Chapter 5 – AP Chemistry Chapter 5 – AP Chemistry

GasesGases

Substances that Exist as GasesSubstances that Exist as Gasesat 1atm and 25 Cat 1atm and 25 C

ElementsElements– HH22

– NN22

– OO22

– FF22

– ClCl22– HeHe– NeNe– ArAr– KrKr– XeXe– RnRn

CompoundsCompounds– HFHF– HClHCl– HBrHBr– HIHI– COCO– COCO22

– NHNH33

– NONO– NONO22

– NN22OO– SOSO22

All Gases have the following All Gases have the following Physical PropertiesPhysical Properties

Assume the volume and shape of their Assume the volume and shape of their containerscontainers

Are the most compressible of the states of Are the most compressible of the states of mattermatter

Will mix evenly and completely when Will mix evenly and completely when confined to the same containerconfined to the same container

Have much lower densities than liquids and Have much lower densities than liquids and solidssolids

Pressure of a GasPressure of a Gas

Gases exert pressure on any surface with Gases exert pressure on any surface with which they come into contactwhich they come into contact

– Example of atmospheric pressureExample of atmospheric pressure The ability to drink liquid through a strawThe ability to drink liquid through a straw

– Sucking air out of the straw reduces the pressure inside the Sucking air out of the straw reduces the pressure inside the strawstraw

– Therefore the greater atmospheric pressure on the liquid Therefore the greater atmospheric pressure on the liquid pushes it up into the straw to replace the air that has been pushes it up into the straw to replace the air that has been sucked outsucked out

Atmospheric PressureAtmospheric Pressure The pressure exerted by Earth’s atmosphereThe pressure exerted by Earth’s atmosphere

Atoms and molecules are subjected to Atoms and molecules are subjected to earth’s gravitational pullearth’s gravitational pull– Therefore, the atmosphere is much denser near Therefore, the atmosphere is much denser near

the surface of the earththe surface of the earth– The denser the air is the greater the pressureThe denser the air is the greater the pressure

AirAir

Gases are fluid Gases are fluid – Pressure exerted on an object in a fluid comes from all Pressure exerted on an object in a fluid comes from all

directionsdirections

Air pressure at the molecular level results from Air pressure at the molecular level results from collisions between air molecules and any surface collisions between air molecules and any surface with which they come into contact withwith which they come into contact with

The magnitude of pressure depends on how often The magnitude of pressure depends on how often and how strongly the molecules impact the surfaceand how strongly the molecules impact the surface

Gas LawsGas Laws

Boyle’s LawBoyle’s Law– PP11VV11 = P = P22VV22

Charles’s LawCharles’s Law– VV11/T/T11 = V = V22/T/T22

Gay – Lussac’s LawGay – Lussac’s Law– PP11/T/T11 = P = P22/T/T22

Combined Gas LawCombined Gas Law– PP11VV11/T/T11 = P = P22VV22/T/T22

Combined Gas LawCombined Gas Law

A small bubble rises from the bottom of a A small bubble rises from the bottom of a lake, where the temperature and pressure lake, where the temperature and pressure are 8 degrees Celsius and 6.4atm, to the are 8 degrees Celsius and 6.4atm, to the water’s surface, where the temperature is 25 water’s surface, where the temperature is 25 degrees Celsius and pressure is 1.0 atm. degrees Celsius and pressure is 1.0 atm. Calculate the final volume of the bubble if its Calculate the final volume of the bubble if its initial volume was 2.1ml. initial volume was 2.1ml.

Ideal Gas LawIdeal Gas Law

PV=nRTPV=nRT– R = .0821L*atm/K*molR = .0821L*atm/K*mol– Volume in LVolume in L– Pressure in atmPressure in atm– n in molesn in moles– Temperature in KelvinTemperature in Kelvin

Sulfur hexafluoride is a colorless, odorless, very Sulfur hexafluoride is a colorless, odorless, very unreactive gas. Calculate the pressure exerted by unreactive gas. Calculate the pressure exerted by 84g of the gas in a steel vessel of a volume 6.09L 84g of the gas in a steel vessel of a volume 6.09L and 55 degrees Celcius.and 55 degrees Celcius.

Ideal Gas LawIdeal Gas Law

Calculate the volume occupied by 14.2g of Calculate the volume occupied by 14.2g of NHNH3 3 atat STP. (Hint 22.4L = 1mol)STP. (Hint 22.4L = 1mol)

Gas StoichiometryGas Stoichiometry

Same a basic reaction chemistry Same a basic reaction chemistry – Please note: 1mole = 22.4L can only be used Please note: 1mole = 22.4L can only be used

when the chemical reaction is at STPwhen the chemical reaction is at STP– If the reaction is at STP, after moles of the If the reaction is at STP, after moles of the

unknown are calculated, then the moles can be unknown are calculated, then the moles can be converted to litersconverted to liters

– If the reaction is NOT at STP then you must use If the reaction is NOT at STP then you must use ideal gas law to solve for volumeideal gas law to solve for volume

Gas Stoich ProblemsGas Stoich Problems

Sodium Azide (NaNSodium Azide (NaN33) is used in some ) is used in some automobile air bags. The impact of a automobile air bags. The impact of a collision triggers the decomposition of NaNcollision triggers the decomposition of NaN33. . The nitrogen gas produced quickly inflates The nitrogen gas produced quickly inflates the bag between the driver and the the bag between the driver and the windshield and dashboard. Calculate the windshield and dashboard. Calculate the volume of Nvolume of N22 generated at 85 degrees generated at 85 degrees Celsius and 812mmHg by decomposition of Celsius and 812mmHg by decomposition of 50g NaN50g NaN33..

Gas StoichGas Stoich

The breakdown on glucose is below. The breakdown on glucose is below. Calculate the volume of COCalculate the volume of CO22 produced at 37 produced at 37

degrees and .5atm when 5.6g of glucose is degrees and .5atm when 5.6g of glucose is used to completion in this reactionused to completion in this reaction

CC66HH1212OO66 + O + O22 → CO → CO22 + H + H22O O

Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

Dalton's lawDalton's law of partial pressuresof partial pressures states states that the total that the total pressurepressure exerted by a exerted by a gaseousgaseous mixture is equal to the sum of the mixture is equal to the sum of the partial pressurespartial pressures of each individual of each individual component in a gas mixture.component in a gas mixture.

Dalton’s Law of Partial PressureDalton’s Law of Partial Pressure Calculation of the moles of each gas based on the partial Calculation of the moles of each gas based on the partial

pressurespressures

– For example two gases in a container (nitrogen and oxygen)For example two gases in a container (nitrogen and oxygen)– P = PP = Pnitrogennitrogen + P + Poxygenoxygen – Then you can use ideal gas law to calculate the moles of each gasThen you can use ideal gas law to calculate the moles of each gas– PPnitrogennitrogen V = n V = nnitrogennitrogen R T R T

– PPoxygenoxygen V = n V = noxygenoxygen R T R T

Please note you can always use the equations backwards. Please note you can always use the equations backwards. Start with moles of each to find partial pressures, then find Start with moles of each to find partial pressures, then find the total pressurethe total pressure

Dalton’s Law ProblemsDalton’s Law ProblemsQuiz ProblemQuiz Problem

Oxygen gas is generated by the Oxygen gas is generated by the decomposition of potassium chlorate in wate decomposition of potassium chlorate in wate vapor chamber. The volume of oxygen gas vapor chamber. The volume of oxygen gas collected at 26 degrees Celsius is 752ml collected at 26 degrees Celsius is 752ml and the atmospheric pressure is and the atmospheric pressure is 1841mmHg. The pressure of the water 1841mmHg. The pressure of the water vapor at 26 degrees Celsius is 25.2 mmHg. vapor at 26 degrees Celsius is 25.2 mmHg. Calculate the mass of oxygen obtained.Calculate the mass of oxygen obtained.

Dalton’s Law of Partial PressuresDalton’s Law of Partial PressuresMole FractionMole Fraction

The mole fraction that express the ratio of The mole fraction that express the ratio of the number of moles of one component to the number of moles of one component to the number of moles of all components the number of moles of all components presentpresent

By using mole fraction you can calculate the By using mole fraction you can calculate the partial pressure of each gas in a systempartial pressure of each gas in a system

PPii = X = XiiPPTT

XXi i -Is the percent of each molar amount of -Is the percent of each molar amount of each gas (part over whole)each gas (part over whole)

Dalton’s Law ProblemsDalton’s Law Problems

A mixture of gases contain 3.85 moles of A mixture of gases contain 3.85 moles of Ne, 0.92 moles of argon and 2.59 moles of Ne, 0.92 moles of argon and 2.59 moles of xenon. Calulate the partial pressure of each xenon. Calulate the partial pressure of each gas if the total pressure is 2.50atm.gas if the total pressure is 2.50atm.

Kinetic Molecular Theory of GasesKinetic Molecular Theory of Gases

A gas consists of a collection of small particles traveling in A gas consists of a collection of small particles traveling in straight-line motion and obeying Newton's Laws. straight-line motion and obeying Newton's Laws.

The molecules in a gas occupy no volume (that is, they are The molecules in a gas occupy no volume (that is, they are points). points).

Collisions between molecules are perfectly elastic (that is, Collisions between molecules are perfectly elastic (that is, no energy is gained or lost during the collision). no energy is gained or lost during the collision).

There are no attractive or repulsive forces between the There are no attractive or repulsive forces between the molecules. molecules.

The average kinetic energy of a molecule is 3The average kinetic energy of a molecule is 3kTkT/2. (/2. (TT is the is the absolute temperature and absolute temperature and kk is the Boltzmann constant.) is the Boltzmann constant.)

Root Mean SquaredRoot Mean Squared

To calculate how fast a molecule moves at any To calculate how fast a molecule moves at any given temperaturegiven temperature

R = 8.314J/K*molR = 8.314J/K*molM = (molar mass in Kg/mol)M = (molar mass in Kg/mol)U = calculated in meters per secondU = calculated in meters per second

Root Mean Squared ProblemRoot Mean Squared Problem

Calculate the root-mean squared speeds of Calculate the root-mean squared speeds of helium atoms and nitrogen molecules in m/s helium atoms and nitrogen molecules in m/s at 25 degrees Celsiusat 25 degrees Celsius

Gas Diffusion and EffusionGas Diffusion and Effusion

Diffusion – the gradual mixing of molecules Diffusion – the gradual mixing of molecules of one gas with molecules of another by of one gas with molecules of another by virtue of their kinetic propertiesvirtue of their kinetic properties

Effusion – the process by which a gas under Effusion – the process by which a gas under pressure escapes from one compartment of pressure escapes from one compartment of a container to another by passing through a a container to another by passing through a small openingsmall opening

Graham’s law of DiffisionGraham’s law of Diffision

A flammable gas made up only of A flammable gas made up only of carbon and hydrogen is found to carbon and hydrogen is found to effuse through a porous barrier in effuse through a porous barrier in 3.5min. Under the same 3.5min. Under the same conditions of temperature and conditions of temperature and pressure, it takes an equal pressure, it takes an equal volume of chlorine gas 7.34 min volume of chlorine gas 7.34 min to effuse through the same to effuse through the same barrier. Calculate the molar mass barrier. Calculate the molar mass of the unknown gas and suggest of the unknown gas and suggest what this gas might be.what this gas might be.

Deviation from Ideal BehaviorDeviation from Ideal Behavior

Van der Waals Van der Waals

AP QuestionsAP Questions 2 H2 H22OO22(aq)(aq) → 2 H → 2 H22OO(l)(l) + O + O22(g)(g) The mass of an aqueous solution of HThe mass of an aqueous solution of H22OO22 is 6.951 g. The H is 6.951 g. The H22OO22 in the in the

solution decomposes completely according to the reaction represented solution decomposes completely according to the reaction represented above. The Oabove. The O22(g)(g) produced is collected in an inverted graduated tube produced is collected in an inverted graduated tube over water at 23.4°C and has a volume of 182.4 mL when the water over water at 23.4°C and has a volume of 182.4 mL when the water levels inside and outside of the tube are the same. The atmospheric levels inside and outside of the tube are the same. The atmospheric pressure in the lab is 762.6 torr, and the equilibrium vapor pressure of pressure in the lab is 762.6 torr, and the equilibrium vapor pressure of water at 23.4°C is 21.6 torr.water at 23.4°C is 21.6 torr.

(a)(a) Calculate the partial pressure, in torr, of O2Calculate the partial pressure, in torr, of O2(g)(g) in the gas- in the gas-collection tube.collection tube.

(b)(b) Calculate the number of moles of OCalculate the number of moles of O22(g)(g) produced in the produced in the reaction.reaction.

(c)(c) Calculate the mass, in grams, of HCalculate the mass, in grams, of H22OO22 that decomposed. that decomposed. (d)(d) Calculate the percent of HCalculate the percent of H22OO22 , by mass, in the original 6.951 g , by mass, in the original 6.951 g

aqueous sample.aqueous sample.

AP QuestionsAP Questions

A rigid 5.00 L cylinder contains 24.5 g of NA rigid 5.00 L cylinder contains 24.5 g of N22(g)(g) and 28.0 g of O and 28.0 g of O22(g)(g) (a)(a) Calculate the total pressure, in atm, of the gas mixture in the cylinder Calculate the total pressure, in atm, of the gas mixture in the cylinder

at 298 K.at 298 K. (b)(b) The temperature of the gas mixture in the cylinder is decreased to 280 The temperature of the gas mixture in the cylinder is decreased to 280

K. Calculate each of the following.K. Calculate each of the following. (i)(i) The mole fraction of NThe mole fraction of N22(g)(g) in the cylinder. in the cylinder. (ii)(ii) The partial pressure, in atm, of NThe partial pressure, in atm, of N22(g)(g) in the cylinder. in the cylinder. (c)(c) If the cylinder develops a pinhole-sized leak and some of the gaseous If the cylinder develops a pinhole-sized leak and some of the gaseous

mixture escapes, would the ratio   in the cylinder increase, decrease, or remain mixture escapes, would the ratio   in the cylinder increase, decrease, or remain the same? Justify your answer.the same? Justify your answer.

A different rigid 5.00 L cylinder contains 0.176 mol of NOA different rigid 5.00 L cylinder contains 0.176 mol of NO(g)(g) at 298 K. A 0.176 mol at 298 K. A 0.176 mol sample of Osample of O22(g)(g) is added to the cylinder, where a reaction occurs to produce is added to the cylinder, where a reaction occurs to produce NONO22(g)(g)..

(d)(d) Write the balanced equation for the reaction.Write the balanced equation for the reaction. (e)(e) Calculate the total pressure, in atm, in the cylinder at 298 K after the Calculate the total pressure, in atm, in the cylinder at 298 K after the

reaction is complete.reaction is complete.

AP QuestionsAP Questions A mixture of HA mixture of H22(g)(g), O, O22(g)(g), and 2 millilitres of H, and 2 millilitres of H22OO(l)(l) is present in a 0.500 is present in a 0.500

litre rigid container at 25°C. The number of moles of Hlitre rigid container at 25°C. The number of moles of H22 and the and the number of moles of Onumber of moles of O22 are equal. The total pressure is 1,146 are equal. The total pressure is 1,146 millimetres mercury. (The equilibrium vapor pressure of pure water at millimetres mercury. (The equilibrium vapor pressure of pure water at 25°C is 24 millimetres mercury.)25°C is 24 millimetres mercury.)

The mixture is sparked, and HThe mixture is sparked, and H22 and O and O22 react until one reactant is react until one reactant is completely consumed.completely consumed.

(a)(a) Identify the reactant remaining and calculate the number of Identify the reactant remaining and calculate the number of moles of the reactant remaining.moles of the reactant remaining.

(b)(b) Calculate the total pressure in the container at the conclusion Calculate the total pressure in the container at the conclusion of the reaction if the final temperature is 90°C. (The equilibrium vapor of the reaction if the final temperature is 90°C. (The equilibrium vapor pressure of water at 90°C is 526 millimetres mercury.)pressure of water at 90°C is 526 millimetres mercury.)

(c)(c) Calculate the number of moles of water present Calculate the number of moles of water present as vaporas vapor in the in the container at 90°C.container at 90°C.