gives the relationship between standard Gibbs energy change for the reaction and its equilibrium constant.
Sample Problems
Problem 1: Determine whether the reaction is spontaneous or non-spontaneous for the given value of Î”H and Î”S. Also, state whether they are exothermic or endothermic.
- Î”H = – 40 kJ and Î”S = +135 JK^{-1} at 300K
- Î”H = – 60 kJ and Î”S = -160 JK^{-1} at 400K
Solution:
- Î”G = Î”H – T Î”S
Î”H = – 40 kJ, Î”S = +135 J K^{-1} = 0.135 kJ K^{-1} and T = 300K
Î”G = -40 (kJ) – 0.135(kJ K^{-1}) Ã— 300(K)
= 80.5 kJ
Because Î”G is negative, the reaction is spontaneous. The negative Î”H value indicates that the reaction is exothermic.
- Î”G = Î”H – T Î”S
Î”H = – 60 kJ, Î”S = – 160 J K^{-1} = – 0.160 kJ K^{-1} and T = 400K
Î”G = -60 (kJ) – 0.160(kJ K^{-1}) Ã— 400(K)
= -60 kJ + 64 kJ = 4kJ
The reaction is non-spontaneous because Î”G is positive and exothermic as Î”H is negative.
Problem 2: For a certain reaction Î”H = -25kJ and Î”S = -40J K^{-1}. At what temperature will it change from spontaneous to non-spontaneous.
Solution;
T = Î”H / Î”S
Î”H = – 25 kJ, Î”S = -40 J K^{-1 }= 0.04 kJ K-1
Hence, T = -25(kJ) / -0.04 kJ K^{-1} = 625K
Because both Î”H and Î”S are negative, the reaction will occur spontaneously at lower temperatures. As a result, the reaction will be spontaneous below 625K and non-spontaneous beyond 625K.
At 625K, the transition from spontaneous to non-spontaneous occurs.
Problem 3: Determine Î”S_{total} and decide whether the following reaction is spontaneous at 298K.
Î”HÂ° = -24.8 kJ, Î”SÂ° = 15 J K^{-1}
Solution:
The heat evolved in the reaction is 24.8 kJ. The same quantity of heat is absorbed by the surroundings.
Hence, Entropy change of the surrounding will be,
Î”S_{surr} = Î”HÂ° / T
= – [(-24.8 (kJ)) / 298 (K)]
= + 83.2 J K^{-1}
Î”S_{total} = Î”S_{System} + Î”S_{Surr}
Î”S_{Sys} = Î”SÂ° = 15 J K^{-1}
= 15(J K^{-1}) + 83.2 (J K^{-1})
= 98.2 J K^{-1}
As Î”S_{total} is positive, the reaction is spontaneous at 298 K.
Problem 4: Determine whether the reaction,
N_{2}â€‹O_{4 }â€‹(g)âŸ¶2NO_{2}â€‹ (g)
is spontaneous at 298 K from the following data.
Î”_{f}HÂ° (N_{2}O_{4}) = 9.16 kJ mol^{-1}, Î”_{f}HÂ° (NO_{2}) = 33.2 kJ mol^{-1}
Solution:
Î”H=âˆ‘Î”fâ€‹HÂ°(products)âˆ’âˆ‘Î”fâ€‹HÂ°(reactants
= 2 Ã— Î”_{f}HÂ° (NO_{2}) – Î”_{f}HÂ° (N_{2}O_{4})
= 2(mol) Ã— 33.2(kJ mol^{-1}) -1(mol) Ã— 9.16(kJ mol^{-1})
= +57.24 kJ
Î”GÂ° = Î”HÂ° – TÎ”SÂ°
57.24(kJ) – 298(K) Ã— 175.8 Ã— 10^{-3} (kJ K^{-1})
= +4.85 kJ.
Because Î”GÂ° is positive, the reaction is non-spontaneous at 298 K.
The temperature at which the reaction changes from spontaneous to non-spontaneous is given by,
T = Î”HÂ° / Î”SÂ°
= 57.24(kJ) / 0.1758(kJ K^{-1})
= 325.6 K
Because Î”HÂ° and Î”SÂ° are both positive, the reaction will be spontaneous at high temperature.
The reaction will be spontaneous above 325.6 K.
Problem 5: Determine K_{p} for the reaction,
2SO_{2}â€‹(g) + O_{2}â€‹(g) âŸ¶ 2SO_{3â€‹}(g)
is 7.1 Ã— 10^{24} at 298 K. Calculate Î”GÂ° for the reaction (R = 8.314 JK^{-1} mol^{-1}).
Solution:
Î”GÂ° = -2.303RT log_{10} K_{p}
K_{p} = 7.1 Ã— 10^{24}
R = 8.314 JK^{-1} mol^{-1}
T = 298K
Hence,
Î”GÂ° = -2.303 Ã— 8.314 Ã— 10^{-3} (kJ K^{-1} mol^{-1}) Ã— log_{10} (7.1 Ã— 10^{24})
= -141.8 kJ mol^{-1}
Problem statement 6: Calculate K_{p} for the reaction at 513 K,
2NOCl (g) âŸ¶ 2NO(g) + Cl_{2}â€‹(g)
with Î”GÂ° = 17.38 kJ mol^{-1}.
Solution:
Î”GÂ° = -2.303 RT log_{10} K_{p}
Î”GÂ° = 17.38 kJ mol^{-1}
R = 8.314 J K^{-1} mol^{-1}
T = 513K
Hence,
log_{10} k_{p} = – Î”GÂ° / 2.303 RT
= – (17380(J mol^{-1}) / 2.303 Ã— 8.314 ( J K^{-1} mol^{-1}) Ã— 513(K))
= -1.769
Hence, K_{p} = antilog(-1.769)
= 0.017
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