**DS, The Entropy Change for the System.**

Measuring and Tabulating Entropy Changes.

DS_{univ} = DS - DH / T (constant P and T)

DH has been tabulated as DHº_{f} so the DH / T term can be calculated under
standard conditions:

DHº_{298} = åm_{}iDHº_{f}(i)

To measure DS we use the observation that **at equilibrium**

DS_{univ} = 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 q_{p}.**

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:

DSº = DHº_{(PC)} / T_{(PC)}

DHº_{(PC)} = latent heat of phase change per mole.

T_{(PC)} = temperature phase change at 100 kPa.

Water:

DHº_{(fusion)} = 6.01 kJ/mol

DHº_{(vap)} = 40.7 kJ/mol

DSº_{(vap)} = DHº_{(vap)} / T

= 40.7 kJ/mol / 373 K

= 0.109 kJ/K.mol

= 109 J/K.mol.

DSº_{(fusion)} = DHº_{(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 1064ºC. Calculate DSº(fusion).

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 = dq_{p} / T

For heating a substance, q_{p} 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 dq_{p} = heat capacity * temperature change = C_{p}dT

and **dS = C _{p}dT/T**.

1. If C_{p} is independent of temperature,

dS = C_{p}dT/T from T_{1} to T_{2} can be evaluated:

DS = C_{p} ln (T_{2}/T_{1}).

Example:

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

DSº = C_{p} ln(T_{2}/T_{1})

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

= 79.3 J/K. * 0.0975

= 7.73 J/K.

Addendum

What is the entropy change when 100 g of hot water (100ºC) and 100 g of
cold water (0ºC) are mixed? Assume C_{p} for water is constant over this
temperature range.

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

for the cold water

DSº = C_{p}*moles*ln(T_{2}/T_{1})

= 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

DSº = C_{p}*mol*ln(T_{2}/T_{1})

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

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

= -60.45

For both:

DSº = (70.63 - 60.45) J/K

= 10.18 J/K

2. If C_{p} 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.**

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

2. Find DSº of fusion at melting point.

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

4. Find DSº of vapourization at boiling point.

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

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

Note that the total entropy change, DSº = Sº_{2} - Sº_{1}.

But Sº_{1} = 0, since it is the entropy of a perfect crystal at 0 K.

Thus DSº = Sº_{2}.

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

Tabulation is of Sº_{298} values for 1 mole of the compound.

Units are J/K.mol.

Example:

CH_{4}(g) + 2 O_{2}(g) CO_{2}(g) + 2 H_{2}O(l)

DSº_{298} = S_{final} - S_{initial}

= Sº_{298}(CO_{2(g)}) + 2 Sº_{298}(H_{2}O_{(l)})

- Sº_{298}(CH_{4(g)}) - 2 Sº_{298}(O_{2(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: DHº_{298} = åm_{i}DHº_{f}(i)

entropy: DSº_{298} = åm_{i}Sº_{298}(i)

2. Sº_{298} values are absolute values, not changes.

3. Because Sº_{298} values are absolute, values for elements are not zero.

4. Sº_{298} values are tabulated at 298 K and 100 kPa.

Use tables to determine if the process:

PbS(s) + CO_{2}(g) + H_{2}O(l) PbCO_{3}(s) + H_{2}S(g)

is spontaneous or not at 298 K and 100 kPa

DS_{univ} = DS - DH / T

DSº_{298} = åm_{i}Sº_{298}(i)

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

= -37.7 J/K

DHº_{298} = åm_{i}DHº_{f}(i)

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

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

= 59.7 kJ

Thus:

DS_{univ} = -37.7 J/K - (59.7/298) kJ/K (watch units!)

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

= -238 J/K

Since DS_{univ} is negative, the reaction is not spontaneous in the direction
written. (It is spontaneous in the reverse direction).

__Free Energy.__

We have seen that DS_{univ} = DS - DH / T

that is DS_{univ} is given in tems of the system variable only.

Definition:

DG / T = -DS_{univ}.

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 = -w_{max}

DG must be **negative** for a spontaneous process.

DG depends on the temperature, T

The equation involving G is usually written:

DG_{T} = DH - TDS.

DG

Calculate DG for the following reaction:

CaCO_{3}(s) | CaO(s) | + CO_{2}(g) |
||

DHº_{f }(kJ/mol) | -1206.9 | -635.1 | -393.5 | |

Sº_{298 }J/K.mol | 92.9 | 39.8 | 213.7 |

To find DG we need to calculate DS and DH.

DHº_{298} = åm_{i}DHº_{f}(i)

DHº_{298} = 1*(-635.1) + 1*(-393.5) - 1*(-1206.9)

= 178.3 kJ.

DSº_{298} = åm_{i}Sº(i)

DSº_{298} = 1*(39.8) + 1*(213.7) - 1*(92.9)

= 160.6 J/K. (NB units!)

DG_{T} = DH - TDS.

DGº_{298} =178.6 kJ - 298 K * 160.6 J/K

=178.6 kJ - 47.8 kJ

= 130.8 kJ

Positive DGº_{298}: reaction non-spontaneous (298 K, 100 kPa)

From tables we can calculate DGº

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

**1. Changing the Temperature, DGº _{T}.**

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 DGº_{T} with temperature is given by the equation: DGº_{T} =
DHº + TDSº

**Example:**

Calculate DGº_{1200} (1200 K) for the calcium carbonate decomposition
reaction.

CaCO_{3}(s) CaO(s) + CO_{2}(g)

DHº_{298}= 178.3 kJ. DSº_{298}= 160.6 J/K.

DGº1200 = 178.3 kJ - 1200 K * 160.6 J/K

= - 14.4 kJ

The reaction is predicted to be spontaneous at 1200 K

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)

G_{2}(i) - G_{1}(i) = n(i)RT ln (P_{2}(i) / P_{1}(i))

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

DG_{2} - DG_{1} = RT å m(i) ln (P_{2}(i) / P_{1}(i))

= RT å ln (P_{2}(i) / P_{1}(i))^{m(i)}

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

DG_{T} - DGº_{T} = RT å ln (P(i) / Pº)^{m(i)}

DG_{T} - DGº_{T} = RT ln Q

where ln Q = å ln (P(i) / Pº)^{m(i)} (called the reaction quotient)

This is usually quoted as

DG_{T} = DGº_{T} + 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) / Pº 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] / []º for an ideal solution only.

[]º = 1 mol / L

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

for an ideal gas: a = P / Pº

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

Q = P a(i)^{m(i)}.

Example for Q:

for the reaction: bB + dD fF + gG

Q = (a_{F}^{f} _{*} a_{G}^{g}) / (a_{B}^{b} _{*} a_{D}^{d})

Or simply put:

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

Calculate DG_{500} for the process when P(CO_{2}) = 51 kPa:

CaCO_{3}(s) CaO(s) + CO_{2}(g)

DHº_{298}= 178.3 kJ. DSº_{298}= 160.6 J/K.

DGº_{500} = 178.3 kJ - 500 K * 160.6 J/K

= 98.0 kJ

DG_{500} = DGº_{500} + RT ln Q

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

Q = P a_{i}^{m(i)}

= a(CO_{2}) * a(CaO) / a(CaCO_{3})

= (P(CO_{2})/Pº) * 1 / 1

= P(CO_{2})/Pº

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

DG_{500}= 98.0 kJ + (8.314 * 500 * ln 0.51) J (Units!)

= 98.0 kJ + (-2800 J)

= 95.2 kJ