Posts Tagged ‘worked examples’

gibbs helmholtz cartoon

History

The Gibbs-Helmholtz equation was first deduced by the German physicist Hermann von Helmholtz in his groundbreaking 1882 paper “Die Thermodynamik chemischer Vorgänge” (On the Thermodynamics of Chemical Processes). In it, he introduced the concept of free energy (freie Energie) and used the equation to demonstrate that free energy – not heat production – was the driver of spontaneous change in isothermal chemical reactions, thereby overthrowing the famously incorrect Thomsen-Berthelot principle.

Although Gibbs was first to state the relations A = U – TS and G = U + PV – TS, he did not explicitly state the Gibbs-Helmholtz equation, nor did he explore its chemical significance. So the honors for this equation really belong more to Helmholtz than to Gibbs.

But from the larger historical perspective, both of these gentlemen can rightly be considered the founders of chemical thermodynamics – Gibbs for his hugely long and insanely difficult treatise “On the Equilibrium of Heterogeneous Substances” (1875-1878), and Helmholtz for his landmark paper referred to above. These works had a significant influence on the development of physical chemistry.

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Derivation

The confusing thing about the Gibbs-Helmholtz equation is that it comes in three different versions, but most physical chemistry texts don’t say why. This is not helpful to students. Chemical thermodynamics is difficult enough already, so CarnotCycle will begin by giving the reason.

It just so happens that the form of calculus used in the Gibbs-Helmholtz equation has the following property:

For a thermodynamic state function (f) and its natural variables (x,y)

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If we choose as our state function (f) the Gibbs Free Energy G and assign its natural variables, temperature T to (x) and pressure P to (y), we obtain:

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Since (∂G/∂T)P = –S, and G ≡ H – TS

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This is the Gibbs-Helmholtz equation. If we apply this equation to the initial and final states of a process occurring at constant temperature and pressure, and take the difference, we obtain:

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where ΔH is the enthalpy change of a process taking place in a closed system capable of PV work; the three equivalent versions of the equation are determined by the properties of calculus:

gh05  (1)

gh06 (2)

gh07  (3)

A further useful relation can be derived from (2) and (3) using the equation ΔG° = –RTlnKp for a gas reaction where each of the reactants and products is in the standard state of 1 atm pressure.

Substituting –RlnKp for ΔG°/T in (2) and (3) yields

gh11  (4)

gh12  (5)

These are equivalent forms of the van ‘t Hoff equation, named after the Dutch physical chemist and first winner of the Nobel Prize in Chemistry, J.H. van ‘t Hoff (1852-1911). Approximate integration yields

gh13  (6)

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Applications of the Gibbs-Helmholtz equation

1. Calculate ΔHrxn from ΔG and its variation with temperature at constant pressure
This application of (1) is useful particularly in relation to reversible reactions in electrochemical cells, where ΔG identifies with the electrical work done –nFE. Scroll down to see the worked example GH1

2. Calculate ΔGrxn for a reaction at a temperature other than 298K
ΔH usually varies slowly with temperature, and can with reasonable accuracy be regarded as constant. Integration of (2) or (3) enables you to compute ΔGrxn for a constant-pressure process at a temperature T2 from a knowledge of ΔG and ΔH at temperature T1. Scroll down to see the worked example GH2

3. Calculate the effect of a temperature change on the equilibrium constant Kp
ΔH usually varies slowly with temperature, and can with reasonable accuracy be regarded as constant. The integrated van ‘t Hoff equation (6) allows the equilibrium constant Kp at T2 to be calculated with knowledge of Kp and ΔH° at T1. Scroll down to see the worked example GH3

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Insight: The Φ function and the meaning of –ΔG/T

By 1897, Hermann von Helmholtz was dead and Max Plank was professor of theoretical physics at the University of Berlin where he published Treatise on Thermodynamics, a popular textbook which ran to several editions. In it, Planck introduced the Φ function (originally deduced by François Massieu in 1869) as a measure of chemical stability:

planck function

The pencilled note in my German copy – found in a charity sale at a downtown church – correctly identifies Φ with –ξ/T, which in modern notation is –G/T

 

S is the entropy of the system, and (U+pV)/T is the enthalpy H of the system divided by its temperature T. Since G ≡ H – TS, we can immediately identify Φ with –G/T.

Planck’s formula indicates that Φ tends to be large when S is large and H is small, i.e. when the energy levels are closely spaced and the ground level is low – the criteria for chemical stability.

The Φ function gets even more interesting when one considers the meaning of ΔΦ.

ΔΦ = –ΔG/T = –ΔH/T + ΔS = ΔSsurroundings + ΔSsystem = ΔSuniverse

ΔΦ and –ΔG/T equate to the increase in the entropy of the universe: a measure of the ultimate driving force behind chemical reactions. The larger the value, the more strongly the reaction will want to go.

A direct association between –ΔG/T and the equilibrium constant K is thus implied, and this can be confirmed by rearranging the relation ΔG° = –RTlnKp to: –ΔG°/T = RlnKp

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Worked Example GH1

An electrochemical cell has the following half reactions:
Anode (oxidation):  Ag(s) + Cl → AgCl(s) + 1e
Cathode (reduction): ½Hg2Cl2(s) + 1e → Hg(l) + Cl

The EMF of the cell is +0.0455 volt at 298K and the temperature coefficient is +3.38 x 10-4 volt per kelvin. Calculate the enthalpy of the cell reaction, taking the faraday (F) as 96,500 coulombs.

Strategy

Use version 1 of the Gibbs-Helmholtz equation

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Substitute for ΔG using the relation ΔG = –nFE

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Calculation

[note that if the EMF is positive, the reaction proceeds spontaneously in the direction shown in the half reactions. If the EMF is negative, the reaction goes in the opposite direction]

The complete cell reaction for one 1 faraday is:
Ag(s) + ½Hg2Cl2(s) → AgCl(s) + Hg(l)
Each mole of silver transfers one mole of electrons (1e) to one mole of Cl ions. So n = 1.
F = 96,500 C
E = 0.0455 V
T = 298 K
(∂E/∂T)P = 3.38 x 10-4 VK-1

ΔHrxn = –1 x 96,500 (0.0455 – 298 (3.38 x 10-4)) joules
Dimensions check: remember that V= J/C, so C x V = J

ΔHrxn = 5329 J = 5.329 kJ

[note that this electrochemical cell makes use of a spontaneous endothermic reaction]

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Insight: The effect of temperature on EMF

From version 1 of the Gibbs-Helmholtz equation

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and the relations ΔG = –nFE and (∂ΔG/∂T)P = –ΔS, it can be seen that

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For many redox reactions that are used to power electrochemical cells, ΔSrxn is typically small (less than 50 JK-1). As a result (∂E/∂T)P is usually in the 10-4 to 10-5 range, and hence electrochemical cells are relatively insensitive to temperature.

winter text

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Worked Example GH2

The Haber Process for the production of ammonia is one of the most important industrial processes: N2(g) + 3H2(g) = 2NH3(g)
ΔG°(298K) = –33.3 kJ
ΔH°(298K) = –92.4kJ
Calculate ΔG° at 500K

Strategy

Use version 3 of the Gibbs-Helmholtz equation

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Making the assumption that ΔH° remains approximately constant, perform integration

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Solve for ΔG°(T2)

Calculation

ΔG°(T1) = –33.3 kJ
ΔH° = –92.4 kJ
T2 = 500K
T1 = 298K

Inserting these values into the integrated equation yields the result ΔG°(500K) ≈ 6.76 kJ. Compared with the negative value of ΔG° at 298K, the small positive value of ΔG° at 500K shows that the reaction has just become unfeasible at this temperature, pressure remaining constant at 1 atm.

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Insight: Reading the Haber process equation

N2(g) + 3H2(g) = 2NH3(g) ΔH°(298K) = –92.4 kJ

The Haber process is exothermic (negative ΔH) and results in a halving of volume (negative ΔS). Since ΔG = ΔH – TΔS, increasing the temperature will drive ΔG in a positive direction, leading to an upper temperature limit on reaction feasibility.

Le Châtelier’s principle shows that the Haber process is thermodynamically favored by low temperature and high pressure. In practice however a compromise has to be struck, since low temperature slows the rate at which equilibrium is achieved while high pressure increases the cost of equipment and maintenance.

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Worked Example GH3

The Haber Process for the production of ammonia is one of the most important industrial processes:
N2(g) + 3H2(g) = 2NH3(g) ΔH°(298K) = –92.4 kJ
The equilibrium constant KP at 298K is 6.73 x 105. Calculate KP at 400K.

Strategy

Use the approximate integral of the van ‘t Hoff equation (6) to solve for KP(T2)

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gh16

Calculation

KP(T1) = 6.73 x 105, ln KP(T1) = 13.42
ΔH°(298K) = –92400 J (mol-1)
R = 8.314 J K-1 mol-1
T2 = 400K
T1 = 298K

Inserting these values into the integrated equation yields the approximate result ln KP(T2) ≈ 3.91, therefore KP(400K) ≈ 49.92. This is reasonably close to the measured value of KP(400K) = 48.91.

Compared to the large value of KP at 298K, the small positive value of KP at 400K shows that the reaction is approaching the point where it will shift to become reactant-favorable rather than product favorable.

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© P Mander August 2015

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The Clapeyron equation

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This equation is exact and applies to equilibrium processes between any two phases, i.e melting, boiling and sublimation. It gives the slope of the coexistence curve at (saturated) vapor pressure P and temperature T.

The original derivation by Clapeyron is a geometrical method in which an expression for the PV work done in a Carnot cycle is equated with Carnot’s principle δW = δQ (δT)/T. This classic proof is given in Appendix I.

Derivation

If two phases A and B of the same pure substance are in equilibrium with each other, and GA and GB are the respective molar Gibbs free energies, then the condition of equilibrium is GA = GB (if this were not the case, substance in the phase with higher molar free energy would pass into the phase with lower molar free energy, which is a non-equilibrium condition).

At equilibrium dG = 0, which is also the condition for the thermodynamically reversible change dGA = dGB. Since dG = VdP – SdT, we can equate the differential expressions for the two phases:

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Since we are dealing with a thermodynamically reversible process at constant pressure and temperature, ΔS =(Qp)/T which we can write ΔS = ΔH/T where ΔH is the molar enthalpy of phase transition. Hence

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Applications of the Clapeyron equation

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1. Calculate the enthalpy of phase transition ΔH
If both dP/dT and ΔV at temperature T are known, ΔH can be calculated. Scroll down to Worked example 1

2. Estimate the effect of pressure change on the boiling point temperature
If both ΔV and ΔH are known at temperature T, dP/dT can be calculated to estimate the effect of pressure change on the boiling point temperature. Scroll down to Worked example 2

3. Estimate the effect of pressure change on the melting point temperature
If both ΔV and ΔH are known at temperature T, dP/dT can be calculated to estimate the effect of pressure change on the melting point temperature. Scroll down to Worked example 3

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Worked Example 1

The vapor pressure of water changes by 27.17 mm Hg from 372.15K to 373.15K, and the corresponding volume change (water>vapor) is 3.0114 x 10-2 m3mol-1. Estimate the enthalpy of vaporization ΔHvap of water at 373.15K.

Strategy

Use the Clapeyron equation to solve for ΔHvap

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The SI units are:
P (vapor) = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
T = K
V = m3mol-1
H = Jmol-1

Calculation

1. At 373.15 K, the vapor pressure of water changes by 27.17 mm Hg per kelvin. This may be taken as the value of dP/dT at 373.15K.

2. Convert dP/dT from mm Hg per kelvin to Nm-2K-1

dP/dT = 27.17 x (105/760) = 3575 Nm-2K-1

3. Compute ΔH = (dP/dT) . TΔV

ΔH = 3575 x 373.15 x 3.0114 x 10-2 = 40172 Jmol-1

The enthalpy of vaporization of water ΔHvap at 373.15 is estimated to be 40.17 kJmol-1.

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Worked Example 2

The enthalpy of vaporization ΔHvap of water at 373.15K is 40,657 Jmol-1 and the corresponding volume change (water>vapor) is 3.0114 x 10-2 m3mol-1. Estimate the boiling point temperature at 770 mm Hg.

Strategy

Use the inverted Clapeyron equation to solve for dT/dP

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The SI units are:
T = K
P = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
ΔV = m3mol-1
ΔH = Jmol-1

Note that P is the vapor pressure. Since T is the boiling point temperature, the vapor pressure and the externally applied pressure will be the same.

Calculation

1. Compute dT/dP

dT/dP = (373.15 x 3.0114 x 10-2)/40657 = 2.76 x 10-4 K(Nm-2)-1

2. Compute the pressure increase in Nm-2

Water boils at 373.15K at a pressure of 760 mm Hg. The pressure increase under consideration is therefore 770-760 = 10 mm Hg.

10 mm Hg = 10 x (105/760) Nm-2 = 1316 Nm-2

3. Put the two facts together

dT/dP tells us that the boiling point is raised by 2.76 x 10-4 K for 1 Nm-2 increase in vapor pressure.

So a 1316 Nm-2 increase in vapor pressure will raise the boiling point temperature by 1316 x 2.76 x 10-4 = 0.36K

The boiling point of water at 770 mm Hg is estimated at 373.15 + 0.36 = 373.51K

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Worked Example 3

At 273.15K the molar enthalpy of fusion of water ΔHfusion is 6 x 103 Jmol-1 and the corresponding volume change (ice>water) is –1.6 x 10-6 m3mol-1. Estimate the melting point of ice at a pressure of 150 atmospheres.

Strategy

Use the inverted Clapeyron equation to solve for dT/dP

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The SI units are:
T = K
P = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
ΔV = m3mol-1
ΔH = Jmol-1

Note that P is the vapor pressure. Since T is the melting point temperature, the vapor pressure and the externally applied pressure will be the same.

Calculation

1. Compute dT/dP

dT/dP = (273.15 x –1.6 x 10-6)/(6 x 103) = –7.28 x 10-8 K(Nm-2)-1

2. Compute the pressure increase in Nm-2

Ice melts at 273.15K at a pressure of 1 atmosphere. The pressure increase under consideration is therefore 150-1 = 149 atmospheres.

149 atmospheres = 1.49×107 Nm-2

3. Put the two facts together

dT/dP tells us that the melting point is lowered by –7.28 x 10-8 K for 1 Nm-2 increase in vapor pressure.

So a 1.49×107 Nm-2 increase in vapor pressure will lower the melting point by

(1.49 x 107) x (–7.28 x 10-8) = –1.08K

The melting point of ice at a pressure of 150 atmospheres is estimated at 273.15 – 1.08 = 272.07K

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The Clausius-Clapeyron equation

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The Clausius-Clapeyron equation is not exact, since approximations are used in its derivation. Its use is confined to processes involving vapor phase equilibrium, but nevertheless the equation is very useful.

Derivation

Beginning with the Clapeyron equation

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use is made of the fact that for sublimation and vaporization processes, there is generally a large difference in molar volume between the two phases. In this case, ΔV can be approximated as the molar volume of the vapor Vvap.

It is then assumed that the behavior of the vapor approximates that of an ideal gas, so that the molar volume Vvap can in turn be replaced with RT/P where P is the vapor pressure, giving

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This is the Clausius-Clapeyron equation, which approximates the slope of the liquid-vapor coexistence curve at vapor pressure P and boiling point temperature T. Using the mathematical identity

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the Clausius-Clapeyron equation can also be written in the form

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If we further assume that ΔHvap is independent of temperature, integration of the above equation can be performed

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This is the integrated form of the Clausius-Clapeyron equation. If the vapor pressure P1 is known at boiling point temperature T1, this equation can be used to estimate the boiling point temperature T2 at another pressure P2. Alternatively, if the boiling point temperatures are known at vapor pressures P1 and P2, the enthalpy of vaporization ΔHvap can be estimated.

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Applications of the Clausius-Clapeyron equation

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or

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and its integrated form

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1. Estimate the enthalpy of vaporization ΔHvap
If both dP/dT and the vapor pressure P are known at boiling point temperature T, the enthalpy of vaporization ΔHvap can be estimated. The value of R needs to be known. Scroll down to Worked example 4

2. Estimate the effect of pressure change on the boiling point temperature
If the vapor pressure P1 is known at boiling point temperature T1, the boiling point temperature T2 at another pressure P2 can be estimated. The values of R and ΔHvap need to be known. Scroll down to Worked example 5

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Worked Example 4

dP/dT is 27.17 mm Hg per kelvin for water at 373.15K. Estimate the enthalpy of vaporization ΔHvap of water at this temperature. The vapor pressure of water at 373.15K is 105 Nm-2, and the gas constant R is 8.3145 JK-1mol-1.

Strategy

Use the Clausius-Clapeyron equation to solve for ΔHvap

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The SI units are:
T = K
P (vapor) = Nm-2 (conversion: 1 atm/760 mm Hg = 105 Nm-2)
R = JK-1mol-1
ΔH = Jmol-1

Calculation

1. Convert dP/dT from mm Hg per kelvin to Nm-2K-1

dP/dT = 27.17 x (105/760) = 3575 Nm-2K-1

2. Compute ΔHvap = 1/P x dP/dT x RT2

ΔHvap = 10-5 x 3575 x 8.3145 x (373.15)2 = 41388 Jmol-1

The enthalpy of vaporization of water ΔHvap at 373.15K is estimated to be 41.39 kJmol-1 (which is about 2% higher than the experimental value of 40.66 kJmol-1).

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Worked Example 5

Water boils at 373.15K at a pressure of 1 atmosphere. At what temperature will water boil in a pressure cooker operating at a pressure of 2 atmospheres? The enthalpy of vaporization ΔHvap of water at 373.15K is 40657 Jmol-1 and the gas constant R is 8.3145 JK-1mol-1.

Strategy

Use the integrated Clausius-Clapeyron equation to solve for T2

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The SI units are:
T = K
R = JK-1mol-1
ΔH = Jmol-1

The units of P1 and P2 are immaterial, so long as they are the same.

Note that P1 and P2 are vapor pressures. Since T1 and T2 refer to boiling point temperatures, the vapor pressures P1 and P2 will be the same as the externally applied pressures.

Calculation

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Water in a pressure cooker operating at a pressure of two atmospheres boils at 394K, or 121°C. This explains why vegetables cook so fast in these devices.

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Appendix 1

The Clapeyron equation – Clapeyron’s proof

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Émile Clapeyron was the first to use the PV diagram as an analytical tool

Émile Clapeyron’s 1834 paper Mémoire sur la Puissance Motrice de la Chaleur (Memoire on the Motive Power of Heat) contains the first analytical representation of the Carnot cycle, depicted above as the quadrilateral ABCD. By assigning infinitely small values to the variations of volume V and vapor pressure P during the four successive operations of the cycle, Clapeyron renders the quadrilateral a parallelogram.

He then demonstrates that the parallelogram ABCD, representing the work done in a complete cycle, is equal in area to BCEF since both stand on the same base BC and lie between the same parallels T and T–δT.

The base-altitude product BF.PQ therefore equates to the work done in a complete cycle.

Now BF=δP is the increase in vapor pressure per δT rise of temperature at constant volume, while PQ=δV equates to δQ/Lv where δQ is the heat absorbed and Lv is the latent heat of expansion per unit increase in volume

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Since by Carnot’s principle the work δW done in a complete cycle is δQ (δT)/T

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This is the famous Clapeyron equation.

In modern notation Lv is ΔH/ΔV and since vapor pressure is independent of volume the subscript can be omitted and the equation written

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