IB Topic 1: Quantitative Chemistry 1.4 (cont): Gaseous Volume Relationships in Chemical Reactions Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Solve problems using the ideal gas equation, PV = nRT Apply Avogadro’s law to calculate reacting volumes of gases Apply the concept of molar volume at standard temperature and pressure in calculations Analyze graphs relating to the ideal gas equation

Gaseous Volume Relationships in Chemical Reactions Kinetic Theory: Tiny particles in all forms of matter are in constant motion Application to Gases 1)A gas is composed of particles that are considered to be small, hard spheres that have insignificant volume and are relatively far apart from one another. Between the particles there is empty space. No attractive or repulsive forces exist between the particles. 2) The particles in a gas move rapidly in constant random motion. They travel in straight paths and move independently of each other. They change direction only after a collision with one another or other objects. 3) All collisions are perfectly elastic. Total kinetic energy remains constant.

Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Variables That Describe a Gas Pressure (P) measured in kPa, mm Hg, atm kPa = 760 mm Hg = 1.00 atm Volume (V) measured in dm 3 or L 1 dm 3 = 1000 cm 3 = 1 L = 1000 mL Temperature (T) measured in K (Kelvin) K = o C Amount of matter (n) measured in moles

Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Boyle’s Law Relates Pressure-Volume As pressure increases, volume decreases if temperature and amount remain constant. Spaces between particles so particles can move close closer together P 1 x V 1 = P 2 x V 2 See pg 335: Sample problem Do practice problems pg 335: 10,11

Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Charles’s Law Relates Temperature-Volume As temperature increases, volume increases if pressure and amount remain constant Particles gain kinetic energy, move farther apart V 1 /T 1 = V 2 /T 2 ; T has to be in Kelvin ; K= o C See pg 337: Sample problem Do practice problems pg 337: 12,13

Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Gay-Lussac’s Law Relates Temperature-Pressure As temperature increases, pressure increases if volume and amount remain constant. Particles gain kinetic energy so they move faster and have more collisions P 1 /T 1 = P 2 /T 2 ; T has to be in Kelvin ; K = o C See pg 338: Sample problem Do practice problems pg : 14,15

Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Combined Gas Law Relates Temperature-Pressure-Volume P 1 x V 1 /T 1 = P 2 x V 2 /T 2 T has to be in Kelvin See pg 340: Sample problem Do practice problems pg 340: 16,17

Solve problems using the ideal gas equation, PV = nRT Gas Laws: Ideal Gas Law Relates Temperature-Pressure-Volume-Amount PV = nRT P = pressure in kPa V = volume in dm 3 or L n = moles R = Gas constant (8.31) T = temperature in K Ideal gas: particles have no volume and are not attracted to each other Pg 342: Sample Practice 22,23 Pg 343: Sample Practice 24,25

Apply Avogadro’s law to calculate reacting volumes of gases Avogadro’s Hypothesis: Equal volumes of gases at the same temperature and pressure contain equal numbers of particles (moles). At STP (standard temperature & pressure: 273 K and kPa) 1 mole of any gas occupies a volume of 22.4 dm 3 (L). Read pg Pg : Practice 31-36

Apply Avogadro’s law to calculate reacting volumes of gases Assuming STP, how many dm 3 of oxygen are needed to produce 19.8 dm 3 SO 3 according to: 2SO 2 (g) + O 2 (g)  2SO 3 (g) Since equal volumes of gases contain the same number of moles, we can use the equation coefficients with the volumes. X dm 3 O 2 = 1 O 2 = 9.90 dm 3 O dm 3 SO 2 2 SO 2

Apply Avogadro’s law to calculate reacting volumes of gases Nitrogen monoxide and oxygen combine to form the brown gas nitrogen dioxide. How many cm 3 of nitrogen dioxide are produced when 3.4 cm 3 of oxygen reacts with an excess of nitrogen monoxide? Assume conditions of STP.

Apply Avogadro’s law to calculate reacting volumes of gases Nitrogen monoxide and oxygen combine to form the brown gas nitrogen dioxide. How many cm 3 of nitrogen dioxide are produced when 3.4 cm 3 of oxygen reacts with and excess of nitrogen monoxide? Assume conditions of STP. 2NO + O 2  2NO 2 X cm 3 NO 2 = 2 NO 2 = 6.8 cm 3 O cm 3 O 2 1 O 2 Pg. 249: Pg. 250: 17,18

Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) a)How many dm 3 of HF are needed to produce 9.40 dm 3 H 2 at STP? b)How many grams of Sn are needed to react with 20.0 dm 3 HF at STP? c)What volume of H 2 is produced from 37.4 g Sn?

Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) a)How many dm 3 of HF are needed to produce 9.40 dm 3 H 2 at STP? x HF = 2 HF = 18.8 dm 3 HF 9.40 dm 3 H 2 1 H 2

Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) b)How many grams of Sn are needed to react with 20.0 dm 3 of HF at STP? 20.0 L HF / 22.4 dm 3 =.893 mol HF x Sn = 1 Sn =.447 mol Sn.893 HF 2 HF.893 mol Sn x = 53.0 g

Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) c)What volume of H 2 at STP is produced from 37.4 g Sn? 37.4 g / =.315 mol Sn x H 2 = 1 H 2 =.315 mol H Sn 1 Sn.315 mol H 2 x 22.4 dm 3 = 7.06 dm 3

Analyze graphs relating to the ideal gas equation

Real gases deviate from ideal behavior at low and high pressures and temperatures. Gas molecules do have some attraction for each other Gas molecules have a volume

IB Topic 1: Quantitative Chemistry 1.5 Solutions Distinguish between the terms solute, solvent, solution and concentration (g dm -3 and mol dm -3 ) Solve problems involving concentration, amount of solute and volume of solution.

Distinguish between the terms solute, solvent, solution and concentration (g dm -3 and mol dm -3 ) Solution: Homogeneous mixtures of two or more substances. Most common is solid, liquid or gas dissolved in a liquid (usually water). These are called aqueous solutions. Can have other solutions such as solid-solid (alloy) or gas-gas (air). Solute: The dissolved particles. Usually the substance in the least amount. Solvent: The dissolving medium. Usually the substance in the greater amount.

Distinguish between the terms solute, solvent, solution and concentration (g dm -3 and mol dm -3 ) Properties of Solutions Solubility Solubility is the amount of solute that dissolves in a given amount of solvent at a given temperature to produce a saturated solution. Units: grams solute/100 g solvent NaCl: solubility of 36.2 g/ 100 g water at 25 o C Any amount less than that is an unsaturated solution. A solution that contains more solute than it should theoretically is supersaturated.

Distinguish between the terms solute, solvent, solution and concentration (g dm -3 and mol dm -3 ) Concentration: Measure of the amount of solute dissolved in a given amount of solvent. g dm -3 (g/dm 3 ) –Is the number of grams of solute dissolved per dm 3 of solution mol dm -3 (mol/dm 3 ) Molarity (M) –is the number of moles of solute dissolved per dm 3 of solution. –M = mol dm -3. –Use [ ] to signify concentration in molarity. –Pg. 511: 8-11

Solve problems involving concentration, amount of solute and volume of solution. Find the concentration in g dm -3 and mol dm -3 of a solution containing 2.00 g sodium hydroxide in 125 cm 3 of solution. g dm cm 3 = dm g/0.125 dm 3 = 16.0 g dm -3 mol dm g NaOH = 2.00 g/ g mol -1 =.0500 mol.0500 mol/0.125 dm 3 = mol dm -3

Solve problems involving concentration, amount of solute and volume of solution. Calculate the amount of hydrochloric acid (in mol & g) present in cm 3 of mol dm -3 HCl(aq)?

Solve problems involving concentration, amount of solute and volume of solution. Calculate the amount of hydrochloric acid (in mol & g) present in cm 3 of mol dm -3 HCl(aq)? Molarity = mol dm -3 so Molarity x dm 3 = mol mol = mol dm -3 x dm -3 = mol HCl grams = mol x g mol -1 =.0862 g HCl

Solve problems involving concentration, amount of solute and volume of solution. What volume of a 1.25 mol dm -3 potassium permanganate solution, KMnO 4 (aq), contains 28.6 grams KMnO 4 ?

Solve problems involving concentration, amount of solute and volume of solution. What volume of a 1.25 mol dm -3 potassium permanganate solution, KMnO 4 (aq), contains 28.6 grams KMnO 4 ? Molarity = mol dm -3 so dm -3 = mol/Molarity mol = 28.6 g / g mol -1 =.181 mol dm -3 =.181 mol/1.25 mol dm -3 =.145 dm 3

Solve problems involving concentration, amount of solute and volume of solution. What will be the concentration of the solution formed by mixing 200 cm 3 of 3.00 mol dm -3 HCl(aq) with 300 cm 3 of 1.50 mol dm -3 HCl(aq)?

Solve problems involving concentration, amount of solute and volume of solution. What will be the concentration of the solution formed by mixing 200 cm 3 of 3.00 mol dm -3 HCl(aq) with 300 cm 3 of 1.50 mol dm -3 HCl(aq)? Find total moles.200 dm -3 x 3.00 mol dm -3 =.600 mol HCl.300 dm -3 x 1.50 mol dm -3 =.450 mol HCl Total moles = mol Find total volume:.200 dm dm 3 =.500 dm 3 Concentration: mol/.500 dm 3 = 2.10 mol dm -3