1501

1502

where Q is the reaction quotient. 1503

where Q is the reaction quotient. For a reaction system at equilibrium and Q = K c 1504

where Q is the reaction quotient. For a reaction system at equilibrium and Q = K c and hence (a very key result) 1505

For a reaction in which all the species are in the gas phase, 1506

It follows directly from the result 1507

It follows directly from the result for K c >

It follows directly from the result for K c > 1 for K c <

It follows directly from the result for K c > 1 for K c < 1 for K c = 1 In this case the position of the equilibrium does not favor either products or reactants forming. 1510

Example: The standard Gibbs energy for the reaction ½ N 2(g) + 3/2 H 2(g) NH 3(g) is kJmol -1. Calculate the equilibrium constant for the reaction at o C. In a certain experiment the initial concentrations are [H 2 ] = M, [N 2 ] = M, and [NH 3 ] = M. Predict the direction of the reaction. 1511

Example: The standard Gibbs energy for the reaction ½ N 2(g) + 3/2 H 2(g) NH 3(g) is kJmol -1. Calculate the equilibrium constant for the reaction at o C. In a certain experiment the initial concentrations are [H 2 ] = M, [N 2 ] = M, and [NH 3 ] = M. Predict the direction of the reaction. From we have 1512

Example: The standard Gibbs energy for the reaction ½ N 2(g) + 3/2 H 2(g) NH 3(g) is kJmol -1. Calculate the equilibrium constant for the reaction at o C. In a certain experiment the initial concentrations are [H 2 ] = M, [N 2 ] = M, and [NH 3 ] = M. Predict the direction of the reaction. From we have 1513

Example: The standard Gibbs energy for the reaction ½ N 2(g) + 3/2 H 2(g) NH 3(g) is kJmol -1. Calculate the equilibrium constant for the reaction at o C. In a certain experiment the initial concentrations are [H 2 ] = M, [N 2 ] = M, and [NH 3 ] = M. Predict the direction of the reaction. From we have 1514

Hence ln K c =

Hence ln K c = therefore K c =

Hence ln K c = therefore K c = 762 (Note almost one significant digit is lost in evaluating the exponential.) 1517

Hence ln K c = therefore K c = 762 (Note almost one significant digit is lost in evaluating the exponential.) To predict the direction of the reaction, use 1518

Now 1519

Now 1520

Now =

Now = 342 Since Q < K c the reaction will move in the direction to produce more NH

Employing 1523

Employing 1524

Employing = – kJ mol kJ mol -1 = – 1.99 kJ mol

Employing = – kJ mol kJ mol -1 = – 1.99 kJ mol -1 Since < 0, the reaction takes place in the forward direction. 1526

Are diamonds forever? 1527

Are diamonds forever? In chemical jargon, is the transition C diamond C graphite spontaneous at room temperature? 1528

Are diamonds forever? In chemical jargon, is the transition C diamond C graphite spontaneous at room temperature? Tabulated data: S 0 (diamond) = 2.44 J K -1 mol -1 S 0 (graphite) = 5.69 J K -1 mol

Are diamonds forever? In chemical jargon, is the transition C diamond C graphite spontaneous at room temperature? Tabulated data: S 0 (diamond) = 2.44 J K -1 mol -1 S 0 (graphite) = 5.69 J K -1 mol -1 for the reaction C graphite C diamond = 1.90 kJ 1530

For the reaction C diamond C graphite 1531

For the reaction C diamond C graphite = 1 mol x 5.69 J K -1 mol -1 – 1 mol x 2.44 J K -1 mol -1 = 3.25 J K

For the reaction C diamond C graphite = 1 mol x 5.69 J K -1 mol -1 – 1 mol x 2.44 J K -1 mol -1 = 3.25 J K -1 Using = x 10 3 J K (3.25 J K -1 ) 1533

For the reaction C diamond C graphite = 1 mol x 5.69 J K -1 mol -1 – 1 mol x 2.44 J K -1 mol -1 = 3.25 J K -1 Using = x 10 3 J K (3.25 J K -1 ) = x 10 3 J J = kJ 1534

For the reaction C diamond C graphite = 1 mol x 5.69 J K -1 mol -1 – 1 mol x 2.44 J K -1 mol -1 = 3.25 J K -1 Using = x 10 3 J K (3.25 J K -1 ) = x 10 3 J J = kJ Conclusion: The reaction C diamond C graphite is spontaneous! 1535

Electrochemistry 1536

Electrochemistry Two Key Ideas in this section: 1537

Electrochemistry Two Key Ideas in this section: (1) Use spontaneous redox chemistry to generate electricity. 1538

Electrochemistry Two Key Ideas in this section: (1) Use spontaneous redox chemistry to generate electricity. (2) Use electric current to drive non-spontaneous redox reactions. 1539

Oxidation-reduction reactions 1540

Oxidation-reduction reactions Oxidation number: A number equal to the charge an atom would have if its shared electrons were held completely by the atom that attracts them more strongly. 1541

Oxidation-reduction reactions Oxidation number: A number equal to the charge an atom would have if its shared electrons were held completely by the atom that attracts them more strongly. Need to be able to distinguish reactions where there is a change in oxidation number (these are redox reactions) from reactions where there is no change in oxidation number. 1542

1543

Example: Na 2 SO 4(aq) + BaCl 2(aq) 2 NaCl (aq) + BaSO 4(s) 1544

Example: Na 2 SO 4(aq) + BaCl 2(aq) 2 NaCl (aq) + BaSO 4(s)

Example: Na 2 SO 4(aq) + BaCl 2(aq) 2 NaCl (aq) + BaSO 4(s) There is no change in oxidation number, there is no electron transfer taking place in this reaction. Example: 2 Na (s) + Cl 2(g) 2 NaCl (s)

Example: Na 2 SO 4(aq) + BaCl 2(aq) 2 NaCl (aq) + BaSO 4(s) There is no change in oxidation number, there is no electron transfer taking place in this reaction. Example: 2 Na (s) + Cl 2(g) 2 NaCl (s)

Example: Na 2 SO 4(aq) + BaCl 2(aq) 2 NaCl (aq) + BaSO 4(s) There is no change in oxidation number, there is no electron transfer taking place in this reaction. Example: 2 Na (s) + Cl 2(g) 2 NaCl (s) In this example, there is a change in oxidation number, so electron transfer is taking place

Half-Equations: 1549

Half-Equations: Na Na + + e - (1) 1550

Half-Equations: Na Na + + e - (1) 2e - + Cl 2 2 Cl - (2) 1551

Half-Equations: Na Na + + e - (1) 2e - + Cl 2 2 Cl - (2)

Half-Equations: Na Na + + e - (1) 2e - + Cl 2 2 Cl - (2) These are called the half-equations

Half-Equations: Na Na + + e - (1) 2e - + Cl 2 2 Cl - (2) These are called the half-equations. Multiply equation (1) by 2 and add equation (2) leads to the overall balanced equation. 2 Na + Cl 2 2 NaCl

Oxidation: A process in which electrons are lost. Na Na + + e

Oxidation: A process in which electrons are lost. Na Na + + e - Reduction: A process in which electrons are gained. 2e - + Cl 2 2 Cl

Oxidation: A process in which electrons are lost. Na Na + + e - Reduction: A process in which electrons are gained. 2e - + Cl 2 2 Cl - Oxidizing Agent: A substance that gains electrons (in the reduction step). It undergoes a decrease in oxidation number. Example: Cl 2 in the above eq. 1557

Oxidation: A process in which electrons are lost. Na Na + + e - Reduction: A process in which electrons are gained. 2e - + Cl 2 2 Cl - Oxidizing Agent: A substance that gains electrons (in the reduction step). It undergoes a decrease in oxidation number. Example: Cl 2 in the above eq. Reducing Agent: A substance that donates electrons in a redox reaction (specifically in the oxidation step). It undergoes an increase in oxidation number. Example: Na in the above oxidation. 1558

Redox reaction: A reaction that can be split apart into an oxidation step and a reduction step. 1559

Redox reaction: A reaction that can be split apart into an oxidation step and a reduction step. Example: 2 Na + Cl 2 2 NaCl 1560