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Balance the following reaction: Cr(s) + CrO42-(aq) Cr(OH)3 (s) • Cr Cr(OH)3 • Balance as if it is in acidic solution • Cr + 3H2O Cr(OH)3 + 3H+ • Balance the H+ by adding equal number of OH- to each side • Cr + 3H2O +3OH- Cr(OH)3 + 3H+ + 3OH- • Balance electrons • Cr + 3H2O +3OH- Cr(OH)3 + 3H+ + 3OH- + 3e-1 • Make every H+ and OH- as H2O on each side and sum up • Cr + 3OH- Cr(OH)3 + 3e-1 (Eq 1) • CrO42-(aq) Cr(OH)3 (s) • Balance as if it is in acidic medium • CrO42-(aq) + 5H+ Cr(OH)3 (s) + H2O • Balance the H+ by adding equal number of OH- to each side • CrO42-(aq) + 5H+ + 5OH- Cr(OH)3 (s) + H2O + 5OH- • Balance the electrons • CrO42-(aq)+5H++5OH- + 3e-1 Cr(OH)3 (s)+H2O +5OH- • Make every H+ and OH- as H2O on each side and sum up • CrO42-(aq)+4 H2O + 3e-1 Cr(OH)3 (s) +5OH- (Eq 2) • Add Eq 1 to Eq 2 • Cr + 3OH- Cr(OH)3 + 3e-1 (Eq 1) • CrO42-(aq)+4 H2O + 3e-1 Cr(OH)3 (s) +5OH- (Eq 2) • Cr(s) + CrO42-(aq) + 4H2O 2Cr(OH)3 (s) + 2OH- Chapter 5 Gases Topics Pressure Gas laws: Boyle, Charles and Avogadro The ideal gas law Gas stoichometry Dalton’s law of partial pressures The kinetic molecular theory of gases Effusion and diffusion Real gases 5.1 Pressure Properties of a gas - Uniformly fills any container. - Mixes completely with any other gas - Exerts pressure on its surroundings. - Compressible - Gas pressure varies with altitudes and storms Measuring atmospheric pressure Torricellian barometer Torricelli (1608-1647) studied the problem using mercury rather than H2O. Mercury is denser than water, so the column wasn’t quite so high. Pascal (SI units) Gas Pressure Force (N) P (Pa) = Force per unit area Area (m2) Pascal and Torricelli Blaise Pascal Evangelista Torricelli (1623-1662) (1608-1647) Torricellian Barometer 760 mmHg P = d·g·h d - density h g - acc. of atmospheric gravity atmospheric pressure pressure Units of Pressure One atmosphere (1 atm) is the average pressure of the atmosphere at sea level is the standard atmospheric pressure Standard Atmospheric Pressure: 1 atm = 76 cm Hg = 760 mm Hg = 760 Torr = 101,325 Pa Very small unit, thus it is not commonly used Example A. What is 475 mm Hg expressed in atm? 485 mm Hg x 1 atm = 0.625 atm 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 103 mmHg 14.7 psi 1.00 atm Manometer Device for Measuring the Pressure of a confined gas 5.2 The Gas Laws of Boyle, Charles and Avogadro Boyle’s Law: PV = const at constant n, T Charles’ Law: V/T = const at constant n, P Avogadro’s Law: V/n = const at constant P, T Boyle’s Law 1 V Slope= 1/k Boyle's Law p 0.012 0.01 Boyle's Law PV =k 1/Pressure (1/Torr) 800 0.008 700 (at constant 0.006 600 T and n) Pressure (Torr) 500 0.004 P1V1 = P2V2 400 0.002 300 0 200 0 50 100 150 200 100 Volume (L) 0 0 50 100 150 200 Volume (L) A Plot of PV Versus P for Several Gases at Pressure Below 1 atm Boyle’s holds Only at very Low pressures A gas strictly obeys Boyle’s law is called Ideal gas Charles’s Law •V/T = b •V = bT (constant P & n) •V1/T1 = V2/T2 Charles' Law 35 30 25 Volume (L) 20 15 10 5 0 0 100 200 300 400 500 Temperature (K) Plots of V Versus T(Celsius) for Several Gases Volume of a gas Changes by 1 273 When the temp. Changes by 1oC. I.e., at -273oC , V=0 ??? All gases will solidify or liquefy before reaching zero volume. • Vn Avogadro’s Law • V = an (constant P& T) V V 1 2 120 Avogadro's Law n n1 2 100 80 Volume (L) 60 40 20 0 0 1 2 moles 3 4 5 5.3 The Ideal Gas Law k V Boyle’s law p V bT Charles's law Universal gas constant V an Avogadro’s law Tn V R( ) PV nRT P The Ideal Gas Law PV = nRT R = 0.0821 atm L mol-1 K-1 V nR P nR V RT T P T V n P The Ideal Gas Law can be used to derive the gas laws as needed! Molar Volume At STP 4.0 g He 16.0 g CH4 44.0 g CO2 1 mole 1 mole 1mole (STP) (STP) (STP) V = 22.4 L V = 22.4 L V = 22.4 L The value of R What is R for 1.00 mol of an ideal gas at STP (0 oC and 1.00 atm)?Given that V of 1 mol of gas at STP= 22.4L PV PV nRT R nT 1.00 atm22.4 L atm. L R 0.0821 R 1.00 mol(273K ) mol.K Example A reaction produces enough CO2(g) to fill a 500 mL flask to a pressure of 1.45 atm at a temperature of 22 oC. How many moles of CO2(g) are produced? PV PV = nRT n RT mol K 0.500 L 1.45 atm 0.0821 atm L 295 K The Ideal Gas Law: Final and initial state problems PV nR Constant T PV PV 1 1 2 2 (Ideal gas equation) T 1 T 2 Ideal Gas Law The ideal gas law is an equation of state. Independent of how you end up where you are at. Does not depend on the path. The state of the gas is described by: P, V, T and n Example A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Data Table Set up Data Table P1 = 0.800 atm V1 = 0.180 L T1 = 302 K P2 = 3.20 atm V2= 90.0 mL ?? T2 = ?? Solution Solve for T2 Enter data T2 = 302 K x atm x mL = K atm mL T2 = K - 273 = °C Calculation Solve for T2 T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T2 = 604 K - 273 = 331 °C Example A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg? Solution T1 = 308 K T2 = ? V1 = 675 mL V2 = 0.315 L = 315 mL P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg T2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL = 178 K - 273 = - 95°C 5.4 Gas Stoichiometry At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occupies 22.4 L. If not at STP, use the ideal gas law to calculate moles of reactant or volume of product. Example A.What is the volume at STP of 4.00 g of CH4? 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L 16.0 g CH4 1 mole CH4 B. How many grams of He are present in 8.0 L of gas at STP? 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 He 1 mole He Example A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure in the cylinder. P = nRT PV = nRT V = (3.74 mol)(0.082L•atm)(297.5K) P= ? (12.25 L) mol•K V = 12.25 L 75.5 g mol = 3.74 mol = 1.009 atm n= 20.18 g = 5670 torr R = 0.082 L•atm mol•K T = 24.5 + 273 = 297.5 K Example 30.2mL of 1.00 M HCl are reacted with excess FeS. What volume of gas is generated at STP? 2 HCl + FeS FeCl2 + H2S(g) # moles HCl = Vol (L) X M 2 HCl + FeS FeCl2 + H2S 1.00 mol HCl mol H 2S 0.0821atm L 0.0302 L 1..00 mol HCl mol H 22S 0.0821atm L 273.15 K 1 00 mol HCl mol H 2S 0..0302 L 1 .00 mol HCl 2 mol HCl mol K 273.15 K 0 .0302 L 0 0302 L L L L L 2 mol HCl mol K 2mol HCl 1.00 atm 0.339 L PV = nRT V = nRT/P Example The decomposition of sodium azide, NaN3, at high temperatures produces N2(g). What volume of N2(g), measured at 735 mm Hg and 26°C, is produced when 70.0 g NaN3 is decomposed. 2 NaN3(s) → 2 Na(l) + 3 N2(g) 2 NaN3(s) → 2 Na(l) + 3 N2(g) Determine moles of N2: 1 mol NaN3 3 mol N2 nN2 = 70 g NaN3 X X = 1.62 mol N2 65.01 g N3/mol N3 2 mol NaN3 Determine volume of N2: nRT (1.62 mol)(0.08206 L atm mol-1 K-1)(299 K) V= = P 1.00 atm (735 mm Hg) 760 mm Hg = 41.1 L Molar mass of a gas PxV= nxRxT m n P x V= mxRxT M M m = mass, in grams M = molar mass, in g/mol Molar mass = m R T PV Density Density(d) is mass divided by volume P x V = mxRxT M P = mxRxT V xM d = m V PM P = dxRxT d M RT Example A glass vessel weighs 40.1305 g when clean, dry and evacuated; it weighs 138.2410 when filled with water at 25°C (d=0.9970 g cm-3) and 40.2959 g when filled with propylene gas at 740.3 mm Hg and 24.0°C. What is the molar mass of polypropylene? PV nRT PV m M RT mRT M Volume of the vessel PV Vflask m H 2O = 138.2410 g – 40.1305 g -3 d (0.9970 g cm ) H 2O = 98.41 cm3 = 0.09841 L mgas = mfilled - mempty = (40.2959 g – 40.1305 g) = 0.1654 g m m RT PV = nRT PV = RT M= M PV (0.6145 g)(0.08206 L atm mol-1 K-1)(297.2 K) M= (0.9741 atm)(0.09841 L) M = 42.08 g/mol Example Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T. P = 1.00 atm T = 273 K Rearrange the ideal gas equation for moles/L d= PXM RxT PXM d RXT g (1.00 atm ) X (32.0 ) d mol 1.43 g / L L.atm (0.0821 )X (273K) mol.K The density of O2 gas at STP is 1.43 grams per liter Example 2.00 g sample of SX6(g) has a volume of 329.5 Cm3 at 1.00 atm and 20oC. Identify the element X. Name the compound P= 1.00 atm 1L V 392.5 Cm3X 0.3295 L 1000Cm3 T = 273+20 = 293K m RT M= PV L.atm (2.00 g )(0.0821 )(293 K ) MM mol.K (1.00atm)(0.3295 L) = 146 g SX6 /mol f Molar mass of (X6 )= 146- 32 = 114 g/mol Molar mass of X = (114 g/mol X6) /6 = 19 X = with a molar mass of 19 = F The compound is SF6 5.5 Dalton’s Law of Partial Pressures For a mixture of gases in a container, the total pressure is the sum of the pressure each gas would exert if it were alone in the container. The total pressure is the sum of the partial pressures. PTotal = P1 + P2 + P3 + P4 + P5 ... nRT For each gas: P V •Partial pressure –Each component of a gas mixture exerts a pressure that it would exert if it were in the container alone PTotal = n1RT + n2RT + n3RT +... V V V Inthe same container R, T and V are the same. PTotal = (n1+ n2 + n3+...)RT V RT Thus, PTotal n ( ) Total V A 250.0 mL flask contains 1.00 mg of He and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres. The total pressure is due to the partial pressures of each of these gases. RT so: Ptotal PHe PH 2 (n He n H 2 ) V For He: 1.00 x 10-3 g He mol = 2.50 x 10-4 mol He _____________________ 4.00 g For H2: 2.00 x 10-3 g H2 mol ______________________ = 9.92 x 10-4 mol H2 2.016 g A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres. so: Ptotal PHe PH (n He n H ) RT V 2 2 For He: 1.00 x 10-3 g He mol = 2.50 x 10-4 mol He _____________________ 4.00 g For H2: 2.00 x 10-3 g H2 mol ______________________ = 9.92 x 10-4 mol H2 2.016 g And: Ptotal = (2.50 x 10-4 + 9.92 x 10-4)(RT/V) = (0.001242 mol)(0.0821 L•atm)(25 + 273)K mol•K (0.2500 L) Ptotal= 0.1216 atm A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres. so: Ptotal PHe PH (n He n H ) RT V 2 2 For He: 1.00 x 10-3 g He mol _____________________ = 2.50 x 10-4 mol He 4.00 g For H2: 2.00 x 10-3 g H2 mol ______________________ = 9.92 x 10-4 mol H2 2.016 g Calculate the pressure due just to He (???): n He RT = 0.0245 atm PHe V and Phydrogen= ? 0.1216 - 0.0245 = 0.0971 atm Magnesium is an active metal that replaces hydrogen from an acid by the following reaction: Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) How many g of Mg are needed to produce 5.0 L of H2 at a temperature of 25 oC and a pressure of 745 mmHg? Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) ?g 5.0 L Hint: find moles of H2 using PV = nRT then work as a stoichiometry problem. mol•K n = PV =____________________________________ 745 mmHg 5.0 L RT 62.4 L•mmHg 298 K n = 0.20 mol The mole fraction Mole fraction: number of moles of one component in a mixture relative to the total number of moles in the mixture symbol is Greek letter chi c n n c 1 = 1 n 1 Total n n n .... 1 2 3 Mole fraction expressed in pressures n n c 1 1 n n n n .... Total 1 2 3 V V n P ( ); n P ( ); .......... 1 1 2 2 RT RT V n P( 1 ) V c 1 RT P( ) 1 1 n V V V c RT P ( ) P ( ) P ( ) ..... Total RT RT RT 1 2 3 1 V ( )( P P P ....) 1 2 3 P P RT c 1 1 ( P P P ....) P 1 n P 1 2 3 Total c 1 1 P cP 1 n P Total Total 1 1 Total Example A 1.00 L sample of dry air at 786 Torr and 25 oC contains 0.925 g N2 plus other gasses (such as O2, Ar and CO2.) a) What is the partial pressure of N2? b) What is the mole fraction of N2? mol PV = nRT 0.925 g 0.0330 mol P = nRT/V 28 .0 g 0.0330 mol 0.0821atm L mol 1 K 1 298 K 0.807 atm 1.00 L PN c N PTotal 760Torr 613 Torr 0.807 atm 613Torr X N 2 0.780 atm 786 Torr Collecting gas over water An insoluble gas is passed into a container of water, the gas rises because its density is much less than that of water and the water must be displaced O2 gas KClO3 Collection of Gases over Water Assuming the gas is saturated with water vapor, the partial pressure of the water vapor is the vapor pressure of the water. Ptotal = Pgas + PH2O(g) Pgas = Ptotal – PH2O(g) Example Oxygen gas generated was collected over water. If the volume of the gas is 245 mL and the barometric pressure is 758 torr at 25oC, what is the volume of the “dry” oxygen gas at STP? (Pwater = 23.8 torr at 25oC) PO2 = PTotal - Pwater = (758 - 23.8) torr = 734 torr P1V1 P2V2 P1V1T2 P1= PO2 = 734 torr; P2= SP = 760. torr V2 P T T1 T2 2 1 V1= 245mL T1= 298K; T2= 273K; V2 = (245mL)(734torr)(273K) V2= ? (298K)(760.torr) =217 mL 5.6 The Kinetic Molecular Theory of Gases It explains why ideal gases behave the way they do. Postulates of the kinetic Theory: Gas particles (atoms or molecules) are so small compared with distances between them, thus we can ignore their volume. The particles are in constant motion and their collisions with walls cause pressure exerted by the gas Kinetic Molecular Theory The particles do not affect each other, neither attracting or repelling The average kinetic energy (KE ) is proportional to the Kelvin temperature. KE = 1/2 mu2 m=mass of gas particle v=average velocity of particles KMT explains ideal gas laws • P&V: P = (nRT) . (1/V) P 1/V – When V decreases # collisions increases • P & T: P = (nR/V).T P T – When T increases, hits with walls become stronger and more frequent • V & T: V=(nR/P).T V T – When T increases hits with walls become stronger and more frequent. To keep P constant, V must increase to compensate for particles speeds • V & n: V= (RT/P). N V n – When n increase P would increase if the volume to be kept constant. V must increase to return P to its original value • Dalton’s law: Individual particles are independent of each other and their volumes are negligible. Thus identities of gas particles do not matter and would be treated as if they belong to one gas Driving the ideal gas law from KMT • The following expression was derived for pressure of an ideal gas: 2 nN A (1 / 2mu 2 ) P [ ] 3 V N Avogadro' s number A m mass of each particle u average of square velocities of particles 2 (KE) N (1/2mu ) avg A 2 2 n(KE) P [ ] avg 3 V PV 2 ( KE ) T avg n 3 PV T n PV RT n The meaning of temperature PV 2 RT = ( KE) avg n 3 ( KE) 3 avg = RT 2 • Thus, Kelvin temperature is an index of the random motion of the gas particles. That is, with higher temperature there is a greater motion. • The same is applicable for liquids and solids Root mean square velocity Combine these two equations (KE)avg = NA(1/2 mu 2 ) (KE)avg = 3/2 RT Where M is the molar mass in kg/mole, and R has the units 8.3145 J/Kmol. The velocity will be in m/s Molecular speed for same gas at two different temperatures 3RT u ( T1 ) 1 1/ 2 M 3RT u ( T2 ) 2 1/ 2 M u T ( ) T1 1 1/ 2 u T T2 2 Molecular speed for two different gases at two different temperatures 3RT u ( T1 ) 1 1/ 2 M 1 3RT u ( T2 ) 2 1/ 2 M 2 u TM ( T1 ) 1 2 1/ 2 u TM T2 2 1 Range of velocities The average distance a molecule travels before colliding with another is called the mean free path and is small (near 10-7) Many collisions among gas molecules produce large number of velocities The actual distribution of molecular velcities is shown in the next slids Effects of Molar Mass on urms At constant T, 273 K, the most probable speed for O2 > CH4 > H2 urms M–1/2 so smaller molar masses result in higher molecular speeds EOS Effects of Temperature on urms At constant mass the most probable speeds for O2 increase with temperature urms T1/2 so higher temperatures result in higher molecular speeds Summary of Behaviors EOS number of particles 273 K Molecular Velocity number of particles 273 K 1273 K Molecular Velocity number of particles 273 K 1273 K 1273 K Molecular Velocity 5.7 Effusion and Diffusion Effusion is the passage of gas through a small hole, into a vacuum. The effusion rate measures how fast this happens. Graham’s Law : the rate of effusion is inversely proportional to the square root of the mass of its particles. Rate of effusion for gas 1 M2 Rate of effusion for gas 2 M1 Deriving Graham’s law The rate of effusion should be proportional to urms Effusion Rate 1 urms for gas 1 = Effusion Rate 2 urms for gas 2 3RT effusion rate 1 u rms 1 M1 M2 effusion rate 2 u rms 2 3RT M1 M2 Diffusion The spreading of a gas through a room. Slow considering molecules move at 100’s of meters per second. Collisions with other molecules slow down diffusions. Best estimate is Graham’s Law. Distance traveled by gas A rms for gas A MB Diatance traveled by gas B rms for gas B MA 5.8 Real Gases Real molecules do take up space and they do interact with each other (especially polar molecules). Correction factors are needed to be added to the ideal gas law to account for these factors. Volume Correction The actual volume available for free movement of molecules is less than the ideal volume because of particle size. More molecules will have more effect. Actual volume V V - nb b is a constant that differs for each gas. The ideal gas equation is modified as follows: nRT Volume of gas particles P Has been taken into V nb account Pressure correction Because the molecules are attracted to each other, the pressure on the container will be less than ideal This effect will depend on the concentration of gas molecules (n/V) since two molecules interact, the effect must be squared: (n/V)2 Pressure correction Thus the pressure corrected for attraction of molecules is given by n 2 Pobserved P a( ) V a is a proportionality constant Corrected Pressure nRT n 2 Pobs a( ) V - nb V Volume correction Pressure correction Volume of container Van der Waals equation n 2 [ Pobs a( ) ] (V - nb) nRT V Corrected pressure Corrected volume Pideal Pideal a and b are determined by experiment. Different for each gas. Bigger molecules have larger b. a depends on both size and polarity. Example Calculate the pressure exerted by 0.5000 mol Cl2 in a 1.000 L container at 25.0ºC Using the ideal gas law. Van der Waal’s equation a = 6.49 atm L2 /mol2 b = 0.0562 L/mol

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