**Free Energy and Equilibrium.**

If a reaction is at equilibrium, then DG for those conditions (which will not be standard conditions) must be zero:

This leads to:

DG_{T} = 0 = DGº_{T} + RT ln Q

but, at equilibrium, Q = K

so

DG_{T} = 0 = DGº_{T} + RT ln K

or

**DGº _{T} = - RT ln K**

This is an important equation!! Do not forget the sign.

There is a direct relationship between the equilibrium constant in terms of
activities, and DGº_{T}.

**ln K = - DGº _{T} / RT**

**Do not forget the sign!!**

Calculate the equilibrium constant at 400 K for the process:

N_{2}(g) | + 3 H_{2}(g) | 2 NH_{3}(g) | ||

DHº_{f} | -46.1 (kJ/mol) | |||

Sº_{298} | 192 | 131 | 192 (J/K.mol) |

DHº = 2 (-46.1) - 1 (0) - 3 (0) = -92.2 kJ

DSº = 2 (192) - 1 (192) - 3 (131) = -201 J/K

DGº_{400} = DHº - 400 DSº = -92200 - 400 (-201)

= -11800 J

(Note unfavourable entropy, favourable enthalpy)

**ln K = - DGº _{T} / RT**

ln K = - (-11800 J) / (8.314 J * 400)

ln K = 3.55

K = 34.8

K is > 1 showing that the equilibrium favours products.

K has no units: it is terms of activities:

K = a(NH_{3})^{2} /a(N_{2}) * a(H_{2})^{3}

From the thermodynamic tables in your text,

1. calculate the normal boiling point of bromine, Br_{2}

2. calculate the vapour pressure of Br_{2} at 25ºC.

The process is: Br_{2}(l) Br_{2}(g)

1. At normal boiling point, the equilibrium is under standard conditions,
so DGº_{T} = 0.

DGº_{T} = DHº - TDSº = 0

so

T = DHº / DSº = 30.9 kJ / 93.3 J.K-1 (watch UNITS)

T = 331 K (58ºC)

2. At equilibrium at 298 K (25ºC)

**ln K = - DGº _{T} / RT**

ln K = -DHº / RT + DSº / R

ln K = -30900 / 8.314*298 + 93.3 / 8.314

(watch the UNITS)

ln K = - 1.25

K = 0.287

K = a(Br_{2(g)})/a(Br_{2(l)}) = {P(Br_{2}) kPa / 100 kPa} / 1

P(Br_{2}) = 28.7 kPa at 25ºC

DGº

DGº_{T} |
K_{300} |
K_{500} |
K_{1000} |

100 | 3.87 x 10^{-18} |
3.57 x 10^{-11} |
0.00000598 |

50 | 1.97 x 10^{-9} |
0.00000598 | 0.00244 |

25 | 0.0000444 | 0.00244 | 0.0495 |

10 | 0.0181 | 0.0902 | 0.300 |

5 | 0.135 | 0.300 | 0.548 |

2 | 0.448 | 0.618 | 0.786 |

1 | 0.670 | 0.786 | 0.887 |

0 | 1 | 1 | 1 |

-1 | 1.49 | 1.27 | 1.13 |

-2 | 2.23 | 1.62 | 1.27 |

-5 | 7.41 | 3.33 | 1.82 |

-10 | 55.2 | 11.1 | 3.33 |

-25 | 22 500 | 410 | 20.2 |

-50 | 508 000 000 | 16 700 | 410 |

DG_{T} = DGº_{T} + RT ln Q

but, DGº_{T} = -RT ln K

so DG_{T} = -RT ln K + RT ln Q

= RT ln {Q/K}

Q is a non-equilibrium quantity. As the reaction proceeds, Q will change,
and **the direction of that change will be towards the value of K**.

Thus:

__if Q > K, then Q must decrease, reaction must go left.__

__if Q < K, then Q must increase, reaction must go right.__

N_{2}(g) + 3 H_{2}(g) 2 NH_{3}(g)

In an experiment, the partial pressures were observed to be:

P(N_{2}) = 150 kPa, P(H_{2}) = 100 kPa, and P(NH_{3}) = 200 kPa.

Predict the way the equilibrium will move and calculate DG_{400} for these
conditions.

1. Obtain a value for Q which can then be compared to K:

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

= a(NH_{3})^{2} * a (N_{2})^{-1} * a(H_{2})^{-3}

= {200kPa/100kPa}^{2}*{150 kPa/100 kPa}^{-1}*{100 kPa/100kPa}^{-3}

= 2^{2} / {1.5 * 1^{3}} = 2.67

2. Compare Q to K: Q < K

so: Q must increase, reaction must go to right (to products).

3. DG_{400} = RT ln {Q/K} = 8.314 * 400 ln {2.67/34.8}

= 8.314 J/K * 400 K * (-2.57)

= -8540 J

(negative DG means process is spontaneous towards products)

DGº_{T} = -RT ln K = DHº - TDSº

so:

ln K = DGº_{T}/RT = DSº/R - DHº/RT

Note that this is the equation which was paralleled by Arrhenius when he proposed the variation of k (rate constant with T):

**ln k = ln A - E _{a}/RT**

The equation:

**ln K = DSº/R - DHº/RT**

shows how K, the equilibrium constant varies with T.

A graphical plot of ln K against 1/T should give a straight line of slope -DHº/R and intercept DSº/R.

Thus DHº and DSº can be obtained graphically from K measured at different temperatures.

In the simplest case, two values of K at different temperatures can be used to determine DHº and DSº.

**Example**:

For the reaction NiO(s) + CO(g) Ni(s) + CO_{2}(g)

K_{936} = 4540 and K_{1125} = 1580.

Calculate DHº and DSº for this process, assuming that both are temperature independent.

ln K_{936} = DSº/R - DHº/R * 936 = ln 4540 = 8.42

ln K_{1125} = DSº/R - DHº/R * 1125 = ln 1580 = 7.37

Subtract:

8.42 - 7.37 = {DHº/R}*{1/1125 - 1/936}

1.05 = {DHº/R}*(-0.0001795)

DHº = -48 600 J (same units as R)

substitute:

8.42 = DSº/8.324 - {(-48 600)/8.314*936)}

DSº = 18.0 J/K.

{Take care to differentiate between DH (non-unit activity) and DHº (unit activity)}

DS(univ) > 0 for a spontaneous process

DS(univ) = 0 at equilibrium.

DS(univ) = DS(system) + DS(surrounds)

DE = q + w = q + PDV

DE = q_{v} (measurement of DE in bomb calorimeter)

DH = DE - PDV = q_{p}

(measurement of DH in open calorimeter)

{note for gases only: PDV = RTDn.}

DS(univ) = DS - DH/T

DS(univ) = -DG/T

DHº_{298} = åm_{i}DHº_{f}(i)

DSº_{298} = åm_{i}Sº_{298}(i)

DG_{T} = DH - TDS

DGº_{298} = åm_{i}DGº_{f}(i)

DG_{T} = DGº_{T} + RT ln Q (Q = Pa(i)^{m(i)})

At equilibrium: DG_{T} = 0 and Q = K

DGº_{T} = - RTln K

DG_{T} = RT ln {Q/K}

ln K_{T} = DSº/R - DHº/RT = - DGº_{T}/RT

Some ways to obtain DH:

by direct measurement: DH = q_{p}

from tables: DHº_{298} = åm_{i}DHº_{f}(i)

from K: ln {K_{1}/K_{2}} = {DHº/R}{1/T_{2} - 1/T_{1}}

Some ways to obtain DS:

from DH at equilibrium: DS = DH/T

from C_{p} on heating: DS = C_{p} ln {T_{2}/T_{1}}

from tables: DSº_{298} = åm_{i}Sº_{298}(i)

from DG and DH: DG = DH - TDS

from a graph of ln K against 1/T (intercept)

Some ways to obtain DG:

from DH and DS: : DG = DH - TDS

from tables: DGº_{298} = åm_{i}DGº_{f}(i)

from Q (Q = Pa(i)^{m(i)}): DG_{T} = DGº_{T} + RT ln Q

from K (= Q at equilibrium): DGº_{T} = -RT ln K_{T}

from K and Q: DG_{T} = RT ln {Q/K}

A note on phase changes:

Phase changes are equilibria.

Normal boiling points and melting points are equilibria under unit
activity. (DGº_{T} = 0.)

Vapour pressure is an equilibrium under non-unit activity conditions. (DG_{T} = 0.)

**From thermodynamic tables, calculate the solubility of silver chloride
in water at 75ºC.**

The equilibrium involved is:

AgCl(s) | Ag^{+}(aq) |
+ Cl^{-}(aq) | ||

DHº_{f} (kJ/mol) |
-127.1 | 105.6 | -167.6 | |

Sº_{298}
(J/K.mol) |
96.2 | 72.68 | 56.5 |

The solubility of a substance is the number of moles/L that will dissolve. (Note solubilty may also be reported in g/100mL of solution)

If s moles/L of AgCl dissolve, then the [Ag^{+}] = [Cl^{-}] = s M

K = a(Ag^{+})*a(Cl^{-})*a(AgCl)^{-1}

K = {[Ag^{+}]/1 M}*{[Cl^{-}]/1M}*1

K = s^{2}

Also ln K_{T} = DSº/R - DHº/RT

Calculate DHº_{298}

= åm_{i}DHº_{f}(i) = {1*105.6 + 1*(-167.6) - 1*(-127.1)} kJ

= 65.1 kJ

Calculate DSº_{298}

= åm_{i}Sº_{298}(i) = {1*72.68 + 1*56.5 - 1*96.2}J/K

= 32.98 J/K

Calculate K_{348 }(watch units!!)

ln K_{348} = 32.98/8.314 - 65100/(8.314*348)

= -18.53

so K = 8.93 x 10^{-9}

and K = s^{2}

so s = 9.45 x 10^{-5}.

The solubility of AgCl in water at 75ºC is predicted to be:

9.45 x 10^{-5} moles/L.

**From thermodynamic tables, estimate the pH of a 1M solution of HF
in water at 25ºC**

The equilibrium involved is:

HF(aq) | + H_{2}O |
H_{3}O^{+}(aq) |
+ F^{-}(aq) | ||

DGº_{f}
(kJ/mol) |
-296.8 | -237.1 | -237.1 | -278.8 |

Known relationships:

pH = - log_{10} a(H_{3}O^{+})

K = a(H_{3}O^{+}) * a(F^{-}) * a(HF)^{-1}

ln K = - DGº_{298}/R*298

Calculation:

DGº_{298} = {1*(-237.1) + 1*(-278.8) - 1*(-296.8) - 1*(-237.1)}kJ

= 18.0 kJ

ln K = - 18000/(8.314*298) UNITS!!

= - 7.265

so K = 0.000700

K = a(H_{3}O^{+})^{2} (since [H_{3}O^{+}] = [F^{-}])

a(H_{3}O^{+}) = 0.0265

so pH = 1.6

1. For a discussion of activity (though it is not called this) see your text, p634, a box: Entropy and Concentration.

2. K_{p} = P P(i)^{m(i)}

3. K_{c} = P [i]^{m(i)}

4. K = P a(i)^{m(i)}

K(for gases) = P {P(i) / Pº} ^{m(i)}

K = {P P(i)^{m(i)}}{ Pº}^{-m(i)}

K = K_{p} * {Pº}^{-m(i)}

If P is measured in atmospheres, Pº = 1 atm, or

if P is measured in bars, where Pº = 1 bar (= 100kPa),

K(for gases) = K_{p} (without the units)

K(for solutions) = P {[i] / 1 M}^{m(i)}

K = K_{c} (without the units)

- Since P(i) = {n/V}RT = [i]RT
- K
_{p}= P P(i)^{m(i)} - K
_{p}= P {[i]RT}^{m(i)} - K
_{p}= {P [i]^{m(i)}}{RT}^{Sm(i)} - K
_{p}= K_{c}* {RT}^{Dn}- (since Sm(i) = Sn(products) - Sn(reactants) = Dn.)