DS, The Entropy Change for the System.

Measuring and Tabulating Entropy Changes.

DSuniv = DS - DH / T (constant P and T)

DH has been tabulated as Df so the DH / T term can be calculated under standard conditions:

D298 = ĺmiDf(i)

To measure DS we use the observation that at equilibrium

DSuniv = 0 = DS - DH / T

So that, at equilibrium:

DS = DH / T

or dS = dH / T if T is not constant

That is, for any process occurring at equilibrium, the entropy change can be measured by measuring the enthalpy change, or qp.

Processes occurring at equilibrium. 1. Phase changes.

A(s) A(l)

A(l) A(g)

Any amount of A(s) and of A(l) will be in equilibrium with each other at the melting temperature.

Heat may be added reversibly (this means in amounts too small to measure) at the molecule level.

The addition or subtraction of heat will not change the equilibrium concentrations since the concentration of a pure liquid and a pure solid are always constant at a given temperature.

For a phase change we can write:

DSş = D(PC) / T(PC)

D(PC) = latent heat of phase change per mole.

T(PC) = temperature phase change at 100 kPa.

Examples of Entropy change calculation for phase changes:

Water:

D(fusion) = 6.01 kJ/mol

D(vap) = 40.7 kJ/mol

D(vap) = D(vap) / T

= 40.7 kJ/mol / 373 K

= 0.109 kJ/K.mol

= 109 J/K.mol.

D(fusion) = D(fusion) / T

= 6.01 kJ/mol / 273

= 0.0220 kJ/K.mol

= 22.0 J/K.mol

Notes:

1. DS values are small compared to DH values and so use joules (J) rather than kilojoules (kJ).

Remember this when using DS and DH together in a single equation: the units do not match!

2. The units for entropy are energy/temp, J/K.

Exercise:

The enthalpy of fusion of gold, Au, is 12.36 kJ/mol and its melting point at 100 kPa is 1064şC. Calculate DSş(fusion).

Processes Occurring at Equilibrium. 2. Temperature Changes.

As noted above, the passing of heat across a systems boundary can be done reversibly and so under equilibrium conditions.

Thus heating a substance can be done reversibly at equilibrium.

For heating a substance under equilibrium conditions:

dS = dH / T (because the temperature is not

constant)

or

dS = dqp / T

For heating a substance, qp is usually measured as the heat capacity.

Definition:

The molar heat capacity of a substance at constant pressure, Cp, is the energy required to raise 1 mole of the substance 1 degree Kelvin (or Celcius) at constant pressure.

Thus dqp = heat capacity * temperature change = CpdT

and dS = CpdT/T.

Calculating Entropy Changes on Heating
.

1. If Cp is independent of temperature, dS = CpdT/T from T1 to T2 can be evaluated:

DS = Cp ln (T2/T1).

Example:

The heat capacity, Cp, of water at 298 K is 79.3 J/K.mol. Assuming that Cp is constant with temperature, calculate the entropy change when 1 mole of water is heated from 20şC to 50şC.

DSş = Cp ln(T2/T1)

= 79.3 J/k.mol * 1 mol * ln (323/293)

= 79.3 J/K. * 0.0975

= 7.73 J/K.

What is the entropy change when 100 g of hot water (100şC) and 100 g of cold water (0şC) are mixed? Assume Cp for water is constant over this temperature range.

Temperature after mixing will be 50şC. (How do we know this?)

for the cold water

DSş = Cp*moles*ln(T2/T1)

= 75.6 J/K.mol*(100/18)mol*ln(323/273)

= 75.6*(100/18)*0.0168 J/K

= 70.63 J/K

for the hot water

DSş = Cp*mol*ln(T2/T1)

= 75.6 J/K.mol*(100/18)mol*ln(323/373)

= 75.6*(100/18)*(-0.144)

= -60.45

For both:

DSş = (70.63 - 60.45) J/K

= 10.18 J/K

Calculating Entropy Changes on Heating

2. If Cp is dependent on temperature, then a graphical evaluation of the integration (i.e. the area under a curve) can be used.

The graphical evaluation of the entropy change on heating adds an important possibility:

although temperatures down to 0 K cannot be achieved, a graph can be extrapolated to this temperature.

Thus, by extrapolation, the entropy change for a substance can be measured from 0 K.

Absolute Entropy Values.

The tabulation of enthalpy values is complicated by the non-existence of a zero of energy.

A zero value for entropy however is defined.

from S = k ln w

S = 0 when w = 1.

A perfectly ordered substance, with only one way of organizing, has an entropy of 0.

Definition: (Third Law of Thermodynamics)

A perfectly ordered crystal at 0 K has an entropy of 0 J/K.

Experimental determination of the absolute entropy of a substance: (proceed along the following only to 298 K)

1. Find graphically the DSş of solid from 0 K to melting point.

2. Find DSş of fusion at melting point.

3. Find graphically the DSş of liquid from melting to boiling.

4. Find DSş of vapourization at boiling point.

5. Find graphically the DSş of gas from boiling to 298 K

6. Total the entropy change from 0 K to 298 K.

Note that the total entropy change, DSş = Sş2 - Sş1.

But Sş1 = 0, since it is the entropy of a perfect crystal at 0 K.

Thus DSş = Sş2.

Thus we have a method of finding absolute entropy values and tabulating these.

Tabulation is of Sş298 values for 1 mole of the compound.

Units are J/K.mol.

Using Tabulated Values of Absolute Entropy
.

Example:

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)

D298 = Sfinal - Sinitial

= Sş298(CO2(g)) + 2 Sş298(H2O(l))

- Sş298(CH4(g)) - 2 Sş298(O2(g)).

= 1 mol*213.74 J/K.mol + 2 mol* 69.91 J/K.mol

- 1 mol*186.26 J/K.mol - 2 mol*205.14 J/K.mol

= - 242.98 J/K.

Notes:

1. The format of the entropy calculation is identical to that for enthalpy and can be summarized in the same way:

enthalpy: D298 = ĺmiDf(i)

entropy: D298 = ĺmi298(i)

2. Sş298 values are absolute values, not changes.

3. Because Sş298 values are absolute, values for elements are not zero.

4. Sş298 values are tabulated at 298 K and 100 kPa.

Calculating the Spontaneity, or not, of a Chemical Process.

Use tables to determine if the process:

PbS(s) + CO2(g) + H2O(l) PbCO3(s) + H2S(g)

is spontaneous or not at 298 K and 100 kPa

DSuniv = DS - DH / T

D298 = ĺmi298(i)

= (1*131.0 + 1*205.8 - 1*91.2 - 1*213.7 - 1*69.9)J/K

= -37.7 J/K

D298 = ĺmiDf(i)

= (1*(-699.1) + 1*(-20.6)

- 1*(-100.4) - 1*(-393.5) - 1*(-285.5))kJ

= 59.7 kJ

Thus:

DSuniv = -37.7 J/K - (59.7/298) kJ/K (watch units!)

= -37.7 J/K - (59 700/298) J/K

= -238 J/K

Since DSuniv is negative, the reaction is not spontaneous in the direction written. (It is spontaneous in the reverse direction).

Free Energy.

We have seen that DSuniv = DS - DH / T

that is DSuniv is given in tems of the system variable only.

Definition:

DG / T = -DSuniv.

where G is called the (Gibbs) free energy.

Notes:

G is an energy term. DG is in fact the maximum work that can be obtained from a change in a system: DG = -wmax

DG must be negative for a spontaneous process.

DG depends on the temperature, T

The equation involving G is usually written:

DGT = DH - TDS.

DGT = DH - TDS.

Calculate DG for the following reaction:

 CaCO3(s) CaO(s) + CO2(g) DHşf (kJ/mol) -1206.9 -635.1 -393.5 Sş298 J/K.mol 92.9 39.8 213.7

To find DG we need to calculate DS and DH.

D298 = ĺmiDf(i)

D298 = 1*(-635.1) + 1*(-393.5) - 1*(-1206.9)

= 178.3 kJ.

D298 = ĺmiSş(i)

D298 = 1*(39.8) + 1*(213.7) - 1*(92.9)

= 160.6 J/K. (NB units!)

DGT = DH - TDS.

D298 =178.6 kJ - 298 K * 160.6 J/K

=178.6 kJ - 47.8 kJ

= 130.8 kJ

Positive D298: reaction non-spontaneous (298 K, 100 kPa)

From tables we can calculate D298.

How can we calculate DG at other temperatures and under non-standard pressure?

1. Changing the Temperature, DT.

This turns out to be very easy, at least to a first approximation:

DH and DS can be considered to be temperature independent.

The variation of DT with temperature is given by the equation: DT = DHş + TD

Example:

Calculate D1200 (1200 K) for the calcium carbonate decomposition reaction.

CaCO3(s) CaO(s) + CO2(g)

D298= 178.3 kJ. D298= 160.6 J/K.

DGş1200 = 178.3 kJ - 1200 K * 160.6 J/K

= - 14.4 kJ

The reaction is predicted to be spontaneous at 1200 K

2. Changing from standard pressure.

Consider just one component in a mixture. How does a change in pressure change the G value of that one component?

dG(i) = dH(i) - TdS(i) (constant T)

dH(i) = dE(i) + PdV(i) + VdP(i) (P and V are not constant)

dE(i) = dq(i) + dw(i) = TdS(i) - PdV(i)

Substituting back:

dG(i) = TdS(i) - PdV(i) + PdV(i) + VdP(i) - TdS(i)

= VdP(i)

Evaluating this: dG(i) = VdP(i) = (n(i)RT)dP(i)/P(i)

G2(i) - G1(i) = n(i)RT ln (P2(i) / P1(i))

This is for one component: calculate for all components in a reaction:

DG2 - DG1 = RT ĺ m(i) ln (P2(i) / P1(i))

= RT ĺ ln (P2(i) / P1(i))m(i)

Let state 1 be standard pressure (100 kPa, P = Pş):

DGT - DT = RT ĺ ln (P(i) / Pş)m(i)

DGT - DT = RT ln Q

where ln Q = ĺ ln (P(i) / Pş)m(i) (called the reaction quotient)

This is usually quoted as

DGT = DT + RT ln Q

The free energy change under any pressure conditions can be calculated from the standard free energy change and the reaction quotient, Q.

The ratio of P(i) / Pş is termed the activity of component, i. Note that it has no units (as no term inside a logarithm does).

Activities for components in solution can also be written in a similar way:

[i] / []ş for an ideal solution only.

[]ş = 1 mol / L

Ideal solutions are dilute solutions only, maximum concentration is 1 M (and that's pushing it!)

Activities defined:

for an ideal gas: a = P / Pş

for an ideal solute a = [i] / 1 M

for solvent in ideal solution a = 1

for a pure solid a = 1

for a pure liquid a = 1

Reaction quotient:

Q = P a(i)m(i).

Example for Q:

for the reaction: bB + dD fF + gG

Q = (aFf * aGg) / (aBb * aDd)

Or simply put:

Q has the same form as K, the equilibrium constant.

Example.

Calculate DG500 for the process when P(CO2) = 51 kPa:

CaCO3(s) CaO(s) + CO2(g)

D298= 178.3 kJ. D298= 160.6 J/K.

D500 = 178.3 kJ - 500 K * 160.6 J/K

= 98.0 kJ

DG500 = D500 + RT ln Q

= 98.0 kJ + 8.314 J/K*500 K*ln Q

Q = P aim(i)

= a(CO2) * a(CaO) / a(CaCO3)

= (P(CO2)/Pş) * 1 / 1

= P(CO2)/Pş

= 51 kPa / 100 kPa = 0.51 (no units!)

DG500= 98.0 kJ + (8.314 * 500 * ln 0.51) J (Units!)

= 98.0 kJ + (-2800 J)

= 95.2 kJ