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The maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant T and p is equal to, that is 1441
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The maximum non-expansion work available from a reversible spontaneous process ( < 0) at constant T and p is equal to, that is 1442
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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
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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
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To prove that, start with a summary of previous results: 1445
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To prove that, start with a summary of previous results: G = H – T S (1) 1446
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To prove that, start with a summary of previous results: G = H – T S (1) H = E + pV (2) 1447
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To prove that, start with a summary of previous results: G = H – T S (1) H = E + pV (2) (3) 1448
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To prove that, start with a summary of previous results: G = H – T S (1) H = E + pV (2) (3) (4) 1449
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To prove that, start with a summary of previous results: G = H – T S (1) H = E + pV (2) (3) (4) (5) 1450
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Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) 1451
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Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) Now take a change in each variable 1452
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Plug Eq. (2) into Eq. (1) so that G = E + pV – TS (6) Now take a change in each variable (7) 1453
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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
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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
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Now fix the conditions: 1456
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Now fix the conditions: (a) constant temperature, so that, 1457
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Now fix the conditions: (a) constant temperature, so that, (b) constant pressure, so that, 1458
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Now fix the conditions: (a) constant temperature, so that, (b) constant pressure, so that, (c) and reversible process, 1459
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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
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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
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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
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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
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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
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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
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The Gibbs Energy and Equilibrium 1466
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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
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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
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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
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Example: Consider a mixture of ice and water at 0 o C and 1 bar. 1470
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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
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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
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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
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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
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Predicting the Outcome of Chemical Reactions 1475
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Predicting the Outcome of Chemical Reactions 1476 Consider the “simple” reaction A B
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Predicting the Outcome of Chemical Reactions 1477 Consider the “simple” reaction A B How do we tell which is the spontaneous direction:
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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 ?
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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.
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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
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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?
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All chemical reactions proceed so as to reach the minimum of the total Gibbs energy of the system. 1482
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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
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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
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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
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Standard Gibbs Energy and the Equilibrium Constant 1488
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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
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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
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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
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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
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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
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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
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Standard Gibbs Energy and the Equilibrium Constant 1495
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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
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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
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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
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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
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