Chapter 11 Molecular Composition of Gases Gay Lussacs

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Chapter 11 Molecular Composition of Gases

Chapter 11 Molecular Composition of Gases

Gay Lussac’s Law of Combining Volumes When measured at the same temperature and pressure,

Gay Lussac’s Law of Combining Volumes When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers. N 2 1 volume + + 3 H 2 → 3 volumes → 2 NH 3 2 volumes

Gay Lussac’s Law of Combining Volumes When measured at the same temperature and pressure,

Gay Lussac’s Law of Combining Volumes When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

Avogadro’s Law Equal volumes of different gases at the same temperature and pressure contain

Avogadro’s Law Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

H 2 + Cl 2 → 2 HCl hydrogen + chlorine → hydrogen chloride

H 2 + Cl 2 → 2 HCl hydrogen + chlorine → hydrogen chloride 1 volume 2 volumes 1 molecule 2 molecules 1 mol Each molecule of hydrogen and each molecule of chlorine contains 2 atoms. 2 mol

Mole-Mass-Volume Relationships

Mole-Mass-Volume Relationships

Mole-Mass-Volume Relationships • Volume of one mole of any gas at STP = 22.

Mole-Mass-Volume Relationships • Volume of one mole of any gas at STP = 22. 4 L. • 22. 4 L at STP is known as the molar volume of any gas.

Find the mass in grams of 2. 80 L of carbon dioxide? • (2.

Find the mass in grams of 2. 80 L of carbon dioxide? • (2. 8 L CO 2)(1 mol CO 2/22. 4 L CO 2)(44 g CO 2 /1 mol CO 2 ) = 5. 5 g CO 2

Density of Gases grams liters

Density of Gases grams liters

Density of Gases depends on T and P

Density of Gases depends on T and P

The molar mass of SO 2 is 64. 07 g/mol. Determine the density of

The molar mass of SO 2 is 64. 07 g/mol. Determine the density of SO 2 at STP. 1 mole of any gas occupies 22. 4 L at STP

Ideal Gas Equation n. T Va P

Ideal Gas Equation n. T Va P

atmospheres n. T Va P

atmospheres n. T Va P

liters n. T Va P

liters n. T Va P

moles n. T Va P

moles n. T Va P

Kelvin n. T Va P

Kelvin n. T Va P

Ideal Gas Constant n. T Va P

Ideal Gas Constant n. T Va P

A balloon filled with 5. 00 moles of helium gas is at a temperature

A balloon filled with 5. 00 moles of helium gas is at a temperature of 25 o. C. The atmospheric pressure is 0. 987 atm. What is the balloon’s volume? Step 1. Organize the given information. Convert temperature to kelvins. K = o. C + 273 K = 25 o. C + 273 = 298 K

A balloon filled with 5. 00 moles of helium gas is at a temperature

A balloon filled with 5. 00 moles of helium gas is at a temperature of 25 o. C. The atmospheric pressure is. 987 atm. What is the balloon’s volume? Step 2. Write and solve the ideal gas equation for the unknown. Step 3. Substitute the given data into the equation and calculate.

Determination of Molecular Weights Using the Ideal Gas Equation

Determination of Molecular Weights Using the Ideal Gas Equation

Calculate the molar mass of an unknown gas, if 0. 020 g occupies 250

Calculate the molar mass of an unknown gas, if 0. 020 g occupies 250 m. L at a temperature of 305 K and a pressure of 0. 045 atm. V = 250 m. L = 0. 250 L g = 0. 020 g T = 305 K P = 0. 045 atm

Determination of Density Using the Ideal Gas Equation • Density = mass/volume D =

Determination of Density Using the Ideal Gas Equation • Density = mass/volume D = MP/ RT

The density of a gas was measured at 1. 50 atm and 27 ºC

The density of a gas was measured at 1. 50 atm and 27 ºC and found to be 1. 95 g/L. Calculate the molar mass of the gas. M = d. RT P • (1. 95 g/L)(0. 08206 L atm/K mol)(300 K) 1. 50 atm = 32. 0 g/mol

Gas Stoichiometry deals with the quantitative relationships among reactants and products in a chemical

Gas Stoichiometry deals with the quantitative relationships among reactants and products in a chemical reaction. • All calculations are done at STP. • Gases are assumed to behave as ideal gases. • A gas not at STP is converted to STP.

Gas Stoichiometry Primary conversions involved in stoichiometry.

Gas Stoichiometry Primary conversions involved in stoichiometry.

What volume of oxygen (at STP) can be formed from 0. 500 mol of

What volume of oxygen (at STP) can be formed from 0. 500 mol of potassium chlorate? • Step 1 Write the balanced equation 2 KCl. O 3 2 KCl + 3 O 2 • Step 2 The starting amount is 0. 500 mol KCl. O 3. The conversion is moles KCl. O 3 moles O 2 liters O 2

What volume of oxygen (at STP) can be formed from 0. 500 mol of

What volume of oxygen (at STP) can be formed from 0. 500 mol of potassium chlorate? 2 KCl. O 3 2 KCl + 3 O 2 • Step 3. Calculate the moles of O 2, using the moleratio method. • Step 4. Convert moles of O 2 to liters of O 2

What volume of oxygen (at STP) can be formed from 0. 500 mol of

What volume of oxygen (at STP) can be formed from 0. 500 mol of potassium chlorate? The problem can also be solved in one continuous calculation. 2 KCl. O 3 2 KCl + 3 O 2

What volume of hydrogen, collected at 30. o. C and 0. 921 atm, will

What volume of hydrogen, collected at 30. o. C and 0. 921 atm, will be formed by reacting 50. 0 g of aluminum with hydrochloric acid? 2 Al(s) + 6 HCl(aq) 2 Al. Cl 3(aq) + 3 H 2(g) Step 1 Calculate moles of H 2. grams Al moles H 2

What volume of hydrogen, collected at 30. o. C and 0. 921 atm, will

What volume of hydrogen, collected at 30. o. C and 0. 921 atm, will be formed by reacting 50. 0 g of aluminum with hydrochloric acid? 2 Al(s) + 6 HCl(aq) 2 Al. Cl 3(aq) + 3 H 2(g) Step 2 Calculate liters of H 2. • Convert o. C to K: 30. o. C + 273 = 303 K

What volume of hydrogen, collected at 30. o. C and 700. torr, will be

What volume of hydrogen, collected at 30. o. C and 700. torr, will be formed by reacting 50. 0 g of aluminum with hydrochloric acid? • Solve the ideal gas equation for V PV = n. RT

For reacting gases at constant temperature and pressure: Volume-volume relationships are the same as

For reacting gases at constant temperature and pressure: Volume-volume relationships are the same as mole-mole relationships. H 2(g) + Cl 2(g) 2 HCl(g) 1 mol H 2 1 mol Cl 2 2 mol HCl 22. 4 L STP 1 volume 2 x 22. 4 L STP 2 volumes

What volume of nitrogen will react with 600. m. L of hydrogen to form

What volume of nitrogen will react with 600. m. L of hydrogen to form ammonia? What volume of ammonia will be formed? N 2(g) + 3 H 2(g) 2 NH 3(g)

Mass- volume problem

Mass- volume problem

Graham’s Law of Effusion • Effusion – the process whereby the molecules of a

Graham’s Law of Effusion • Effusion – the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container. • Rate (like diffusion) depends on the relative velocities of gas molecules. • Lighter molecules move faster than heavier molecules at the same temperature.

Graham’s Law of Effusion • The rates of effusion of gases at the same

Graham’s Law of Effusion • The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. • Rate of effusion of A = MB Rate of effusion of B MA

Calculate the ratio of the effusion rates of H 2 & UF 6, a

Calculate the ratio of the effusion rates of H 2 & UF 6, a gas used in the enrichment process to produce fuel for nuclear reactors. • Rate of effusion of H 2 Rate of effusion of UF • Square root of molar mass of UF 6 Square root of molar mass of H 2 Square root of 352. 02/2. 016 = 13. 2