The maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant T and p is equal to, that is 1441

The maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant T and p is equal to, that is 1442

The maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant T and p is equal to, that is This is a second important application of. 1443

The maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant T and p is equal to, that is This is a second important application of. The key constraints are indicated in blue type. 1444

To prove that, start with a summary of previous results: 1445

To prove that, start with a summary of previous results: G = H – T S (1) 1446

To prove that, start with a summary of previous results: G = H – T S (1) H = E + pV (2) 1447

To prove that, start with a summary of previous results: G = H – T S (1) H = E + pV (2) (3) 1448

To prove that, start with a summary of previous results: G = H – T S (1) H = E + pV (2) (3) (4) 1449

To prove that, start with a summary of previous results: G = H – T S (1) H = E + pV (2) (3) (4) (5) 1450

Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) 1451

Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) Now take a change in each variable 1452

Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) Now take a change in each variable (7) 1453

Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) Now take a change in each variable (7) Plug Eq. (4) into Eq. (3) and insert the result into Eq. (7): 1454

Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) Now take a change in each variable (7) Plug Eq. (4) into Eq. (3) and insert the result into Eq. (7): (8) 1455

Now fix the conditions: 1456

Now fix the conditions: (a) constant temperature, so that, 1457

Now fix the conditions: (a) constant temperature, so that, (b) constant pressure, so that, 1458

Now fix the conditions: (a) constant temperature, so that, (b) constant pressure, so that, (c) and reversible process, 1459

Now fix the conditions: (a) constant temperature, so that, (b) constant pressure, so that, (c) and reversible process, then Eq. (8) simplifies to (9) 1460

Now fix the conditions: (a) constant temperature, so that, (b) constant pressure, so that, (c) and reversible process, then Eq. (8) simplifies to (9) which simplifies using Eq. (5) to yield 1461

Now fix the conditions: (a) constant temperature, so that, (b) constant pressure, so that, (c) and reversible process, then Eq. (8) simplifies to (9) which simplifies using Eq. (5) to yield (10) 1462

Now fix the conditions: (a) constant temperature, so that, (b) constant pressure, so that, (c) and reversible process, then Eq. (8) simplifies to (9) which simplifies using Eq. (5) to yield (10) For a reversible change, hence 1463

Now fix the conditions: (a) constant temperature, so that, (b) constant pressure, so that, (c) and reversible process, then Eq. (8) simplifies to (9) which simplifies using Eq. (5) to yield (10) For a reversible change, hence 1464

A true reversible process takes an infinite amount of time to complete. Therefore we can never obtain in any process the amount of useful work predicted by the value of. 1465

The Gibbs Energy and Equilibrium 1466

The Gibbs Energy and Equilibrium When a system goes from an initial to a final state, a indicates a spontaneous change left to right, a indicates a non-spontaneous process, the reaction is spontaneous right to left. 1467

The Gibbs Energy and Equilibrium When a system goes from an initial to a final state, a indicates a spontaneous change left to right, a indicates a non-spontaneous process, the reaction is spontaneous right to left. It is possible that, and hence 1468

The Gibbs Energy and Equilibrium When a system goes from an initial to a final state, a indicates a spontaneous change left to right, a indicates a non-spontaneous process, the reaction is spontaneous right to left. It is possible that, and hence When, the system is at equilibrium, there is no net change. 1469

Example: Consider a mixture of ice and water at 0 o C and 1 bar. 1470

Example: Consider a mixture of ice and water at 0 o C and 1 bar. Neither freezing nor melting is spontaneous, provided no heat is added or removed from the system. 1471

Example: Consider a mixture of ice and water at 0 o C and 1 bar. Neither freezing nor melting is spontaneous, provided no heat is added or removed from the system. There is a dynamic equilibrium: 1472

Example: Consider a mixture of ice and water at 0 o C and 1 bar. Neither freezing nor melting is spontaneous, provided no heat is added or removed from the system. There is a dynamic equilibrium: ice water 1473

Example: Consider a mixture of ice and water at 0 o C and 1 bar. Neither freezing nor melting is spontaneous, provided no heat is added or removed from the system. There is a dynamic equilibrium: ice water The ice lattice is broken down to form liquid water and water freezes to form ice at every instant. At equilibrium, and therefore the amount of useful work that can be extracted from the system is zero. 1474

Predicting the Outcome of Chemical Reactions 1475

Predicting the Outcome of Chemical Reactions 1476 Consider the “simple” reaction A B

Predicting the Outcome of Chemical Reactions 1477 Consider the “simple” reaction A B How do we tell which is the spontaneous direction:

Predicting the Outcome of Chemical Reactions 1478 Consider the “simple” reaction A B How do we tell which is the spontaneous direction: A B or B A ?

Predicting the Outcome of Chemical Reactions 1479 Consider the “simple” reaction A B How do we tell which is the spontaneous direction: A B or B A ? Examination of for each reaction gives the answer.

Predicting the Outcome of Chemical Reactions 1480 Consider the “simple” reaction A B How do we tell which is the spontaneous direction: A B or B A ? Examination of for each reaction gives the answer. Suppose A B is spontaneous

Predicting the Outcome of Chemical Reactions 1481 Consider the “simple” reaction A B How do we tell which is the spontaneous direction: A B or B A ? Examination of for each reaction gives the answer. Suppose A B is spontaneous – will the reaction B A take place to any extent?

All chemical reactions proceed so as to reach the minimum of the total Gibbs energy of the system. 1482

All chemical reactions proceed so as to reach the minimum of the total Gibbs energy of the system. Always between the total Gibbs energy of the products and the total Gibbs energy of the reactants, there will be some point where the total Gibbs energy of a mixture of reactants and products has a minimum Gibbs energy. 1483

All chemical reactions proceed so as to reach the minimum of the total Gibbs energy of the system. Always between the total Gibbs energy of the products and the total Gibbs energy of the reactants, there will be some point where the total Gibbs energy of a mixture of reactants and products has a minimum Gibbs energy. The minimum indicates the composition at equilibrium, i.e. A B. 1484

take place to some extent. It is necessary to keep in mind that all reactions for which is positive in the forward direction, take place to some extent. However the extent of the reaction may be extremely small (particularly for many typical inorganic reactions). 1485

1486

1487

Standard Gibbs Energy and the Equilibrium Constant 1488

Standard Gibbs Energy and the Equilibrium Constant The Gibbs energy for a species X which is not in its standard state is given by 1489

Standard Gibbs Energy and the Equilibrium Constant The Gibbs energy for a species X which is not in its standard state is given by 1490

Standard Gibbs Energy and the Equilibrium Constant The Gibbs energy for a species X which is not in its standard state is given by where a X is the activity of species X. 1491

Standard Gibbs Energy and the Equilibrium Constant The Gibbs energy for a species X which is not in its standard state is given by where a X is the activity of species X. Recall that. 1492

Standard Gibbs Energy and the Equilibrium Constant The Gibbs energy for a species X which is not in its standard state is given by where a X is the activity of species X. Recall that. In a number of situations the activity coefficient satisfies, so that, 1493

Standard Gibbs Energy and the Equilibrium Constant The Gibbs energy for a species X which is not in its standard state is given by where a X is the activity of species X. Recall that. In a number of situations the activity coefficient satisfies, so that, so that the above result simplifies to 1494

Standard Gibbs Energy and the Equilibrium Constant 1495

Standard Gibbs Energy and the Equilibrium Constant If a reaction is run under conditions such that all of the reactants and products are not in their standard states – then for a reaction 1496

Standard Gibbs Energy and the Equilibrium Constant If a reaction is run under conditions such that all of the reactants and products are not in their standard states – then for a reaction a A + b B c C + d D 1497

Standard Gibbs Energy and the Equilibrium Constant If a reaction is run under conditions such that all of the reactants and products are not in their standard states – then for a reaction a A + b B c C + d D is given by = c G C + d G D – a G A – b G B 1498

Standard Gibbs Energy and the Equilibrium Constant If a reaction is run under conditions such that all of the reactants and products are not in their standard states – then for a reaction a A + b B c C + d D is given by = c G C + d G D – a G A – b G B = + – – 1499

1500