1. Lecture 31/26/05

2. Initial 1.00 M 0 Change (  ) Equil. 0.925 M What if 1.00 mol of SO 2 and 1.00 mol of O 2 put into 1.00-L flask at 1000 K. At equilibrium, 0.925 mol of SO 3 have been formed. Calculate K 2SO 2 (g) + O 2 (g) ↔ 2SO 3 (g)

3. Initial 1.00 M 0 Change (  ) Equil. 0.925 M What if 1.00 mol of SO 2 and 1.00 mol of O 2 put into 1.00-L flask at 1000 K. At equilibrium, 0.925 mol of SO 3 have been formed. Calculate K 2SO 2 (g) + O 2 (g) ↔ 2SO 3 (g)

4. Initial 1.00 M 0 Change (  ) + 0.925 M Equil. 0.925 M What if 1.00 mol of SO 2 and 1.00 mol of O 2 put into 1.00-L flask at 1000 K. At equilibrium, 0.925 mol of SO 3 have been formed. Calculate K 2SO 2 (g) + O 2 (g) ↔ 2SO 3 (g)

5. Initial 1.00 M 0 Change (  ) - 0.925 M- 0.925/2 M+ 0.925 M Equil. 0.925 M What if 1.00 mol of SO 2 and 1.00 mol of O 2 put into 1.00-L flask at 1000 K. At equilibrium, 0.925 mol of SO 3 have been formed. Calculate K 2SO 2 (g) + O 2 (g) ↔ 2SO 3 (g)

6. Initial 1.00 M 0 Change (  ) - 0.925 M- 0.4625 M+ 0.925 M Equil. 0.075 M0.5375 M0.925 M What if 1.00 mol of SO 2 and 1.00 mol of O 2 put into 1.00-L flask at 1000 K. At equilibrium, 0.925 mol of SO 3 have been formed. Calculate K 2SO 2 (g) + O 2 (g) ↔ 2SO 3 (g)

8. Skills - recap Need to be able to: Balance a reaction Equilibrium constant expression Make ICE table

9. Example: Finding equilibrium concentrations H 2 and I 2 combine to form HI. If you place 1 mole each of H 2 and I 2 into a 0.5 L flask at 699 K, what are all the equilibrium concentrations? K = 55.17 at 699 K.

10. Step 1: Balance equation H 2 and I 2 combine to form HI. If you place 1 mole each of H 2 and I 2 into a 0.5 L flask at 699 K, what are all the equilibrium concentrations? K= 55.17 at 699 K H 2 + I 2 ↔ HI H 2 + I 2 ↔ 2HI

11. Step 2: Equilibrium expression H 2 and I 2 combine to form HI. If you place 1 mole each of H 2 and I 2 into a 0.5 L flask at 699 K, what are all the equilibrium concentrations? K= 55.17 at 699 K

12. Step 3: ICE table H 2 and I 2 combine to form HI. If you place 1 mole each of H 2 and I 2 into a 0.5 L flask at 699 K, what are all the equilibrium concentrations? K= 55.17 at 699 K H 2 + I 2 ↔ 2HI Initial 2.00 M 0 Change (  ) Equil.

13. Step 3: ICE table H 2 and I 2 combine to form HI. If you place 1 mole each of H 2 and I 2 into a 0.5 L flask at 699 K, what are all the equilibrium concentrations? K= 55.17 at 699 K H 2 + I 2 ↔ 2HI Initial 2.00 M 0 Change (  ) - X + 2X Equil. From stoichiomety

14. Step 3: ICE table H 2 and I 2 combine to form HI. If you place 1 mole each of H 2 and I 2 into a 0.5 L flask at 699 K, what are all the equilibrium concentrations? K= 55.17 at 699 K H 2 + I 2 ↔ 2HI Initial 2.00 M 0 Change (  ) - X + 2X Equil. 2.00 – X 2X

15. Step 4: Substitute into equilibrium expression H 2 and I 2 combine to form HI. If you place 1 mole each of H 2 and I 2 into a 0.5 L flask at 699 K, what are all the equilibrium concentrations? K= 55.17 at 699 K

17. Step 5: Calculate equilibrium concentrations using X H 2 and I 2 combine to form HI. If you place 1 mole each of H 2 and I 2 into a 0.5 L flask at 699 K, what are all the equilibrium concentrations? K= 55.17 at 699 K X=1.58 H 2 = 2 - X = 2-1.58 = 0.42 M I 2 = 2 - X = 2-1.58 = 0.42 M HI = 2X = 2(1.58) = 3.16 M

18. For the reaction: CO (g) + H 2 O (g) ↔ CO 2 (g) + H 2 (g) If a reaction vessel at 673 K is charged with an equimolar mixture of CO and steam such that P CO =P H2O =2.00 atm. What are the partial pressures at equilibrium? K p = 10 at 673 K