English, asked by Mrparadox, 11 months ago

what is entropy, and relation between gibbs energy​

Answers

Answered by singhishaan010
0

Answer:

In statistical mechanics, entropy is an extensive property of a thermodynamic system. It is closely related to the number Ω of microscopic configurations that are consistent with the macroscopic quantities that characterize the system

Gibbs free energy combines enthalpy and entropy into a single value.

Gibbs free energy is the energy associated with a chemical reaction that can do useful work. It equals the enthalpy minus the product of the temperature and entropy of the system.

G = H – TS

At constant temperature

ΔG = ΔH – TΔS

ΔG predicts the direction of a chemical reaction. If ΔG is negative, then the reaction is spontaneous. If ΔG is positive, then the reaction is non-spontaneous.

Answered by rehan1741
0

Answer:

Gibbs energy is the energy of a chemical reaction that can be used to do non-mechanical work. Gibbs Energy is described as

G=H−TS(4.6.1)

Where H is enthalpy, T is temperature, and S is entropy. ΔG is used to predict spontaneity within a system by

ΔGsys=ΔHsys–Δ(TS)sys(4.6.2)

Gibbs energy was developed in the 1870’s by Josiah Willard Gibbs. He originally termed this energy as the “available energy” in a system. His paper published in 1873, “Graphical Methods in the Thermodynamics of Fluids,” outlined how his equation could predict the behavior of systems when they are combined.

At a constant temperature and pressure, the Gibbs Energy of a system can be described as

ΔGsys=ΔHsys–TΔSsys(4.6.3)

This equation can be used to determine the spontaneity of the process.

If ΔGsys ≤ 0, the process is spontaneous.

If ΔGsys = 0, the system is at equilibrium.

If ΔGsys > 0, the process is not spontaneous.

Furthermore,

If ΔGsys < 0, the process is exergonic

If ΔGsys > 0, the process is endergonic.

Gibbs Energy is a useful tool to describe in what manner the reaction is conducted. If ΔH >> TΔS, the reaction is enthalpy driven. However, if TΔS >> ΔH, the reaction is driven by entropy. The Clausius-Clapeyron Equation is an application derived from Gibb's energy:

ln(P2P1)=ΔvapHR(T2−T1T2T1)(4.6.4)

Another important application of Gibb's energy is the Maxwell relations (also available in a link at the end of the wiki page.)

Temperature & Pressure Dependence

When the pressure & temperature of a reaction are not held constant,

G=H−TS(4.6.5)

For an infinitesimal process,

ΔG=ΔH–Δ(TS)(4.6.6)

ΔG=ΔH–TΔS–SΔT(4.6.7)

For a reaction where temperature is held constant,

ΔG=ΔH–TΔS(4.6.8)

From the First Law of Thermodynamics, we know

H=U+PV(4.6.9)

ΔH=ΔU+PΔV+VΔP(4.6.10)

Since

ΔU=TΔS–PΔV(4.6.11)

We find that

ΔG=VΔP−SΔT(4.6.12)

showing the obvious dependence of ΔG on temperature and pressure. To observe the change in Gibbs energy due to temperature change alone (pressure held constant) the equation becomes

ΔG=−SΔT(4.6.13)

Solving for S and plugging it into Eq. (1), the Gibbs-Helmholtz Equation is found:

(∂(ΔG/T)∂T)p=−ΔHT2(4.6.14)

The Gibbs-Helmholtz Equation is very important because it relates the change in Gibbs energy to its temperature dependence, and the position of equilibrium to change in enthalpy.

To observe the change in Gibbs energy due to pressure change alone (temperature held constant) the equation becomes

ΔG=VΔP(4.6.15)

If the gas is assumed to be ideal then

ΔG=nRTln(P2P1)(4.6.16)

for an initial and final pressure (P1 and P2) at a constant T.

Standard-State Free Energy of Formation

Gibbs Energy is defined as a state function (a property that depends only on conditions describing the system, not how the change occurs as in a path function.) This is because each component of the equation (H,T, and S) are all state functions. Therefore, we can know the change in Gibbs energy without knowing every detail of the process. In a process that takes place at constant temperature and pressure (298 K, 1 atm) the standard molar free energy of formation can be determined by the change in free energy from the reactants and products. Using predetermined values, Eqn. (17) can be used

ΔGo)f=∑(CoefficientproductsG∘products)−∑(CoefficientreactantsG∘reactants)(4.6.17)

Standard-State Free Energy of Reaction

Gibbs Energy can be found at standard-state conditions using

ΔG°=ΔH°−TΔS°(4.6.18)

ΔH° and ΔS° values can be found in the appendix of any general chemistry textbook, or using this link.

Free Energy of Reaction

Gibbs energy can be found at any conditions by relating it to the standard-state free energy of reaction, using

ΔG=ΔG°+RTlnQ(4.6.19)

Where Q is the reaction quotient. Very rarely does chemistry actually occur at the given "standard-state" conditions. Using the above equation and standard-state values, chemists can determine the overall Gibb's energy for the system, regardless of the conditions.

References

Olmsted, J. Chemistry, 4th ed.; John Wiley & Sons, Inc: Hoboken 2006.

Chang, R. Physical Chemistry for the Biosciences; University Science Books: Sausalito, 2005.

Christian, S.D. Gibbs-Duhem equation and free energy calculations. J. Chem. Educ. 1962, 39 (10) 521-524.

Porter, S.K. The volume-entropy-energy surface of J.W. Gibbs. J. Chem Educ. 1971, 48(4) 231-234.

Problems

Calculate the standard-state free energy of formation for the H2O2(l) from H2 and O2, given the following values:

ΔfG (H2): 0 kJ/mol

ΔfG (O2): 0 kJ/mol

ΔfG (H2O2(l)): -120.4 kJ/mol

Hint: (product*coefficient) - [(Reactant1*coefficient)+(Reactant2*coefficient)]

Consider the following reaction:

CaCO3CaO+CO2(4.6.20)

At what temperature will this reaction become favorable? Note: Assume ΔHr°and ΔSr° are temperature independent.

Given values:

CaCO3

-1206.9 kJ/mol

92.9 J/K mol

CaO

-635.6 kJ/mol

39.8 J/K mol

CO2

-393.5 kJ/mol

213.6 J/K mol

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