Content Curator | Updated On - May 8, 2024
Helmholtz free energy is a thermodynamic term that measures the work of a closed system with constant temperature and volume. The Helmholtz function is the second thermodynamic potential that is used to calculate work done in a closed system with constant temperature and volume. Internal energy, enthalpy, and Gibbs free energy are the other three thermodynamic potentials.
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Key Terms: Helmholtz Free Energy, Helmholtz Free Energy Formula, Gibbs Free Energy, Helmholtz Free Energy Partition Function, Joules, Kelvin
What is Helmholtz Free Energy
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Helmholtz free energy is a thermodynamic term that measures the work of a closed system with constant temperature and volume. It may be expressed as the following equation:
| F = U − TS |
Where,
- F → The Helmholtz free energy in Joules
- U → System's internal energy in Joules.
- T → Absolute temperature of the environment in Kelvin.
- S → system's entropy in joules per Kelvin.
What is Helmholtz Function?The Helmholtz function is a thermodynamic function defined as the function's decline and equal to the greatest amount of work available during a reversible isothermal process. Which of the following is correct?[A = Helmholtz function (Helmholtz Free Energy), G = Gibbs function (Gibbs Free Energy), U = Internal energy, H = Enthalpy, T = Absolute temperature, S = Entropy]
There are eight properties of a system, namely pressure (p), Volume (V), temperature (T), internal energy (u), enthalpy (h), entropy (s), Helmholtz function (f) and Gibbs function (g) and h, f and g are referred to as thermodynamic potentials.
Gibbs free energy function (G)
Gibbs free energy functions is a property comprises enthalpy, temperature, and entropy. Gibbs free energy is a measure of the amount of energy available to do work in an isothermal and isobaric (constant temperature and pressure) thermodynamic system.
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Significance of Free Energy
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The significance of Free Energy is elaborated below:
- When compared to closed systems, real-life-based open systems exist in thermodynamics as the process of exchanging heat and work with respect to the environment. For example, biological systems that achieve an intrinsic reduction in entropy as they age.
- The Helmholtz function is the second thermodynamic potential that is used to calculate work done in a closed system with constant temperature and volume. Internal energy, enthalpy, and Gibbs free energy are the other three thermodynamic potentials.
- To maintain the integrity of thermodynamic rules in open systems, parameters like energy dimensions, known as thermodynamic potentials, are specified in the system. These formulae use Helmholtz and Gibbs Free Energy references to explain the overall work done in the system.
Gibbs Free Energy
Gibbs' Free Energy is the energy available in a thermodynamically-closed system at constant pressure and temperature. The key difference between Gibbs and Helmholtz Free Energy is that Gibbs free energy is defined under constant pressure whereas Helmholtz free energy is defined under constant volume.
When a system is disconnected from its heat reservoir, it is no longer a component of the system, and therefore the energy change (U) may be expressed as:
| ΔU = Q - W |
Where,
- Q → Amount of heat absorbed
- W → Amount of work done
They are not state functions and are path-dependent. However, in the case of a reversible isothermal process, heat would be directly proportional to maximal work.
| -Wmax = ΔU - TΔS |
Where,
- S → Entropy change in a system kept at a constant temperature, such as a battery.
- The entropy change of the surroundings is not addressed in this system, therefore the U, T, and S are state functions.
Solved Examples on Gibbs Free EnergyExample: Estimate the temperature where ΔG = 0 for the following reaction: (Given: ΔH = -176 kJ and ΔS = -284.5 J/K) Solution: NH3(g) + HCl(g) ---> NH4Cl(s) Using the equation above, we seperate T to be on one side, giving us (G+H)/S = T. Then we subsitute the values in the equation. As how the question states it, Beware of units! 1 J = 1000kJ, therefore -284.5 J/K/1000 gives us -0.2845 kJ. By subsituting this value in the equation for S, you should be able to obtain the answer 619K. Example: The standard Gibbs free energy ΔGo is related to equilibrium constant KP as: Solution: For a reaction, the Gibb's free energy equation is: ΔG=ΔG0+RTlnKP At equilibrium, ΔG=0 ∴ΔG0=−RTlnKP ∴KP=e−ΔG0/RT |
Relationship Between Helmholtz Free Energy & Gibbs Free Energy
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Relationship Between Helmholtz Free Energy & Gibbs Free Energy is that:
“Gibbs Free Energy is based on constant pressure, while Helmholtz Free Energy is based on constant volume.”
The overall work of the preceding equation can be regarded as smaller than that of the ideal maximum of W ≤ −ΔF. The entire process can be reversible, thereby restricting the case.
When the Helmholtz free energy reaches its lowest value, the system enters an equilibrium where no more work can be derived. As a result, this maximum entropy equilibrium condition is required for the least Helmholtz free energy at a constant temperature.
Helmholtz Free Energy concept occurs within a container with solid walls. Instead of a steady temperature, the whole process is carried out under continuous pressure. A vegetable growing in a garden occurs under normal air pressure and at varying temperatures.
| -Wmax = ΔU – TΔS |
If the system is a natural process that involves a large volume change (such as steaming water), the work done by atmospheric water vapour to push through the surrounding pressure is defined as PV, which may be added to the total work done:
| Wmax = W’max + PΔV (atmospheric work) |
Where,
- W'max → Greatest work done in a system with constant Temperature and Pressure
- PV → Atmospheric Work
When PV is substituted, we get:
| -Wmax = ΔU + PΔV - TΔS |
which leads to the basis of Gibbs Energy:
| G = U + PV - TS |
which can alternatively be described as the following equation for isothermal systems under constant pressure:
| ΔG = ΔU + PΔV – TΔS |
Except for the atmospheric work done, the negative number reflects the reversible maximum work that can be taken from the system. When the Gibbs energy hits its minimum, the system enters an equilibrium state from which no additional work can be taken, similar to the Helmholtz Free Energy scenario.
Entropy can fall further here as long as it is balanced by the combined value of ΔU + PΔV. As a result, for open systems with constant temperature and pressure, the direction for spontaneous change is in the state of lower free energy, which is commonly depicted as a ball rolling downhill.
Applications of Helmholtz Free Energy
Real World Example: Relation Between Helmholtz Free Energy & Gibbs Free Energy
| A real-world example is a car battery that has the highest Gibbs Energy when charged and the lowest Gibbs Energy when discharged. The charge that can be collected from the battery under constant temperature and pressure may be used to determine the total work between the two states. Gibbs Free Energy is expressed as: G = U - TS + PV Where,
The Gibbs free energy change in the process of glucose oxidation may be estimated as 2870 kJ, which is critical for energy production in all living cells. |
Read More: Ionization Enthalpy
Difference between Helmholtz Free Energy and Gibbs Free Energy
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The differences between Helmholtz Free Energy and Gibbs Free Energy are tabulated below.
| Helmholtz Free Energy | Gibbs Free Energy |
|---|---|
| Helmholtz free energy is defined as the beneficial work produced by a certain system. | Gibbs free energy is defined as the greatest reversible work gained from a certain system. |
| It is the amount of energy necessary to maintain a system at a constant temperature and volume. | It is the amount of energy needed to establish a system with constant pressure and temperature. |
| Helmholtz free energy is less commonly used since the volume of the system should remain constant. | Because the pressure in the system remains constant, Gibbs free energy is more useful. |
Applications of Helmholtz Free Energy
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Some of the applications of Helmholtz free energy are as follows.
- In the equation of state: Helmholtz function is used to characterize pure fluids with great precision (such as industrial refrigerants) as a sum of an ideal gas and residual terms.
- In auto-encoder: An auto-encoder is a type of artificial neural network that is used to efficiently encode data. Helmholtz energy is utilized to calculate the total coding cost and reconstructed code in this case.
- Helmholtz free energy is used in explosives research as explosive reactions are caused by their nature to induce pressure changes.
- Helmholtz free energy defines fundamental equations of the state of pure substances.
Things to Remember
- Helmholtz free energy is a thermodynamic term that measures the work of a closed system with constant temperature and volume.
- Internal energy, enthalpy, Gibbs free energy and Helmholtz free energy are the thermodynamic potentials.
- The Helmholtz function is a thermodynamic function defined as the function's decline and equal to the greatest amount of work available during a reversible isothermal process.
- Helmholtz free energy is given by the equation: F = U − TS
- When the Helmholtz free energy reaches its lowest value, the system enters an equilibrium where no more work can be derived.
- The Gibbs free energy change in the process of glucose oxidation may be estimated as 2870 kJ, which is critical for energy production in all living cells.
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Sample Questions
Ques. For a reaction at 298 K
2A + B → C
ΔH = 400 KJ mol-1 and AS = 0.2 KJ K-1 mol-1
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range? (2 marks)
Ans. As per the Gibbs Helmholtz equation:
ΔG = Δ H- TΔ S
For ΔG=0 ; ΔH=TΔS or T=ΔH/ΔS
T = (400 KJ mol-1)/(0.2 KJ K-1 mol-1) = 2000 k
Thus, the reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.
Ques. What is Gibbs Helmholtz equation? (2 marks)
Ans. ΔG = ΔH – TΔS
Where ΔG = free energy change.
ΔH = enthalpy change.
ΔS = entropy change.
Ques. Give reason for the following:
(a)Neither q nor w is a state function but q + w is a state function.
(b)A real crystal has more entropy than an ideal crystal. (2 marks)
Ans. (a) q + w = Δu
As Δu is a state function hence, q + w is a state function.
(b) A real crystal has some disorder due to the presence of defects in its structural arrangement whereas an ideal crystal does not have any disorder. Hence, a real crystal has more entropy than an ideal crystal.
Ques. (a)What is a spontaneous process? Mention the conditions for a reaction to be spontaneous at constant temperature and pressure.
(b) Discuss the effect of temperature on the spontaneity of an exothermic reaction. (2 marks)
Ans. (a) A process is said to be spontaneous if it takes place on its own or under some condition.
ΔG gives criteria for spontaneity at constant temperature and pressure.
(b) If the temperature is so high that TΔS > ΔH in magnitude, ΔG will be positive and the process will be non-spontaneous.
If the temperature is made low so that TΔS < ΔH in magnitude, ΔG will be negative and the process will be spontaneous.
Ques. Why standard entropy of an elementary substance is not zero whereas the standard enthalpy of formation is taken as zero? (2 marks)
Under what conditions will the reaction occur, if
(i) both ΔH and ΔS are positive
(ii) both ΔH and ΔS are negative
Ans. (a) A substance has a perfectly ordered arrangement of its constituent particles only at absolute zero. When the element forms from itself, this means no heat change.
Thus, Δf H = 0
(i) If both AH and AS are positive ΔG can be – ve only if TΔS > ΔH in magnitude. Thus, the temperature should be high.
(ii) If both AH and AS are negative ΔG can be negative only if TΔS < ΔH in magnitude. Thus, the value of T should be low.
Ques. Why standard entropy of an elementary substance is not zero whereas the standard enthalpy of formation is taken as zero? (2 marks)
Ans. A substance has a perfectly ordered arrangement only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from its constituents means no heat change.
Ques. Many thermodynamically feasible reactions do not occur under ordinary conditions. Why? (2 marks)
Ans. Under ordinary conditions, the average energy of the reactants may be less than the threshold energy. They require some activation energy to initiate the reaction.
Ques. Define the following: (2 marks)
(i) First law of thermodynamics.
(ii) Standard enthalpy of formation.
Ans. (i) First law of thermodynamics: It states that energy can neither be created nor be destroyed. The energy of an isolated system is constant.
Δu = q + w
(ii) It is defined as the amount of heat evolved or absorbed when one mole of the compound is formed from its constituent elements in their standard states.
Ques. 1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation C(graphite) + 02 (g) → C02 (g) During the reaction, the temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm? (3 marks)
Ans. Suppose q is the quantity of heat from the reaction mixture and Cv is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.
q = Cv/ΔT
The quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
q = – Cv x ΔT
= – 20.7 kJ/ K x (299 – 298) K
= – 20.7 kJ
(Here, the negative sign indicates the exothermic nature of the reaction). Thus, AU for the combustion of the 1 g of graphite = – 20.7 KJ K-1. For combustion of 1 mol of graphite,
\(= \frac{12.0g\;mol^{-1} \times(-2.07 KJ)}{1g}\)
\(= -2.48 \times 10^2 KJ \; mol^{-1}. \;since \; \triangle ng = 0,\)
\(\triangle H=\triangle U = -2.48 \times 10^2\;KJ \; mol^{-1}\)
Previous Year Questions Based on Thermodynamics
- Thermodynamic processes are indicated in the following diagram… [NEET 2017]
- A thermodynamic system undergoes a cyclic process… [KCET 2019]
- A thermodynamical system is changed from state… [BITSAT 2007, BITSAT 2013]
- In an open system… [COMEDK UGET 2008]
- Heat of neutralization of any strong acid and strong base… [COMEDK UGET 2008]
- An intensive property is… [COMEDK UGET 2011]
- For an adiabatic change in a system, the condition which is applicable will be… [COMEDK UGET 2015]
- An endothermic reaction is found to have +ve entropy change… [COMEDK UGET 2015]
- For one mole of NaCl(s) the lattice enthalpy is… [COMEDK UGET 2015]
- For a spontaneous reaction, free energy… [COMEDK UGET 2006]
- In the reaction, C(s)+O2(g)→CO2(g)… [COMEDK UGET 2006]
- A spontaneous reaction is impossible if… [COMEDK UGET 2010]
- Gibb's free energy G is defined as… [COMEDK UGET 2010]
- The enthalpy of neutralization of HCI and NaOH… [COMEDK UGET 2012]
- The temperature of one mole of an ideal gas increases from… [COMEDK UGET 2012]
- The heat of formation of H2O(l) is -286 kJ… [COMEDK UGET 2012]
- A reaction is spontaneous at all temperatures when… [COMEDK UGET 2009]
- Which of the following is an intensive property… [COMEDK UGET 2009]
- A process is spontaneous at a given temperature if… [COMEDK UGET 2007]
- The enthalpy of the formation of CO2 and H2O are… [UPSEE 2017]
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