- +1-530-264-8006
- info@tutorsglobe.com

18,76,764

Questions

Asked

21,311

Experts

9,67,568

Questions

Answered

Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

Submit Assignment
## Free energy functions, Chemistry tutorial

:IntroductionWe have studied the application of the first law of thermodynamics in computing the enthalpy and internal energy changes of the reactions. These energy computations don't state us whether a given reaction is feasible or not. The concept of entropy was introduced via the second law of thermodynamics; we studied that the net entropy value of the system and the surroundings can assist us in deciding the spontaneity of the reaction. However the main difficulty is that it is not for all time possible to predict the entropy change of the Surroundings.

:Spontaneous and Non-Spontaneous processesIn this part, we will describe how spontaneous and non-spontaneous processes differ. Let us try to comprehend what a spontaneous process is. We are familiar that:

1) Water flows down the hill spontaneously.

2) A gas expands spontaneously to vacuum.

3) The heat is conducted spontaneously from the hot end of the metal bar to the colder end till the temperature of the bar is similar all through

4) A gas diffuses spontaneously to the other gas.

However, the reverse of the above changes doesn't take place spontaneously. All the natural processes take place spontaneously. Therefore we can state that a spontaneous or a natural process takes place in a system by itself. No external force is needed to make the process continue. On the other hand a non-spontaneous process will not take place unless some external force is continuously applied. The chemist is for all time interested in knowing whether, under a given set of conditions, a reaction or a process is feasible or not. According to the second law of thermodynamics, dS ≥ dq/T in which the equality refers to a system undergoing the reversible process and the inequality refers to the irreversible process. For an isolated system for which dq = 0, the above equation is reduced to dS ≥ 0. In an isolated system, the irreversible change is for all time spontaneous. This is because in such systems, no external force can interact by the system. Therefore, the tendency of entropy of an isolated system to increase can be employed as a criterion for the spontaneous change. Is this for all time trueΔ Let us look at this question. Water freezes to crystalline ice spontaneously at around 263 K. Ice is in a more ordered state than water and the entropy reduces in freezing. How do we illustrate the above processΔ The answer to this question lies in the fact that we should for all time consider the net entropy change, that is, the entropy change of the universe. This entropy change is equivalent to the sum of the entropy changes of the system and the surroundings.

ΔS

_{Total}= ΔS_{Universe}= ΔS_{System }+ ΔS_{surroundings }The equation above is very difficult to apply for testing the spontaneity of a process. This is so as for computing the total entropy change, we have to calculate ΔS for the surroundings as well. In most of the cases, such computations are very difficult and at times not even practical. Thus, it will be beneficial if we can redevelop the criteria for spontaneity in such a way that only changes in the properties of the system are considered.

:Helmholtz free energy and Gibbs free energyHelmholtz free energy (A) and Gibbs free energy (G) are stated by the expressions:

A = U - TS

And, G = H - TS

As, H = U + pV

G = U + pV - TS

As U, H, p, S, V and T are the state functions, A and G are as well based only on the state of the system. In simpler words, a system in a particular state has definite values of A and G.

Physical importance of A and G:

On differentiating the equation A = U - TS, we get

dA = dU - TdS - SdT

At constant temperature (dT = 0), we get

dA = dU - TdS

We are familiar that for a reversible process:

TdS = dq

_{rev }As well for a reversible process provides:

dU = dq

_{nv }+ dw_{rev }Replacing the values for TdS and dU in Eq. (dA = dU - TdS), we get,

dA = (dq

_{rev}+ dw_{rev}) - dq_{rev }Or dA = dw

_{rev }Or - dA = - dw

_{rev}As the procedure is carried out reversibly, - dw

_{rev }symbolizes the maximum work done via the system. This is a significant conclusion which defines that the change in Helmholtz; free energy is equivalent to the amount of reversible work done on the system; or reduction in the Helmholtz free energy (- dA) is the maximum amount of work which can be obtained from the system (-dw_{rev}) throughout the given change. As a outcome, the function A is at times as well termed to as the work function.In a similar manner, the differentiation of equation G = U + pV - TS yields,

dG = dU + pdV+ Vdp - SdT - TdS

At constant temperature (dT = 0) and pressure (dp = 0), the equation above is reduced to:

dG = dU + pdV - TdS

Once again if the procedure is carried out reversibly, the by using the equation (dU = dq

_{rev }+ dw_{rev}) and TdS = dq_{rev}, we getdG = dq

^{rev}+ dw^{rev}+ pdV - dq^{rev}Or dG = dw

_{rev}+ pdVNow dw

_{rev }comprises of expansion work (- pdV) and some; other type of work know as the helpful or the total work dw_{net }done on the system. Substituting dw_{rev}by - pdV + dwnet in the above equation, we obtaindG = - pdV + dw

_{net }+ pdVOr dG = dw

_{net }Or - dG = - dw

_{net}Therefore decrease in the Gibbs free energy (- dG) is a measure of the maximum helpful work which can be obtained from the system at constant temperature and pressure. Most of the experiments in the laboratory are taken out under such conditions. Therefore the property 'G' or the change related by it (ΔG) is very significant.

:Changes in A and GWhenever a system goes from state 1 to state 2 at constant temperature, the change in A can be obtained from the equation (dA = dU - TdS) via integration between the limits A

_{1}→ A_{2}, U_{1 }→ U_{2}and S_{1}→ S_{2}_{A1}∫^{A2}dA =_{U1}∫^{U2}dU - T_{S1}∫^{S2}dSOr A

_{2 }- A_{1}= U_{2 }- U_{1}- T (S_{2}- S_{1})Or ΔA = ΔU - TΔS

The change in G

_{2}if a system goes from a state 1 to state 2 can be obtained by integrating dG = dU + pdV - TdS at constant temperature and pressure among the limits G_{1}→ G_{2}, U_{1}→ U_{2}, V_{1}→_{V2 }and S_{1}→ S_{2}._{G1}∫^{G2}dG =_{U1}∫^{U2 }dU + P_{V1}∫^{V2}dV - T_{S1}∫^{S2}dSOr G

_{2}- G_{1}= U_{2}- U_{1}+ p(V_{2}- V_{1}) - T(S_{2}- S_{1})Or ΔG = ΔU + pΔV - TΔS

We are familiar that at constant pressure, ΔU + pΔV = ΔH

Therefore, ΔG = ΔH - TΔS

Variation of A and G with Temperature:

From the equation dA = dU - TdS - SdT

As TdS = dq

dA = dU - SdT - dq

Replacing from the first law of thermodynamics, we have

dU = dq - pdV in the equation dA = dU - SdT - dq , we obtain

dA = dq - pdV - SdT - dq

Or dA = - pdV - SdT

From the equation above at constant volume (dV = 0), we encompass

(∂A)

_{V }= - S(∂T)_{V}Or (∂A/∂T)

_{V}= -SAt constant temperature (dT = 0), the equation (dA = - pdV - SdT) is reduced to

(∂A)

_{T}= - p(∂V)_{T}Or (∂A/∂V)

_{V}= - pThe equations alike to (∂A/∂T)

_{V}= - S and (∂A/∂V)_{V}= - p can be obtained in the case of Gibbs free energy. We have the equation dG = dU + pdV+ Vdp - SdT - TdSAs dq = TdS from the Second law of thermodynamics and dU = dq - pdV from the first law of thermodynamics,

dG = dq - pdV + pdV + Vdp - SdT - dq

Or dG = Vdp - SdT

At constant pressure (dp = 0), the equation above is reduced to,

(∂G)

_{p }= -S(∂T)_{p }Or (∂G/∂T)

_{P}= - SAt constant temperature (dT = 0), we have from the equation dG = Vdp - SdT

(∂G)

_{T}= V(∂p)_{T}Or (∂G/∂p)

_{P }= VIf G

_{1}and G_{2 }are the free energies of the system in the initial and final states, correspondingly, then it constant temperature, the free energy change (ΔG) is represented by integrating the equation (∂G)_{T}= V(∂p)_{T}ΔG =

_{G1}∫^{G2 }dG =_{P1}∫^{P2}V dpHere, p

_{1}and p_{2}are the initial and the final pressures, correspondingly.For n mol of an ideal gas

pV = nRT or V = nRT/p

Thus, ΔG =

_{G1}∫^{G2}nRT (dp/p) = nRT_{P1}∫^{P2}(dp/p)= nRT ln (P

_{2}/P_{1}) = 2.303 nRT log (P_{2}/P_{1})As pressure is inversely proportional to the volume for an ideal gas at constant temperature, we encompass,

ΔG = 2.303 nRT log (P

_{2}/P_{1}) = 2.303 nRT log (V_{1}/V_{2})Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online Chemistry tutoring. Chat with us or submit request at info@tutorsglobe.com