Chemistry Chapter 11 Molecular Composition of Gases

Volume and mass Gay-Lussac examined gas volume in reactions Noted: 2 L H 2 and 1 L O 2 can form 2 L water vapor 2:1:2 volume relationship of H:O: water Simple definite proportions hold true for other gases in reactions this lead to ….

Gay-Lussac’s Law of Combining Gas Volumes At constant temperature and pressure the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers

Avogadro (again) Combining volumes seemed to challenge the indivisibility of the atom… Avogadro posited that some molecules might contain more than one atom (ex. O 2 explains the 2:1:2 H:O:water ratio) Avogadro's Law: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

Implications of Avogadro At the same temperature and pressure the volume of a gas varies directly with the number of molecules Avogadro believed that some elements must exist in diatomic form (H 2, O 2, N 2 )

Avogadro continued H, O and water illustrate this well 2 volumes H volume O 2 = 2 volumes H 2 O Leads to the balanced equation… 2H 2 + O 2  2H 2 O Which confirms the diatomic molecule hunch nicely!

Avogadro in algebra Gas volume is directly proportional to the amount of gas (number of particles) at a given temperature and pressure give us: V = kn Where – V is volume – n is the amount of gas (in moles) – k is a constant

Molar Volumes One mole of gas at STP will occupy 22.4 L 1 mole/ 22.4 L of ________gas can be used as a conversion factor to find number of particles, mass, or volume of a gas at STP Practice! Problems 1-3 page 337

Ideal Gas Law A mathematical relationship among pressure, volume, temperature and number of moles To derive (see p. 341) Ideal gas law: V= nRT/P OR PV = nRT V is volume, P is pressure, T is temperature, n is number of moles, and R is a constant 11-2

The Ideal Gas Constant R= (1 atm) (22.4 L)/(1 mol) ( K) Or R= L x atm / mol x K (round to.0821) USE ONLY when units are appropriate! For any other units see chart on p. 342 Practice Problems! p. 345

Finding Molar Mass or Density Use V= nRT/P but remember that n (number of moles) is equal to mass (m) /molar mass (M) Substituting gives: PV= mRT/M or M= mRT/PV Density is just mass (m) per unit volume (V) Substituting gives: M= DRT/P or D= MP/RT Practice problems! 1-4 page 346

Stoichiometry of gases Volume ratios of gases in reactions can be used exactly as mole ratios are in standard mass-mole, mass-mass, mole mole etc. problems Practice! 1-2 pg

More Stoich! When given a volume for a reactant and a mass for a reactant… Go through moles Need conditions (temp, pressure, etc) for each gas Ideal gas law works well for this Practice problems! #1-2, p.349 # 1-2 p.350

Effusion and Diffusion Rates of either can be calculated! Remember! KE= ½ m v 2 And… for any two gases (A & B, lets say) at the same temp KE A = KE B so… ½ M A v A 2 = ½ M B v B 2 Where M is molar mass and v is molecular velocity You can multiply by 2 to clean up and get M A v A 2 = M B v B 2

Effusion and diffusion (con’t) Recall: M A v A 2 = M B v B 2 Rearrange: v A 2 / v B 2 = M B / M A Take square roots: v A / v B = √M B / √M A Because rate of effusion is directly proportional to molecular velocity we can say that: rate of effusion A / rate of effusion B = √M B / √M A

Grahams Law of effusion Rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses So? Density varies directly with molar mass so… rate of effusion A / rate of effusion B = √D B / √D A (Where D is density)

Grahams… Can be used to find density or molar mass of gases effusing. Practice problems! #1-3 p.355

That’s all folks!