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Warm-Up Problem!. What volume of 0.175 M solution of KOH is needed to titrate 30.0 mL of 0.200 M H 2 SO 4 ?. Chapter 10 : Gases. Units of Pressure. mm Hg or torr These units are literally the difference in the heights measured in mm ( h ) of two connected columns of mercury. Atmosphere

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## Warm-Up Problem!

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**Warm-Up Problem!**What volume of 0.175 M solution of KOH is needed to titrate 30.0 mL of 0.200 M H2SO4?**Units of Pressure**• mm Hg or torr These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury. • Atmosphere 1.00 atm = 760 torr**Manometer**Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.**Boyle’s Law**The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure. P1V1 = P2V2**Charles’s Law**• The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. V1 = V2 T1 T2 A plot of V versus T will be a straight line.**Avogadro’s Law**• The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Relates V and moles**nT**P V Ideal Gas Equation V 1/P (Boyle’s law) VT (Charles’s law) Vn (Avogadro’s law) PV = nRT**Warm-Up!**A sample of gas occupies a volume of 1248 ft3 at 0.988 atm and 28 °C. Calculate the pressure of the gas if its volume is decreased to 978 ft3 while its temperature is held constant.**P**RT dRT P d = = n V P RT m V P RT = = Calculating Density and Molecular Weight! rearrange to solve for MW (x M=) (m/v=d)**Cumulative Problem!**An organic compound had the following analysis: C, 55.8% by mass; H, 7.03%; O, 37.2%. A 1.500 g sample was vaporized and was found to occupy 530 cm3 at 100oC and 740 torr. What is the molecular formula of the compound?**AP Question 1972**A 5.00 gram sample of a dry mixture of potassium hydroxide, potassium carbonate, and potassium chloride is reacted with 0.100 liter of 2.0 molar HCl solution. (a) A 249 milliliter sample of dry CO2 gas, measured at 22°C and 740 torr, is obtained from the reaction. What is the percentage of potassium carbonate in the mixture? (b) The excess HCl is found by titration to be chemically equivalent to 86.6 milliliters of 1.50 molar NaOH. Calculate the percentages of potassium hydroxide and of potassium chloride in the original mixture.**Warm –Up!**A Scuba divers tank contains 0.29 kg of oxygen compressed into a volume of 2.3 L. Calculate the pressure inside the tank at 9 °C. What volume would this oxygen occupy at 26 °C and 0.95 atm.**Agenda for Today**• Go back to 1972… • Partial Pressure discussion • Partial Pressure Practice Problems!**KOH + HCl KCl + H2O**K2CO3 + 2HCl 2KCl + H2O + CO2 KCl + HCl no reaction • Find % K2CO3 in original 5 g total solid • Use PV=nRT to find moles of CO2 • .249 L * .974 atm = 0.01 moles CO2 • .0821Latm/molK(295K) • 0.01 moles CO2 x 1molK2CO3 = 0.01 mol K2CO3 • 1 mol CO2 • 0.01 mol K2CO3x 138g/mol = 1.38 g • Find percent mass = 1.38g K2CO3/5 g total solid = 27.6% 2. Find moles of original HCl 0.1L x 2 mol/L = 0.2 mols HCl total 3. Find moles of HCl 0.1 moles K2CO3 x 2 mol HCl = 0.02 moles HCl reacted with K2CO3 1 mol K2CO3 4. Use NaOH titration to find unreacted HCl: 0.0866L NaOH x 1.5 mol = 0.13 moles NaOH = 0.13 moles HCl (b/c 1:1) L 5. Find moles of HCl that reacted with KOH from steps 2, 3, and 4: 0.2 moles total – 0.02 moles reacted with K2CO3 - 0.13 moles excess = 0.05 moles HCl 6. Find grams of KOH: 0.05moles HCl x 1 mol KOH x 56.1 g KOH = 2.81g KOH 1 mol HCl 1 mol KOH 7. Find mass % KOH 2.81g KOH = 56.2 % 5g total 8. Find % KCl 100 - 56.2 - 27.6 = 16.1 % KCl**Dalton’s Law of Partial Pressures**• The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. • Ptotal = P1 + P2 + P3 + …**Partial Pressure Practice (say that 3 times fast)**What is the total pressure exerted by a mixture of 2.00 g of hydrogen and 8.00 g of nitrogen at 273 K in a 10.0-L vessel?**Partial Pressure = Mole Fraction x Ptot**• What’s a mole fraction? • The relationship between the moles of each gas to the total moles of the gas mixture n1/ntotal = c1 Use the mole fraction and total pressure to find partial pressures! P1 = c1 (Ptotal) What is the mole fraction of oxygen in air? Air = 78% nitrogen, 21% oxygen, 1% water vapor, 0.9% argon, and .04% carbon dioxide**Partial Pressures**• A mixture of 9.00 g of oxygen, 18 g of argon, and 25 g of carbon dioxide exert a pressure of 2.45 atm. What is the partial pressure of Argon in the mixture? • What is the partial pressure of oxygen in air? (Atmospheric pressure = 760 torr)**AP Multiple Choice Warm Up!!**• A student pipetted five 25.00-milliliter samples of hydrochloric acid and transferred each sample to an Erlenmeyer flask, diluted it with distilled water, and added a few drops of phenolphthalein to each. Each sample was then titrated with a sodium hydroxide solution to the appearance of the first permanent faint pink color. The following results were obtained. Volumes of NaOH Solution First Sample..................35.22 mLSecond Sample..............36.14 mLThird Sample.................36.13 mLFourth Sample..............36.15 mLFifth Sample..................36.12 mL Which of the following is the most probable explanation for the variation in the student's results? (A) The burette was not rinsed with NaOH solution. (B) The student misread a 5 for a 6 on the burette when the first sample was titrated.(C) A different amount of water was added to the first sample.(D) The pipette was not rinsed with the HCl solution.(E) The student added too little indicator to the first sample.**Collecting Gases over Water**Ptotal= Pgas + PH2O 2KClO3(s) 2KCl(s) + 3O2(g) The volume of gas collected is 0.250 L at 26°C and 765 torr total pressure. (a) How many moles of O2 are collected? (b) How many grams of KClO3 were decomposed? (The pressure of water vapor at 26°C = 25 torr .)**Kinetic – Molecular Theory**• Gases consist of large numbers of molecules that are in continuous, random motion. • The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained. • Attractive and repulsive forces between gas molecules are negligible. • The collisions between molecules are perfectly elastic. • The average kinetic energy of the molecules is proportional to the absolute temperature.**Pressure is caused by collisions of molecules with the walls**of the container! • Magnitude of pressure is related to how often and how forcefully the particles strike the walls. • Temperature of a gas is a measure of the average kinetic energy of its molecules! • Motion increases with increasing temperature. • Why does pressure increase with increasing temperature?**KE = 1/2mu2**√ r1 = M2 r2M1 √ t1 = M1 t2M2**Effusion Problem**• At a particular T and P, neon gas effuses at a rate of 16 mol/s. • (a.) What is the rate at which Ar effuses under the same conditions? • (b.) Under a different set of conditions, 3.0 mol of Ar effuse in 49.0 seconds. How long will it take an equal amount of helium to effuse?**Deviation from Ideal Behavior**Real gases at high pressure are like the mall at christmas. Too many people in a small space, people are on top of one another! Compared to the space in the checkout line, the people in front of you and behind you have significant volume. Ugh… If molecules are crowded in a container, as pressure increases attractive forces take over. Also the volume of the gas particles are significant compared to the volume of the container. NOT IDEAL!**(P +**) (V − nb) = nRT n2a V2 Deviation from Ideal Behavior At low temperature there is not enough energy to overcome attractive forces and the gas is forced to condense into a liquid. Ice is NOT an ideal gas… The van der Waals Equation – corrects for non-ideal behavior**Non-Ideal Practice**Arrange the following gases in order of increasing deviation from ideality: H2O, CH4, Ne WHY???

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