Thermodynamics

Spontaneous - A reaction or process that will eventually achieve equilibrium without outside help by reacting from left to right as written. A product favored reaction is one where at equilibrium products are favored over reactants. A product favored reaction would be spontaneous although the reaction may be very slow.

System – the part of the universe that is singled out for observation. The universe is made up of the system and the surroundings.

Thermodynamic Functions

DE = internal energy = q + w (remember q is heat and w is work)

DH = enthalpy, heat energy transferred at a constant pressure.

The functions written with capital letters are state functions, these functions only depend on the initial and final state, not on how the change is done. They do not depend on the path.

1^st Law of Thermodynamics – Energy cannot be created or destroyed – the total energy of the universe is constant.

A mathematical statement of this is DE(universe) = DE(system) + DE(surroundings) = 0 for any change.

Since DE(universe) = 0, then DE(system) = -DE(surroundings)

A result of the 1st Law is, we can measure the change of energy inside a system by measuring the energy change in the surroundings. For example, by measuring how much the temperature of a known mass of water increases, we know how much energy was given off by a reaction.

Enthalpy (DH) – the quantity of thermal energy (heat) transferred when a process takes place at constant temperature and pressure. We cannot measure the total enthalpy of something, but we can measure the change in enthalpy. The enthalpy of formation (DH_fº) of elements in their standard state at 25ºC is defined to be zero, similar to how we define Celsius temperature as the freezing point of water at 1 atmosphere of pressure to be 0ºC. The enthalpy of formation of a compound will be the change in enthalpy of the compound minus the enthalpy of the elements.

DH = H(final state) – H(initial state)

For a chemical reaction

2 H_{2 (g)} + O_{2 (g)} → 2 H₂O _(L)

DHº (rxn) = [2 DH_fº (H₂O_(L))] – [ 2 DH_fº(H₂) + DH_fº(O₂) ]

Reversible process - a process that can go back and forth between reactants and products along exactly the same path; a system at equilibrium is reversible because it can be reversed by an infinitesimal modification of a variable such as T. For example, the melting of solid water at 0ºC can be reversed by the removal of a tiny amount of heat.

Entropy, S - a measure of the probability of disorder of a system, discovered by Rudolf Clausius. S is a state function. The more disordered or random a system, the larger its entropy.

For a reversible process at constant temperature and pressure

DS = DH / T(K)

DS_rxn = S m S (products) - S n S (reactants) ; Where m and n are stoichiometric coefficients.

S > q/T irreversible process (units of J/K mol)

S = q/T reversible process

Substance(1 mole) S° (entropy at 25°C)

H₂O (L) 69.91 J/ K mol

H₂O (g) 188.83

H₂ (g) 130.58

H (g) 114.60

O₃ (g) 238.93

O₂ (g) 205.0

O (g) 161.0

CH₄ (g) 186.3

C₂H₆ (g) 229.5

NaCl (s) 72.33

NaCl (aq) 115.5

MgCl₂ (s) 89.62

HCl (g) 186.69

HCl (aq) 56.5

These numbers illustrate some points

1. A phase change from solid to liquid to gas involves an increase in S.

S (H₂O (g)) - S (H₂O (L)) = 188.83 - 69.91 = 118.92 J/K mol

2. A reaction where the number of moles of gaseous substance increases involves an increase in S.

H₂O (g) ® H₂ + 1/2 O₂

DS = [130.58 + 1/2 (205.0)] - 188.83 = 44.3 J/K mol

3. As molecule size increases, S increases (Look at O vs O₂ vs O₃, CH₄ vs C₂H₆, etc).

4. Dissolving a solid or liquid in a solvent usually involves an increase in S. Dissolving a gas in a solvent involves a decrease in S.

NaCl (s) ® NaCl (aq)

DS = 115.5 - 72.33 = 43.2 J/K mol

HCl (g) ® HCl (aq)

DS = 56.5 - 186.69 = - 130.2 J/K mol

Hydrating Ions with multiple charges sometimes involve a decrease in entropy due to the ordering of water molecules around the charge.

Third Law of Thermodynamics

Number 3 comes about because as molecules get more complex, they have more ways they can store energy, which increases their entropy. Nonlinear Molecules can store 3 translational modes of energy (x, y, z directions), 3 rotational modes (rotation about these axis), and increasing number of vibrational modes. The textbook illustrates 3 different types of vibrations for a water molecule. A H₂O will have 6 types of vibrations, and a CH₄ will have 9, (i.e., 3N-6 = # vibrations, where N is the number of atoms.)

As T increases, a molecules S increases because it will be translating, rotating and vibrating more, so it will have a greater S. As T decreases, it will have less energy and it will be translating, rotating and vibrating less. As T reaches 0K, the molecule has 0 energy and it will stop moving and then it has 0 entropy. So, the absolute entropy, different from internal energy, E, and enthalpy, H, can be measured. Note the tables in the back of the text show DH and DG, but they show S without the delta. This shows the entropy of one mole of a substance at 25°C. The third Law of Thermodynamics states that the entropy of a pure, perfect crystalline material at 0K will be zero. Heisenberg's Uncertainty principle keeps us from getting to 0K, but we can see that S approaches 0 (see graph on page 727).

Second Law of Thermodynamics

The second law states that the entropy of the universe increases in a spontaneous process.

DS_system + DS_surroundings = DS_universe > 0 Spontaneous

DS_unv < 0 Nonspontaneous (but the reverse process will be spontaneous)

DS_unv = 0 Reversible (at equilibrium)

Consider the reaction (reference Kotz and Purcell, 1991)

H₂ + 1/2 O₂ ® H₂O (g) DH° = -241.82 kJ, DS = -44.3 J/K

The entropy of the system decreased. How did the entropy of the universe change? Well, the exothermic reaction releases energy into the surrounding, causing the S of the surrounding to increase. In this case

-DH°_sys = DH_surr = +241.82 kJ

DS_surr = DH_surr/T = 241.82 kJ/ 298.15 K = 811.07 J/K

DS_unv = DS_sys + DS_surr = -44.3 + 811.07 = 766.8 J/K > 0, So the reaction is spontaneous.

We would like to use just system variables to determine if a rxn is spontaneous, instead of having to determine the DS of the entire universe. So for a process at constant T and P

DS_unv = DS_sys + DS_{surr =} DS_sys - DH_sys/T (multiply both sides by -T)

** -TDS_unv = DH_sys - TDS_sys

Remember DS_unv > 0 implies a reaction is spontaneous. So

-TDS_unv < 0 implies a reaction will be spontaneous.

-TDS_unv can be calculated using only system variables. This quantity is called the Gibbs Free Energy, the Gibbs function, or just free energy and the symbol used is DG. It is named after Josiah Willard Gibbs, an American who developed much of the theoretical foundations of chemical thermodynamics. So we have at constant T and P

DG = DH - TDS

So

If DG < 0, the reaction is spontaneous as written from left to right

If DG > 0, the reaction is non-spontaneous, (but the reverse reaction will be spontaneous).

If DG = 0, the reaction is reversible and at equilibrium

What is DG for the following reaction at 25°C

H₂ + 1/2 O₂ ® H₂O (g)

DG = DH - TDS = -241.8 kJ/mol - (298K) (-44.3 J/K mol) (1 kJ/1000 J)

= -228.6 kJ/mol

This is < 0, so indeed it is a spontaneous reaction at this temperature.

We cannot measure the absolute Gibbs free energy, similar to enthalpy and internal energy. We can only measure changes in G. So, similar to enthalpy, measurements have been made of the change in G going from elements in their standard state at 25°C, to compounds at 25°C. So using tabulated data,

DG°_rxn = S m DG°_f (products) - S n DG°_f (reactants)

Temperature effects on Spontaneity of Reactions

(Section 19.6 and pay close attention to table 19.4)

DG = DH - TDS

Case 1 (see line 1 on the graph below). If DH < 0 the reaction is exothermic, heat leaves the system and the surrounding disorder increases. If DS>0, the system is becoming more disorganized. When DS>0, then -TDS<0. In this example, both the system enthalpy and entropy are making DG < 0 and the reaction will be spontaneous at any temperature. The following reaction has DH < 0 and DS > 0

2 O₃ ® 3 O₂

The bonds in diatomic oxygen are strong enough that there is a net release of energy, and 3 molecules (or 3 moles of molecules) of gas has more dissorder or entropy (S) than 2 molecules. DG is always <0 and the reaction will be spontaneous at any T.

Line 2 is for the reverse reaction.

3 O₂ ® 2 O₃

Here DH > 0, DS<0, and so DG is always >0 and the reaction is non-spontaneous at any temperature.

DG = -TDS + DH

Line 3 is for the a reaction like that producing water or ammonia from the elements

H₂ + 1/2 O₂ ® H₂O (g)

N₂ (g) + 3 H₂ ® 2 NH₃ (g)

Here DH<0, and DS<0, so the reaction will be spontaneous at low T (DG<0, and become non-spontaneous at higher T. This reaction is said to be enthalpy driven.

Line 4 is typical for the following reactions where both S and H are > 0.

NH₄NO₃ (s) ® NH₄ ⁺(aq) + NO₃^-(aq)

H₂O (s) ® H₂O (L)

These reactions are entropy driven. They are non-spontaneous at low T and become spontaneous at a higher T. This is why most liquids and solids have an increasing solubility and you more disorganized phases with increasing T.

 Gibbs Free Energy and Equilibrium

The DGº that we have been using is the free energy change of a reaction going to completion at standard state. The following expression relates the change in Gibbs free energy for a reaction at any state.

DG = DGº + RT LN Q

R is the gas constant (.008314 kJ/K mole), T is the Kelvin temperature, and Q is the reaction quotient. For example, the following reaction

2 NO₂ (g) ® N₂O₄ (g) DGº = -5.4 kJ/mol AT 25ºC

So if 2 moles of nitrogen dioxide at 25ºC and 1 atmosphere of pressure are converted into 1 mole of dinitrogen tetroxide, 5.4 kJ of energy could be used as work. But what if we have .25 atm of NO₂ and .60 atm of N₂O₄. Will the reaction proceed in the forward direction?

DG = -5.4 kJ/mol + (.008314 kJ/K mol) (298 K) LN (.6 / .25²) = .200 kJ/mol

The positive value tells us the forward reaction is not spontaneous. Thus the backward reaction is spontaneous. And the reaction will go backward until DG = 0, when we are at equilibrium.

At equilibrium, Q = K and

DGº = -RT LN K

K = exp (-DG/RT)

In the above example, K = 8.8 and Q = .6/ .25² = 9.6 > K, so the reaction will go backward to reach equilibrium.

Making the Non-spontaneous, Spontaneous – Coupling Reactions

The following reaction is actually 1 step used in producing lead.

2 PbO (s) ® 2 Pb (s) + O₂ (g) DGº _RXN = -2DG_{f, PbO} = 375.8 kJ/mol

DH > 0 and DS < 0, so using DG = DH -TDS, we would predict this reaction to be non-spontaneous at any temperature. We can make the reaction spontaneous by coupling it with the combustion of charcoal, (graphite).

2 PbO (s) ® 2 Pb (s) O₂ (g) DGº = 375.8 kJ/mol

+ C(s) + O₂ (g) ® CO₂ (g) DGº= -394.4 kJ/mol

2 PbO (s) + C(s) ® 2 Pb (s) + CO₂ (g) DGº= -18.6 kJ/mol

The free energy from the graphite combustion can make the non-spontaneous reduction of lead spontaneous.

In our bodies, the free energy from glucose metabolism is stored in ATP. ATP is then used to provide energy for other reactions.

ADP^3- + H₂PO₄^- ® ATP + H₂O DGº’ = 30.5 kJ/mol (DGº’ refers to standard biological state at 37ºC, and [H⁺] = 10^-7M.

The metabolism of one mole of glucose (through a series of steps) converts 32 moles of ADP into 32 moles of ATP, which can then be used for other reactions.

C₆H₁₂O₆ + 6 O₂ ® 6 CO₂ + 6 H₂O DGº’ = -2870 kJ/mol

32 ADP^3- + 32 H₂PO₄^- ® 32 ATP + 32 H₂O DGº’ = 976. kJ/mol

DGº’ = -1894 kJ/mol

So, out of 2870 kJ of energy, 976 is stored, with the remaining energy lost as heat, (maintaining body temperature). This gives us an efficency, of 976/2870 X 100% = 34 %. The energy stored in the ATP can be used to join amino acids to form a peptide bond to build proteins (3 ATP/ peptide bond).