Chemical Equilibrium………..  Let’s look at problem #12 in the homework….  The reaction of elemental hydrogen and fluorine to form hydrofluoric acid has an equilibrium constant of 1.15 x 10 2 at a certain temperature. In one experiment, 3.00 mol of all 3 components, including the product, was added to a 1.50-liter flask. Calculate the equilibrium concentrations of all species.  Write the reaction:  H 2 (g) + F 2 (g) > 2 HF (g)  This is an odd problem, because there is actually some of the product present before the reaction actually begins! This is not common – product isn’t usually present until the reaction happens!

Chemical Equilibrium………..  We want to set up an ICE:  1 H 2 (g) + 1 F 2 (g) > 2 HF (g) I C E Remember, I mean initial, C means change, and E means end or equilibrium! And we use concentration – not moles, or amount! What are the initial concentrations of each chemical?

Chemical Equilibrium………..  What does our ICE look like?  1 H 2 (g) + 1 F 2 (g) > 2 HF (g) I 2M 2M 2M C -x -x +2x E 2 - x 2 - x 2 + 2x  K c 115 = [HF] 2 = [2 + 2x] 2 1 [H 2 ] 1 [F 2 ] 1 [2-x] 1 [2-x] 1 Remember, we insert the concentrations into our K at equilibrium! We want to solve for the amounts at equilibrium – this is going to require some algebra!

Chemical Equilibrium………..  K c 115 = [HF] 2 = [2 + 2x] 2 1 [H 2 ] 1 [F 2 ] 1 [2-x][2-x] We have to expand this…..  115 = 4 + 8x + 4x x + x – 460x + 115x 2 = 4 + 8x + 4x 2 Rearranging, 0 = 456 – 468x + 111x 2 or 111x 2 – 468x = 0 How do we solve this?

Chemical Equilibrium……….. 111x 2 – 468x = 0 Use the quadratic formula! You can use the program on your calculator to solve this You can go to and enter in the equation – it will solve it for you! What are the solutions for x? X = and One of these solutions can’t work…. Which one? Why?

Chemical Equilibrium………..  H 2 (g) + F 2 (g) > 2 HF (g) I 2M 2M 2M C -x -x +2x E 2 - x 2 - x 2 + 2x  X = and  It is impossible to lose when you only start with 2!  This is the non-real solution –  So x =  What are the amounts at equilibrium then?

Chemical Equilibrium………..  H 2 (g) + F 2 (g) > 2 HF (g) I 2M 2M 2M C -x -x +2x E Let’s plug these values back into K to see if they equal 115/1!  K c 115 = [HF] 2 = [5.0568] /1 1 [H 2 ] 1 [F 2 ] 1 [.4716] 1 [.4716] 1

We can set up a ratio of pressures instead of concentrations too! This is called K p Let’s look at number 24 – Hydrogen gas and iodine vapor react to form hydrogen iodide gas. The equilibrium constant, or K c, is 1.00 x 10 2 at 25 0 C. Suppose hydrogen at 5.0 x atm, iodine at 1.0 x atm, and hydrogen iodide at 5.0 x atm are added to a 5.0 L flask. Calculate the equilibrium pressures of all species. What is the reaction?

Hydrogen gas and iodine vapor react to form hydrogen iodide gas. The equilibrium constant, or K c, is 1.00 x 10 2 at 25 0 C. Suppose hydrogen at 5.0 x atm, iodine at 1.0 x atm, and hydrogen iodide at 5.0 x atm are added to a 5.0 L flask. Calculate the equilibrium pressures of all species. H 2 (g) + I 2 (g) > 2 HI (g) What would K p look like? K p = [P HI] 2 [P H 2 ] 1 [P I 2 ] 1 Where P is the pressure, instead of concentration, of the gas! But the problem gives us the K c, and not the K p – are they the same? No!

K p = K c (RT)  n What is delta n? It is the moles of gas produced minus the moles of gas you start with – In this reaction, you make 2 moles of HI, and start with one mole of H 2, and one mole of I 2 So,  n = 2 – 2 = 0 K p = 100 [(.0821)(298)] 0 K p = 100 In this case, the K p actually did equal the K c ! So…..

 H 2 (g) + I 2 (g) > 2 HI (g) I.5 atm.1 atm.5 atm C -x -x +2x E.5 – x.1 – x.5 + 2x Same exact setup as before – but we are using pressures instead of concentrations, so we use a K p and not a K c ! K p = [.5+2x] 2 = 100 [.5-x] 1 [.1-x] 1 Can you do the algebra?

Try and do problems 1-31 for homework this weekend…. I will do a homework check on Tuesday…. Remember, if concentrations are given, use a K c – if pressures are given, use a K p – To convert between them: K p = K c (RT)  n