VIEWS: 3 PAGES: 35 POSTED ON: 3/21/2013
Kinetic Molecular Theory of Matter Know the points of KMT points p 115 1. gases are mainly empty space 2. gas molecules are in constant straight line motion until they collide with something gas pressure = caused by the collisions of gas molecules with the sides of the container 3. collision are totally elastic no energy is lost Description of Gases Need 4 variables to describe a gas system Volume Amount of gas present = moles Temperature Pressure Volume and Amount of Gas Vol = space the gas occupies Units = L, mL or cm3 1 mL = 1 cm3 Gases fill any container diffusion Gas vol can change Gases are compressible Amount of a gas is measured in moles Avogadro's # of particles If given grams convert to moles Temperature Use only Kelvins in this chapter!!! Units ºC or Kelvins ºC + 273 = K Temp is a meas of average kinetic energy of particles Ek = ½ mass x (velocity)2 temp movement kinetic energy Temp tells us the direction of energy flow Pressure Pressure is caused by gas molecules colliding with the sides of a container Pressure is measured with a meter Barometer Used to meas press of an open sample of gas Manometer Used to meas the press in a closed system Barometer Measures the pressure of the atmosphere based on the height of a column of mercury that air will support Measured in mm of mercury Mm Hg Barometer Manometer Measures the pressure in a closed system by using a mercury plug between the gas sample and the atmosphere. The press of the atmosphere is known The press of the gas is related to the press of the atmosphere Manometer Units of Pressure mm Hg (also called torr) atmospheric pressure ~ 760 mm Hg 1 atmosphere = 760 mm Hg 101.3 Kilopascal (KPa) = 760 mm Hg 1.0 pascal = 1 newton/m2 Basic SI unit of pressure Be able to convert from one unit to another Ideal Gas Law: PV = nRT The ratio of the (pressure x vol) of a contained sample of a gas, to the (number of moles x pressure) is a constant R = PV P = press V = vol nT n = mole T = temp R has specific units L x atm the 4 variables must be mol x Kelvins measured in these units R = 0.0821 L atm/mol K Gases Laws Charles’ Law = vol and temp Vi = Vf = k direct Ti Tf determine 0 Kelvins Boyle’s Law = vol and pressure PiVi = PfVf = k indirect Avogadro’ Law = vol and moles Vi = Vf = k direct ni n f Guy-Lusaac’s Law = press and temp Ti = Tf = k direct Pi Pf DIRECT AND Inverse relationships inverse direct direct direct PV = nRT 4 types of problem: Calcs of P, V, n, or T Initial and final state problems Molar mass determinations Density determinations Calcs of P, V, n, or T Given 3 of the 4 variables deter the other PV = nRT (R = 0.0821 Latm/molK) A 2.50 g sample of SF6 is placed in a 500.0 mL vessel at 83ºC. What pressure in mm Hg is developed? (make a table of given variables) n = 2.50 g SF6 x 1 mol SF6 = 0.0171 mol SF6 146.1 g SF6 V = 500.0 mL x 1L = 0.5000 L 1000 mL T = 83ºC + 273 = 356 K P = nRT = (0.0171 mol)(0.0821 L atm/mol)(356 K) = 1.00 atm V (0.5000 L) Initial and final state problems Given a change in conditions (P, V, n, or T) of a gas, determine the new conditions (P, V, n, or T) Units are not as important (except T) Units must be consistent PiVi = Ri = Rf = PfVf niTi nfTf Cancel out any values that remain constant. In the following example n is constant ni = nf and Ti = Tf they cancel out PiVi = Ri = Rf = PfVf niTi nfTf Find the volume of a gas at 800.0 mm Hg and 40.0ºC, if it's vol at 720.0 mm Hg and 15.0ºC is 6.84 L? Make a table: initial final P 800 mm Hg 720 mm Hg n is held constant T 313 K 288 K V Vi 6.84 L PiVi = PfVf Vi = PfVfTi niTi nfTf Tf Pi Vi = PfVfTi = (720 mmHg)(6.84 L)(313 K) = 6.69 L nfTfPi (288 K)(800 mmHg) Calculation of Molar Mass Mol = grams mm Subbing into Ideal gas law PV = nRT PV = gRT mm Therefore: mm = gRT PV If you know the grams of a gas at a specific temp, press, and vol you can determine it’s molar mass Determination of Density Density = mass/vol PV = nRT n = mol = g/mm PV = gRT or g = P(mm) = density mm V RT Example: Determine the density, in g/L, of ammonia (NH3), at 97ºC and 755 mm Hg? P = (755/760 atm) = 0.993 atm mm = (17.0 g/mol) T = (97ºC + 273) = 370 K D = Pmm = (0.993 atm)(17.0 g/mol) RT (0.0821 L atm/mol K)(370 K) = 0.556 g/L Gas Stoichiometry What is the volume of CO2 produced at 37.0˚C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) g C6H12O6 mol C6H12O6 mol CO2 V CO2 1 mol C6H12O6 6 mol CO2 5.60 g C6H12O6 x x = 0.187 mol CO2 180 g C6H12O6 1 mol C6H12O6 L•atm 0.187 mol x 0.0821 x 310.15 K nRT mol•K V= = = 4.76 L P 1.00 atm Gas Stoichiomerty PROBLEM: A laboratory method for reducing a metal oxide is to heat it with H2. The pure metal and H2O are products. What volume of H2 at 765 torr and 225˚C is needed to form 35.5g of Cu from copper (II) oxide? PLAN: Since this problem requires stoichiometry and the gas laws, we have to write a balanced equation, use the moles of Cu to calculate mols and then volume of H2 gas. SOLUTION: CuO(s) + H2(g) Cu(s) + H2O(g) mol Cu 1 mol H2 35.5g Cu = 0.559mol H2 63.55g Cu 1 mol Cu PV = nRT L atm 0.559 mol H2 x 0.0821 x 498K V = nRT mol K = 22.6L P 765 atm 760 Dalton's Law of Partial Pressure Two ways of looking at this law 1 - The total pressure of a mixture of gases is equal to the sum of the pressures of the individual gases if alone in the same volume Ptotal = Pgas1 + Pgas2 + Pgas3 … Used a lot for gases collected under water Water vapor pressure must be accounted for in the collected gas sample Ptotal = Pgas a - Pwater Gases Collected over H2O A 152 mL sample of O2 gas is collected under water at 25°C (vpH2O= 23.76 mm). If the O2 exerts a pressure of 758 mm Hg determine: a. the partial pressure of the O2 Ptotal = PH2O + PO2 Ptotal - PH2O = PO2 758 - 23.76 = 734 mm Hg b. the grams of O2 collected nH2 = PH2V = (734 / 760 atm)(0.152 L) = 0.00600 mol O2 RT (0.0821 L atm/mol K)(298K) = 0.00600 mol O2 x 32.0 g = 0.192 g O2 mol Dalton's Law of Partial Pressure 2-The pressure of the components of a gas mixture is relative to their percents or mole fraction Mol frac = mole gas A total moles gases I know it's a decimal, but it's called mole fraction!!! Mol fraction is the same as percent PgasA = Ptotal x MFgasA or PgasA = Ptotal x PercentgasA(as a decimal) Dalton's Law of Partial Pressure A gas sample contains 2.0 mol O2, 1.0 mol N2, and 3.0 mol He. Determine The mol fraction of each gas. MFO2 = 2.0/6.0 = 0.33 MFN2 = 1.0/6.0 = 0.17 MFHe = 3.0/6.0 = 0.50 The partial press of each gas if the total press on the system = 950 mm Hg PPO2 = 950 x 0.33 = 314 mm Hg PPO2 = 950 x 0.17 = 162 mm Hg PPO2 = 950 x 0.50 = 475 mm Hg Dalton's Law of Partial Pressure A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 Xpropane = = 0.0132 8.24 + 0.421 + 0.116 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm Velocity of Gas Molecules Root mean speed = u U = 3RT ½ mm Thermodynamics = diff value for R R = 8.31 x 103 g m2 / s2 mol K Calc the root mean speed, u, of N2 at 25°C u = (3)(8.31 x 103 g m2 / s2 mol K)(298K) ½ (28.0 g / mol) = 515 m/s Graham's Law of Effusion Effusion = movement of gas through a small opening or pore Lighter gases effuse faster than heavier gases at the same T and P ua = ratea = mmb ½ = timeb ub rateb mma timea Determination of Molar Masses An unknown gas effuses 1.66 times slower than CO2., Determine the molar mass of the gas rateCO2 = mmx ½ ratex mmCO2 1.66 = mmx ½ 44.0 g/mol mmx = 121 g/mol Rates of Effusion Consider H2 and O2 gases at the same temp and pressure. Which gas effuses faster? How much faster? H2 will effuse faster as it is lighter rateH2 = mmO2 ½ = 32.0 g/mol ½ rateO2 mmH2 2.0 g/mol rateH2 = 4.0 H2 will effuse 4.0 times faster rateO2 Rates of Effusion Hydrogen gas, H2 effuses 4.5 times faster than gas X. Determine the molar mass of gas X. rateH2 = mmX ½ rateX mmH2 4.5 = X ½ 1.0 2.0 20.25 x 2.0 = X = 40.5 g/mol Boltzmann’s Distribution Real vs. Ideal Gases Ideal gases obey the gas laws PV/nT = constant Occurs at room conditions of T and P Gases don't always obey the gas laws PV/nT is not a constant Called "real gas" Occurs under extreme conditions Gases act in an ideal or real manner Depends on conditions Real vs. Ideal Gases Under extreme conditions the gases don't follow the gas laws Low temp Gas molecules stay by each other long enough to be attached together – V (and PV/RT) is too small At high pressure (200-400 atm) Gas molecules close enough to attract each other – V (and PV/RT) is too small Extreme press (above 400 atm) Gas molecules squeezed together and e- clouds repel – V (and PV/RT) too large Van der Waal's Equation Amends the IGL for real behavior a and b are constants for each diff gas a corrects for attractive forces – Occurs at low T or high P b corrects for V of gas molecules – Occurs at extreme P