1. Sunday is Groundhog Day!

2. Enjoy the Super Bowl on Sunday!

3. Thursday, Jan. 30 th : “A” Day Friday, Jan. 31 st : “B” Day Agenda  Homework Questions/Problems/Collect  Finish Section 12.3: “Molecular Composition of Gases” Gay-Lussac’s law of combining volumes, Dalton’s law of partial pressure, gas stoichiometry  Homework: Practice Pg. 442: #1-3 Section 12.3 review, pg. 442: #1-12 Concept Review: “Molecular Composition of Gases” Quiz over section 12.3 next time!

4. Homework  “Chemistry: Practice Problems for the Gas Laws” worksheet  Questions/Problems  Turn In

5. Gas Reactions Allow Chemical Formulas to be Deduced  In 1808, Joseph Gay-Lussac made an important discovery: if the pressure and temperature are kept constant, gases react in volume proportions that are small whole numbers.  Gay-Lussac’s law of combining volumes: The law that states that the volumes of gases involved in a chemical change can be represented by the ratio of small whole numbers.

6. Gas Reactions Allow Chemical Formulas to be Deduced  Hydrogen molecules combine with chlorine molecules in a 1:1 volume ratio to produce twice the volume of hydrogen chloride, HCl.

7. Dalton’s Law of Partial Pressure  In 1805, John Dalton showed that in a mixture of gases, each gas exerts a certain pressure as if it were alone with no other gases mixed with it.  Partial pressure: the pressure of each gas in a mixture.  Dalton’s law of partial pressure: The law that states that the total pressure of a mixture of gases is the sum of the partial pressures of the component gases.

8. Dalton’s Law of Partial Pressure P total = P A + P B + P C  P total is the total pressure, and P A, P B, and P C are the partial pressures of each gas.

9. Dalton’s Law of Partial Pressure  Dalton’s law of partial pressure is explained by the kinetic molecular theory.  All the gas molecules are moving randomly, and each has an equal chance to collide with the container wall.  Each gas exerts a pressure proportional to its number of molecules in the container. The presence of other gas molecules does not change this fact.

10. Gas Stoichiometry  Ratios of gas volumes will be the same as mole ratios of gases in balanced equations.  Avogadro’s law shows that the mole ratio of two gases at the same temperature and pressure is the same as the volume ratio of the two gases.  This greatly simplifies the calculation of the volume of products or reactants in a chemical reaction involving gases.

11. Gas Stoichiometry  For example, consider the Haber process for the production of ammonia gas… 3 H 2 (g) + N 2 (g) → 2 NH 3 (g)  3 L of H 2 react with 1 L of N 2 to form 2 L of NH 3, and no H 2 or N 2 is left over.  If we know the number of moles of a gaseous substance, we can use the ideal gas law to calculate the volume of that gas. And that’s what we’re going to do………..

12. Sample Problem G, pg. 441 How many liters of hydrogen gas will be produced at 280.0 K and 96.0 kPa if 1.74 mol of sodium react with excess water according to the following equation? 2 Na(s) + 2 H 2 O(l) → 2 NaOH(aq) + H 2 (g) 1.74 mol L = ?  Use stoichiometry: change mol Na mol H 2  Use mol H 2 in the ideal gas law: PV = nRT  (96.0 kPa) V = (.870 mol) (8.314) (280.0 K) V = 21.1 L H 2

13. Additional Practice A student wishes to prepare oxygen by using the thermal decomposition of potassium chlorate, KClO 3. Given that the gas will have a temperature of 700˚C and a pressure of 98.6 kPa, how much potassium chlorate will be necessary to produce 125 mL of oxygen? 2 KClO 3 3 O 2 + 2 KCl  Use PV=nRT to find moles of O 2  (98.6 kPa) (.125 L) = n(8.314)(973 K) =.0015 mol O 2  Use stoichiometry to find moles of KClO 3.  Change mol of KClO 3 to g of KClO 3. =.122 g KClO 3

14. Additional Practice #2 Methane has a volume of 0.65 L when under 100 kPa of pressure at a temperature of 305 K. Using the balanced equation below, find the mass of oxygen that is needed to use up all of the methane. CH 4 + 2 O 2 CO 2 + 2 H 2 O  Use PV = nRT to find moles of methane  (100 kPa) (0.65 L) = n (8.314) (305 K)  n =.026 mol methane  Use stoichiometry and molar mass to find mass of oxygen needed 1.66 g O 2

15. Homework  Practice, pg. 442: #1-3  Section 12.3 review, pg. 442: #1-12  Concept Review: “Molecular Composition of Gases” Next Time: 12.3 Quiz Lab Write-up Mini Lab!