The ICE table is as follows: H 2 I 2 HI 601

The ICE table is as follows: H 2 I 2 HI 602

The ICE table is as follows: H 2 I 2 HI – y – y y 603

The ICE table is as follows: H 2 I 2 HI – y – y y 604

Now think about any possible simplification. 605

Now think about any possible simplification. There is none! 606

Math Aside: The quadratic equation 607

Math Aside: The quadratic equation A quadratic equations looks like: 608

Math Aside: The quadratic equation A quadratic equations looks like: 609

Math Aside: The quadratic equation A quadratic equations looks like: The solutions (there are two of them) take the form 610

Math Aside: The quadratic equation A quadratic equations looks like: The solutions (there are two of them) take the form 611

Math Aside: The quadratic equation A quadratic equations looks like: The solutions (there are two of them) take the form Note that in an equilibrium problem, only one of the solutions will generate a physically acceptable solution to the problem. 612

We now need to solve 613

We now need to solve 614

We now need to solve 615

We now need to solve 616

We now need to solve Now solve the quadratic equation using with a = 50.3, b = , and c = 8.99x

618

y = M or y = M. 619

y = M or y = M. The first answer is physically impossible. 620

y = M or y = M. The first answer is physically impossible. Why? 621

y = M or y = M. The first answer is physically impossible. Why? The concentration of I 2 = – y = M which is physically impossible. We cannot have a negative concentration. 622

y = M or y = M. The first answer is physically impossible. Why? The concentration of I 2 = – y = M which is physically impossible. We cannot have a negative concentration. So the solution we seek is y = M. 623

At equilibrium: 624

At equilibrium: [H 2 ] = – = M 625

At equilibrium: [H 2 ] = – = M [I 2 ] = – = M 626

At equilibrium: [H 2 ] = – = M [I 2 ] = – = M [HI] = x = M 627

At equilibrium: [H 2 ] = – = M [I 2 ] = – = M [HI] = x = M Check: 628

At equilibrium: [H 2 ] = – = M [I 2 ] = – = M [HI] = x = M Check: 629

There is a simple way to tell which way the reaction will proceed to equilibrium. 630

There is a simple way to tell which way the reaction will proceed to equilibrium. Consider H 2(g) + I 2(g) 2 HI (g) and suppose all three gases are present at the start of the reaction. 631

There is a simple way to tell which way the reaction will proceed to equilibrium. Consider H 2(g) + I 2(g) 2 HI (g) and suppose all three gases are present at the start of the reaction. The reaction quotient is where the initial concentrations are employed. 632

There is a simple way to tell which way the reaction will proceed to equilibrium. Consider H 2(g) + I 2(g) 2 HI (g) and suppose all three gases are present at the start of the reaction. The reaction quotient is where the initial concentrations are employed. If Q c > K c there must be too much HI in the mixture for equilibrium to exist. Consequently, the reaction will proceed from right to left, until Q c becomes equal to K c. 633

There is a simple way to tell which way the reaction will proceed to equilibrium. Consider H 2(g) + I 2(g) 2 HI (g) and suppose all three gases are present at the start of the reaction. The reaction quotient is where the initial concentrations are employed. If Q c > K c there must be too much HI in the mixture for equilibrium to exist. Consequently, the reaction will proceed from right to left, until Q c becomes equal to K c. If Q c < K c, the reaction will proceed left to right. 634

In many equilibrium problems, the equilibrium constant is very small. This allows a simplification of the mathematics, so that the general quadratic equation does not have to be solved. 635

In many equilibrium problems, the equilibrium constant is very small. This allows a simplification of the mathematics, so that the general quadratic equation does not have to be solved. Example 3: 1.00 moles of N 2 is placed in a 1.00 liter vessel along with 1.10 moles of O 2 at 25 o C. Calculate the amount of NO formed at 25 o C. K c = 4.8 x at 25 o C for the reaction N 2(g) + O 2(g) 2 NO (g). 636

The ICE table is as follows: N 2 O 2 NO 637

The ICE table is as follows: N 2 O 2 NO 638

The ICE table is as follows: N 2 O 2 NO 639

The ICE table is as follows: N 2 O 2 NO 1.00 – y 1.10 – y 2y 640

The ICE table is as follows: N 2 O 2 NO 1.00 – y 1.10 – y 2y 641

Now 642

Now At this point, always stop and look for any possible simplifications. 643

Now At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. 644

Now At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. In this case there is a mathematical simplification on the right-hand side of the equation – 645

Now At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. In this case there is a mathematical simplification on the right-hand side of the equation – what is it? 646

Now At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. In this case there is a mathematical simplification on the right-hand side of the equation – what is it? Notice that K c is very small, which means very little product will be formed. 647

Since y is very small, we try the approximations: 1.00 – y – y

Since y is very small, we try the approximations: 1.00 – y – y 1.10 We will check momentarily the quality of these approximations. The equilibrium constant expression simplifies to 649

Since y is very small, we try the approximations: 1.00 – y – y 1.10 We will check momentarily the quality of these approximations. The equilibrium constant expression simplifies to K c 650

Since y is very small, we try the approximations: 1.00 – y – y 1.10 We will check momentarily the quality of these approximations. The equilibrium constant expression simplifies to K c 651

Since y is very small, we try the approximations: 1.00 – y – y 1.10 We will check momentarily the quality of these approximations. The equilibrium constant expression simplifies to K c So y = 3.6 x M. 652

Since y = 3.6 x M the approximations 1.00 – y – y 1.10 are justified. At equilibrium: [NO] = 2 y = 7.2 x M 653

When can approximations be made to simplify equilibrium constant expressions? 654

When can approximations be made to simplify equilibrium constant expressions? The only places where approximations can be made are (where A and B are concentrations): 655

When can approximations be made to simplify equilibrium constant expressions? The only places where approximations can be made are (where A and B are concentrations): (A – y) A where A >> y 656

When can approximations be made to simplify equilibrium constant expressions? The only places where approximations can be made are (where A and B are concentrations): (A – y) A where A >> y and (B + y) B where B >> y 657

When can approximations be made to simplify equilibrium constant expressions? The only places where approximations can be made are (where A and B are concentrations): (A – y) A where A >> y and (B + y) B where B >> y Never in places like y (A – y) or 658

Le Châtelier’s Principle 659

Le Châtelier’s Principle We will examine the effect of the following on the equilibrium constant. 660