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 Gases1) 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, atm101.3 kPa = 760 mm Hg = 1.00 atm

• Volume (V) measured in dm3 or L1 dm3 = 1000 cm3 = 1 L = 1000 mL

• Temperature (T) measured in K (Kelvin) K = oC + 273

• 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

• P1 x V1 = P2 x V2

• See pg 335: Sample problem 12-1. 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• V1/T1 = V2/T2 ; • T has to be in Kelvin ; K= oC + 273• See pg 337: Sample problem 12-2. 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

• P1/T1 = P2/T2 ; T has to be in Kelvin ; K = oC + 273

• See pg 338: Sample problem 12-3. Do practice problems pg 338-339: 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

• P1 x V1/T1 = P2 x V2/T2

T has to be in Kelvin• See pg 340: Sample problem 12-4. 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 dm3 or L• n = moles• R = Gas constant (8.31)• T = temperature in KIdeal gas: particles have no volume and are not attracted to

each otherPg 342: Sample 12-5. Practice 22,23Pg 343: Sample 12-6. 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 101.3 kPa) 1 mole of any gas occupies a volume of 22.4 dm3 (L).

Read pg.347-348. Pg. 348-349: Practice 31-36

Apply Avogadro’s law to calculate reacting volumes of

gasesAssuming STP, how many dm3 of oxygen are

needed to produce 19.8 dm3 SO3 according to:

2SO2(g) + O2(g) 2SO3(g)

Since equal volumes of gases contain the same number of moles, we can use the equation coefficients with the volumes.

X dm3 O2 = 1 O2 = 9.90 dm3 O2

19.8 dm3 SO2 2 SO2

Apply Avogadro’s law to calculate reacting volumes of

gasesNitrogen monoxide and oxygen combine to form the

brown gas nitrogen dioxide. How many cm3 of nitrogen dioxide are produced when 3.4 cm3 of oxygen reacts with an excess of nitrogen monoxide? Assume conditions of STP.

Apply Avogadro’s law to calculate reacting volumes of

gasesNitrogen monoxide and oxygen combine to form the

brown gas nitrogen dioxide. How many cm3 of nitrogen dioxide are produced when 3.4 cm3 of oxygen reacts with and excess of nitrogen monoxide? Assume conditions of STP.

2NO + O2 2NO2

X cm3 NO2 = 2 NO2 = 6.8 cm3 O2

3.4 cm3 O2 1 O2

Pg. 249: 15-16. 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) SnF2(s) + H2(g)

a) How many dm3 of HF are needed to produce 9.40 dm3 H2 at STP?

b) How many grams of Sn are needed to react with 20.0 dm3 HF at STP?

c) What volume of H2 is produced from 37.4 g Sn?

Apply the concept of molar volume at standard temperature and

pressure in calculationsTin(II) fluoride, formerly found in many kinds of

toothpaste, is formed by this reaction: Sn(s) + 2HF(g) SnF2(s) + H2(g)

a) How many dm3 of HF are needed to produce 9.40 dm3 H2 at STP?

x HF = 2 HF = 18.8 dm3 HF9.40 dm3 H2 1 H2

Apply the concept of molar volume at standard temperature and

pressure in calculationsTin(II) fluoride, formerly found in many kinds of

toothpaste, is formed by this reaction: Sn(s) + 2HF(g) SnF2(s) + H2(g)

b) How many grams of Sn are needed to react with 20.0 dm3 of HF at STP?20.0 L HF / 22.4 dm3 = .893 mol HF x Sn = 1 Sn = .447 mol Sn.893 HF 2 HF.893 mol Sn x 118.69 = 53.0 g

Apply the concept of molar volume at standard temperature and

pressure in calculationsTin(II) fluoride, formerly found in many kinds of

toothpaste, is formed by this reaction: Sn(s) + 2HF(g) SnF2(s) + H2(g)

c) What volume of H2 at STP is produced from 37.4 g Sn?37.4 g / 118.69 = .315 mol Sn x H2 = 1 H2 = .315 mol H2

.315 Sn 1 Sn

.315 mol H2 x 22.4 dm3 = 7.06 dm3

Analyze graphs relating to the ideal gas equation

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 SolutionsSolubility

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 oC

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/dm3)– Is the number of grams of solute dissolved per dm3 of

solution

• mol dm-3 (mol/dm3) Molarity (M) – is the number of moles of solute dissolved per dm3 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 cm3 of solution.

g dm-3

• 125 cm3 = 0.125 dm3

• 2.00 g/0.125 dm3 = 16.0 g dm-3

mol dm-3

• 2.00 g NaOH = 2.00 g/ 40.01 g mol-1 = .0500 mol• .0500 mol/0.125 dm3 = 0.400 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 23.65 cm3 of 0.100 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 23.65 cm3 of 0.100 mol dm-3 HCl(aq)?

Molarity = mol dm-3 so Molarity x dm3 = mol

mol = 0.100 mol dm-3 x .02365 dm-3 = .002365 mol HCl

grams = .002365 mol x 36.46 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, KMnO4(aq), contains 28.6 grams KMnO4?

Solve problems involving concentration, amount of solute and volume of solution.

What volume of a 1.25 mol dm-3 potassium permanganate solution, KMnO4(aq), contains 28.6 grams KMnO4?

Molarity = mol dm-3 so dm-3 = mol/Molarity

mol = 28.6 g / 158.04 g mol-1 = .181 mol

dm-3 = .181 mol/1.25 mol dm-3 = .145 dm3

Solve problems involving concentration, amount of solute and volume of solution.

What will be the concentration of the solution formed by mixing 200 cm3 of 3.00 mol dm-3 HCl(aq) with 300 cm3 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 cm3 of 3.00 mol dm-3 HCl(aq) with 300 cm3 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 HClTotal moles = 1.050 mol

Find total volume: .200 dm3 + .300 dm3 = .500 dm3

Concentration: 1.050 mol/ .500 dm3 = 2.10 mol dm-3