Chem 1211/1203 Lecture Notes Unit 4 PDF

Chem 1211/1203 Lecture Notes Unit 4 Unit Four: Gases Readings from Silberberg, Amateis, Venkateswaran and Chen, Canadian Edition, 3/e: 4.1 Overview of the Physical States of Matter 4.2 Gas Pressure and Its Measurement 4.3 The Gas Laws an...

Chem 1211/1203 Lecture Notes Unit 4 Unit Four: Gases Readings from Silberberg, Amateis, Venkateswaran and Chen, Canadian Edition, 3/e: 4.1 Overview of the Physical States of Matter 4.2 Gas Pressure and Its Measurement 4.3 The Gas Laws and Their Experimental 4.4 Rearrangements of the Ideal Gas Law Foundations Assigned Questions from Silberberg, Amateis, Venkateswaran and Chen, Canadian Edition, 3/e: Chapter 4: Sample Problems (in‒chapter): 1−12. End−of−Chapter Problems (red coloured): 1−60, 83, 88, 90, 92, 93, 106, 111. Particularly Interesting Questions (PIQ, these should “pique” your interest….get it?!?! These are questions that are not easy and should relate chemistry to real life): 94, 97 (very good question!), 102, 128, 129, 148. Related Websites/Movies: https://youtu.be/v12xG80KcZw: brief description of the three common states of matter. https://youtu.be/uifMSZNjJxg: O2 gas collection in a pneumatic trough https://youtu.be/7KOnCfj0_JY: H2 gas collection https://youtu.be/gmN2fRlQFp4: great video on all the gas laws with animations ·4–1· Preview Chem 1211/1203 Lecture Notes Unit 4 Overview of Physical States Solids and liquids we know are, under normal conditions, incompressible. That is, we can increase the external pressure on an ice cube an it’s volume will not change appreciably. Can the same be said of, say, steam? Gases have some unique properties:  Uniformly fills up any closed container  Easily compressed  Mixes completely with other gases  Exerts pressure on its surroundings Gases are atoms or molecules that move around and bump into things: – each other – the walls of the vessel containing the gas(es) It is true that liquids share three of the above four properties, but they are not easily compressed. Pressure What is the relationship between the amount of air in your lungs and the pressure inside the lung? The less pressure means less air. More on this in a few moments… The air in the balloon (or your lung) exerts a force on the inside area of the ballon: ·4–2· Pressure Chem 1211/1203 Lecture Notes Unit 4 𝑓𝑜𝑟𝑐𝑒 𝑁 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑃 = = 2 = 𝑃𝑎 𝑎𝑟𝑒𝑎 𝑚 The SI unit for pressure is Pascals (on The Weather Network they use thousands of Pa, or kPa when discussing air pressure). Scientists often talk about the air pressure of the Earth at sea level, i.e. the force that the constituent molecules of the air are hitting your skin… 1 standard atmospheric unit = 1 atm = 101.3 kPa A common pressure unit is based on the height of mercury in a column that the gas pressure can support, usually in mm: barometer: a device used to measure pressure atm 760 mm Hg Conversions between common pressure units (don’t memorize, all are on your data sheet): 1 atm = 101.3 kPa = 760 mm Hg = 760 torr ·4–3· Pressure Chem 1211/1203 Lecture Notes Unit 4 Often scientists study the behavior of gases under a standard set of conditions in order to normalize their results. Standard Temperature 0 0°C C (or 273F K) 8 273 and Pressure (STP) 1 atm (101.3 kPa) latm 01.3hPa Now when scientists say they perform an experiment at STP (or normalize their results to STP) they don’t need say that P and T are constant, we know they are! The (Simple) Gas Laws Please watch the video on the Simple Gas: https://vimeo.com/171129683. Fill in the blanks at home after watching the video. Over the centuries, many studies have been performed on behavior of gases with four easily manipulated variables: temperature (T) pressure (P) amount of gas (n or mol) volume (V) We want to see how one of these variables relates to another while keeping the other two constant. All the combinations are called the simple gas laws and are named after the scientist who performed the observations. I’ll summarize them here (you should read the derivations of them on your own). All of them are (or should be) intuitive… Relationship between V and P (Boyle’s Law) Think of a balloon. If a filled balloon is squished (i.e. a reduction in volume) what often happens to the balloon? ·4–4· The Gas Laws Chem 1211/1203 Lecture Notes Unit 4 It bursts! This is because the pressure inside the balloon has increased so much that the walls of the balloon can no longer contain the gas inside. So what is the relationship between V and P empirically? if V ↑ then P ↓ 1 V  thus V and P are inversely related: P we must remove the “proportional to” relation and put a constant in to make it an “equals to” relationship: a V = (T & n constant) P Note, a V vs. P graph is NOT linear (y = mx + b). Relationship between V and T (Charles’ Law) Instead of changing the pressure inside the balloon, what will happen if we heat it up a bit? Heating up a balloon will make the gas inside expand, thus increasing the volume it occupies. Thus, if V ↑ then T ↑ V  T or V = bT thus V and T are directly related: ( P & n constant) Relationship between V and n (Avogadro’s Law) What will happen to the balloon if we blow twice as much air into it? The volume will double! V  n or V = cn ( P & T constant) ·4–5· The Gas Laws Chem 1211/1203 Lecture Notes Unit 4 if V ↑ then n ↑ thus V and n are directly related: Notice that for all the Laws, two of the four parameters must be held constant (otherwise we’d need a 3D graph to relate the variables) though clearly all four are related to each other. Combining all of the simple gas laws we can relate V to T, P and n and come up with what is called the ideal gas law: 𝑃𝑉 = 𝑛𝑅𝑇 common value for R = 0.08206 L∙atm∙mol −1∙K −1 other R: 8.314 L∙kPa∙mol−1∙K −1 Make sure you use the R with the right pressure units! 1 L·kPa = 1 J, thus R = L∙kPa∙mol−1∙K −1 (useful in Chem 212) Some notes on PV = nRT:  it is empirical, and thus an approximation  gases approach ideal behaviour at low P, high T  ideal gases are like Santa Claus, they don’t really exist  you cannot use PV = nRT 1 for solids or liquids  is used to measure properties of a gas under one set of conditions (i.e. a way to measure moles of gas in the way that molar mass and mass are used to get moles) i.e. the ideal gas law is an equation of state for a gas, its P, V, T and n are all related, and only three of these variables are needed to determine the fourth. ·4–6· The Gas Laws  Chem 1211/1203 Lecture Notes Unit 4 Q.1a How many mol are present in a sample of CO2 (MM = 44.0 g·mol‒1) in a 10.0 L flask at 25°C and 1.00 atm? Notice we have one set of experimental conditions, so we need only apply the ideal gas law once: PV nRT ⇒ 𝑃𝑉 = 𝑛𝑅𝑇 𝑛= 𝑃𝑉 = 1 00 atm (1.00 𝑎𝑡𝑚)(10.0 10.02 𝐿) not 0.409 = 0.409 𝑚𝑜𝑙 R L.atm.no 2981f (0.08206 𝑅𝑇 08206 𝐿 ⋅ 𝑎𝑡𝑚 ⋅ 𝑚𝑜𝑙 −1 ⋅ 𝐾 −1 )(298 𝐾 )  is used to measure the change in two (or more) properties if the other properties are held constant. Q.1b The above sample of CO2 is compressed and heated to a final temperature and volume of 55 °C and 5.00 L respectively. Calculate the final pressure. In this case, we have two sets of experimental conditions, i.e. we are changing some of the variables. We know that n and R will remain constant: 1 𝑃1 𝑉1 𝑃2 𝑉2 nR constant 𝑇1 ur and = 𝑛𝑅 Bff 𝑇2 = 𝑛𝑅 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 So we can equate these two functions and isolate the quantity we want to measure, the final pressure P2: 𝑃P 1 𝑉1 𝑃2 𝑉2 = 8 𝑇1 𝑇2 𝑇2 𝑃1 𝑉1 (3281.00 𝐾 )(1.0040.04 atm 𝑎𝑡𝑚)(10.0 𝐿) 628k B If ∴ 𝑃2 = 𝑇1 𝑉2 = (298 298k 5 02 𝐾 )(5.0 𝐿 ) 2.20= atm 2.20 𝑎𝑡𝑚 ·4–7· The Gas Laws Chem 1211/1203 Lecture Notes Unit 4 Prove Avogadro’s Law here, i.e. show that the mol of CO2 at the 2nd set of T, P and V is the same as under the 1st set of parameters. What should the answer be? Same as Q.1a only using different T, P and V values. Applications of The Ideal Gas Law Molar Mass of Gases The IGL can be rewritten to solve for a gas’ MM: 𝑚 𝑚𝑅𝑇 𝑛= 𝑀𝑀 ⇒ 𝑃𝑉 = 𝑛𝑅𝑇 = 4 𝑀𝑀 this equation can be rearranged yet again, to give: 𝑚𝑅𝑇 𝑚𝑎𝑠𝑠 𝑃= 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑑 = 𝑉(𝑀𝑀) 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑅𝑇 𝑑𝑅𝑇 P𝑃 = 𝑀𝑀 𝑜𝑟 𝑀𝑀 MM 4ft = dRpI or 𝑃 A Dumas bulb is shown on the next page. Hot water (or oil) is often used with a bulb of known volume and mass. The liquid sample evaporates and the vessel is open to the air to let the vessel vent. The bulb is cooled (gas liquid) and the mass of the sample is determined. ·4–8· Application of IGL Chem 1211/1203 Lecture Notes Unit 4 Once all the excess gas escapes the vessel, what is Pgas? Patm Patm We now know all the variables needed to get the molar mass of the gas: m and V (density), T (of the water/oil bath) and P (atmospheric) Q.2 A chemist takes an unknown liquid and performs a Dumas analysis of the liquid. The following results are obtained: Vflask = 192 mL T = 100°C Patm = 735 mm Hg mflask = 85.869 g mflask + mgas = 86.214 g What is the molar mass of the liquid, in g·mol–1? 1st, we need the density of the gas: d𝑑 = 𝑚 = (𝑚𝑓𝑙𝑎𝑠𝑘+𝑔𝑎𝑠 −mflask 85.869g kf.tt0.192 86.214g 𝑚𝑓𝑙𝑎𝑠𝑘 ) 86.214 − 85.869 𝑔 f𝑉 9𝐿 = 0.1922 0.192 𝐿 1 ⋅ 𝐿−1 2 1.80 = 1.80 𝑔 g 𝑑𝑅𝑇 MM = ART 𝑀𝑀 𝑃 p k 373 k 80g 21 G 08206 2 atm mol 56.7g molt 735 mmHg ·4–9· mmtlg Application of IGL Chem 1211/1203 Lecture Notes Unit 4 (1.80 𝑔 ⋅ 𝐿−1 )(0.08206 𝐿 ⋅ 𝑎𝑡𝑚 ⋅ 𝑚𝑜𝑙 −1 ⋅ 𝐾 −1 )(373 𝐾 ) = (1 𝑎𝑡𝑚) (735 𝑚𝑚 𝐻𝑔) ( ) 760 𝑚𝑚 𝐻𝑔 = 56.7 𝑔 ⋅ 𝑚𝑜𝑙 −1 Gas Stoichiometry The rules of stoichiometry apply to gases as well as pure liquids, solids and solutions learned previously. The only “twist” is we might have to use the IGL to get mol rather than MM and mass or volume of solution and concentration of solute. Let us attempt a question that involves not only three reactants instead of two, but one where all states of matter are present and one that incorporates all the stoichiometric rules we’ve discussed so far:1 Note this question could also ask you to name any reactants/products and even balance the reaction (it is a Red/Ox reaction, after all!) and include % yield as well! Q.3 next shows an example of a limiting reagent problem with not only three reagents (you’ll see this doesn’t make the question any more difficult!) but all three reagents in a different state, and thus a different way to determine mol for each compound. 1 IMPORTANT TIP: If you’re given amounts of all the reactants and asked to find amount(s) of product(s), it is likely (but not guaranteed!) a limiting reagent must be determined. · 4 – 10 · Application of IGL  Chem 1211/1203 Lecture Notes Unit 4 Q.3a Hydrogen cyanide (HCN, MM = 27.02 g·mol‒1) gas can be produced by reacting an aqueous solution of ammonia (NH3, MM = 17.03 g·mol‒1) with carbon disulfide (CS2, MM = 76.14 g·mol‒1) and oxygen gas which will also produce pure elemental sulfur (S8, MM = 256.5 g·mol‒1): 4 NH3 (𝑎𝑞 ) + 2 O2 (𝑔) + 4 CS2 (𝑙 ) → 4 HCN(𝑔) + S8 (𝑠) + 4 H2 O(𝑙 ) If 15.0 g of CS2 is added to 125 mL of 0.100 M NH3 solution in a 2.0 L vessel filled to 1.10 atm O2 at 25°C, what is the amount of elemental sulfur that can be prepared? To determine limiting reagent, need mol of all reactants, then determine have:need: 5 1 not 62 mol S2 1 𝑚𝑜𝑙 CS0.197 𝑚𝑜𝑙 CS5.0g 2 2 = (15.0 𝑔 CS2 )1( 2 ) = 0.197 𝑚𝑜𝑙 CS2 76.14 𝑔 CS2 mol NH3 C 1252501 3 0.0125 months 𝑛 𝑚𝑜𝑙 NH3 = (0.125 𝐿 𝑠𝑜𝑙 ) ( 5 0.100 𝑚𝑜𝑙 NH3 𝑛 ) mol 022 1.10 atm 2 02 1 𝐿 𝑠𝑜𝑙 Pr = 0.0125 𝑚𝑜𝑙 NH3 atm not E 298k 0.082062 𝑃𝑉 0.0900 mol(1.10 02 𝑎𝑡𝑚)(2.0 𝐿) 𝑚𝑜𝑙 O2 = = 𝑅𝑇 (0.08206 𝐿 ⋅ 𝑎𝑡𝑚 ⋅ 𝑚𝑜𝑙 −1 ⋅ 𝐾 −1 )(298 𝐾 ) 0 197m01 Csa 49 = 0.0900 𝑚𝑜𝑙 moles 0.0492 O2 ℎ𝑎𝑣𝑒 0.197 𝑚𝑜𝑙 CS2 is ammonia⇒ the Limiting=reagent 0.0492 𝑛𝑒𝑒𝑑 4 𝑚𝑜𝑙 CS2 · 4 – 11 · Application of IGL Chem 1211/1203 Lecture Notes Unit 4 0.0125 𝑚𝑜𝑙 NH3 0.0900 𝑚𝑜𝑙 O2 = 0.0031 = 0.045 4 𝑚𝑜𝑙 NH3 2 𝑚𝑜𝑙 O2 ammonia is the limiting reagent. ∴ammonia The ICF table has states omitted (we only care about ratios after all anyway) and the mol converted to mmol to make the table easier to read: 4 NH3 + 2 O2 + 4 CS2 → 4 HCN + S8 + 4 H2 O I (mmol) 12.5 90.0 197 0 00 0 12.5 90.0 197.0 0 0 C (mmol) ‒12.5 ‒6.2 ‒12.5 +12.5 +3.12 +12.5 12.5 6 2 12.5 12.5 3.15 12.5 F (mmol) 0 83.8 84.5 12.5 3.12 12.5 0 83.85 184.5 12.5 3.12 12.5 1 𝑚𝑜𝑙 S8 256.5 𝑔 S8 𝑚𝑎𝑠𝑠 S8 = (12.5 𝑚𝑚𝑜𝑙 NH3 ) ( )( ) 12.5mmol NH 4 𝑚𝑜𝑙 NH3 artists 1 𝑚𝑜𝑙 S8 = 802 𝑚𝑔 S8 = 0.802 𝑔 S8 802mg58 0.802958 Dalton’s Law of Partial Pressures Simple gas laws were, for the most part, based on experiments with air as the gas…what is air? Roughly: 78% N2(g), 20% O2(g), 1% H2O(l), 1% Ar(g). Thus far, we’ve only dealt with pure gases, but what about real– life mixtures? For a mixture of n gases partial pressure of x: Px 𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃𝐴 + 𝑃𝐵 + 𝑃𝐶 + ⋯ + 𝑃𝑛 · 4 – 12 · Application of IGL Chem 1211/1203 Lecture Notes Unit 4 In other words, a gas is a vacuum to all other gases present in a mixture. The sum of these partial pressures of all the components of a mixture will be equal to the total pressure. We can substitute the ideal gas law into the above equation, taking constant temperature into account, and get the following: 𝑛𝑡𝑜𝑡𝑎𝑙 𝑅𝑇 𝑛𝐴 𝑅𝑇 𝑛𝐵 𝑅𝑇 𝑛𝐶 𝑅𝑇 𝑛𝑛 𝑅𝑇 = + + + ⋯+ 𝑉𝑡𝑜𝑡𝑎𝑙 𝑉𝐴 𝑉𝐵 𝑉𝐶 𝑉𝑛 ∴ 𝑉𝑡𝑜𝑡𝑎𝑙 = 𝑉𝐴 + 𝑉𝐵 + ⋯ and 𝑛𝑡𝑜𝑡𝑎𝑙 = 𝑛𝐴 + 𝑛𝐵 + ⋯ Thus, volume adds and moles add in the same way pressure does. Now we can define what is called mole fraction, x: 𝑛𝐴 𝑛𝐵 𝜒𝐴 = 𝜒𝐵 = 𝑉 or 𝑃 can be used instead of 𝑛 𝑛𝑡𝑜𝑡𝑎𝑙 𝑛𝑡𝑜𝑡𝑎𝑙 What should the sum of all the mole fractions be in any mixture? 11 𝑃Ptotal 𝑡𝑜𝑡𝑎𝑙 = 𝜒𝐴 𝑃𝑡𝑜𝑡𝑎𝑙 + 𝜒𝐵 𝑃𝑡𝑜𝑡𝑎𝑙 + 𝜒𝐶 𝑃𝑡𝑜𝑡𝑎𝑙total +⋯ Ptotal XaBPtotal xcP Factor out 𝑃𝑡𝑜𝑡𝑎𝑙 and you get: 𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃𝑡𝑜𝑡𝑎𝑙 (𝜒𝐴 + 𝜒𝐵 + 𝜒𝐶 + ⋯ ) = Ptotal 𝑃𝑡𝑜𝑡𝑎𝑙 i.e. the pressure of a gas in a mixture is directly proportional to the percentage composition of that gas.2 When a mixture of gases are present, “Pressure” refers to total pressure, and “partial pressure” refers to the pressure of each component of the mixture. 2 Think back to weighted averages again…it’s a very similar expression · 4 – 13 · Application of IGL Chem 1211/1203 Lecture Notes Unit 4 Thus, in a mixture of gases, saying “increase the pressure” means increasing Ptotal. In other words, there is a difference between saying “increase the pressure” and “increase the partial pressure”! Q.3b What will be the total pressure inside the vessel from Q. 3a after the reaction is complete? What are the mole fractions, x, and partial pressures of each of the gases? The only gases are O2 and HCN. 𝑚𝑜𝑙 O2 83.8 𝑚𝑚𝑜𝑙 83.8 𝑚𝑚𝑜𝑙 𝑥O2 = = = = 0.870 𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙 83.8 + 12.5 𝑚𝑚𝑜𝑙 96.3 𝑚𝑚𝑜𝑙 𝑚𝑜𝑙 HCN 12.5 𝑚𝑚𝑜𝑙 𝑥HCN = = = 0.130 𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙 96.3 𝑚𝑚𝑜𝑙 Note: the sum of mole fractions is 1. 𝑛O2 𝑅𝑇 𝑃O2 = 𝑉 𝐿 ⋅ 𝑎𝑡𝑚 (0.0838 𝑚𝑜𝑙) (0.08206 ) (298 𝐾) = 𝑚𝑜𝑙 ⋅ 𝐾 = 1.02 𝑎𝑡𝑚 2.0 𝐿 𝑛HCN 𝑅𝑇 𝑃HCN = 𝑉 𝐿 ⋅ 𝑎𝑡𝑚 (0.0125 𝑚𝑜𝑙) (0.08206 ) (298 𝐾) = 𝑚𝑜𝑙 ⋅ 𝐾 = 0.153 𝑎𝑡𝑚 2.0 𝐿 𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃O2 + 𝑃HCN = 1.02 + 0.153 𝑎𝑡𝑚 = 1.17 𝑎𝑡𝑚 · 4 – 14 · Application of IGL Chem 1211/1203 Lecture Notes Unit 4 Application of Dalton’s Law: Many chemical reactions produce gas…how can that gas be trapped quantitatively? The “old school” method: O2 and 02 and collecting gas over H2O 120 water Pneumatic trough Decomposition of KClO3: 2 KClO3 (𝑠) → 2 KCl(s) + 3 O2 (𝑔) What gases are present over the water? O2(g) and HH2O 2O(g) 02cg and Thus, according to Dalton’s Law, the atmospheric pressuregis exactly balanced by the pressure above the water, or: 𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑒 = 𝑃O2 + 𝑃H2O Pestan 2f.pt barometer Once 𝑃O2 is determined, it can be used in stoichiometric calculations, along with PV = nRT. We need to know the pressure of water at a given temperature…look up in table (see Formula Sheet) · 4 – 15 · Application of IGL  Chem 1211/1203 Lecture Notes Unit 4 nuematictoophsetup Q.5 Oxygen gas is generated via the setup shown on page 4– 15. The atmospheric pressure and temperature is 742 mm Hg and 30ºC respectively. The amount of O2(g) collected was 128 mL. What is the mass of the collected oxygen gas? How much KClO3 (MM = 122.6 g∙mol–1), in mg, was decomposed (assume 100% yield)? From Table 9 on Formula Sheet: 𝑃H2O (30 °C) = 31.82 𝑚𝑚 𝐻𝑔 = 0.04187 𝑎𝑡𝑚 𝑃𝑡𝑜𝑡𝑎𝑙 = Patm Po = 𝑃Piezo 𝑃𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑒 O2 + 𝑃H2 O 1 𝑎𝑡𝑚 742mm (742 𝑚𝑚 ) Hg 𝐻𝑔 (760 𝑚𝑚 𝐻𝑔 Po) = 𝑃0O2 04187 atm𝑎𝑡𝑚 + 0.04187 g PosO2 0.9763 atm 0.04187 atm 0.9344atm 𝑃 = 0.9763 𝑎𝑡𝑚 − 0.04187 𝑎𝑡𝑚 = 0.9344 𝑎𝑡𝑚 Convert P into mol using ideal gas law: 0.9344 atm 𝑎𝑡𝑚 0.1282 p n 𝑃𝑉 (0.9344 )(0.128 𝐿) 𝑛= = 03 not−1 ⋅E𝐾 −1 )( K C 082062 𝑅𝑇 (0.08206 atm⋅ 𝑚𝑜𝑙 𝐿 ⋅ 𝑎𝑡𝑚 303 𝐾) = 0.00481 0 00481 𝑚𝑜𝑙mol = 4.81 4.81 𝑚𝑚𝑜𝑙 mmol O2 02 Do we need the stoichiometric coefficient here? NO! I 0 We measured the volume of O2 directly (as opposed to indirectly using limiting reagent). 32.0 𝑔 O2 4.81mmol 𝑚𝑔 02 O2 = (4.81 𝑚𝑚𝑜𝑙 O2 ) (34 902 42 154mg 𝑔 2 ) = 154 O2 1 𝑚𝑜𝑙 O2 We DO need stoichiometric ratios to calculate the amount of KClO3, because we did not measure it directly: · 4 – 16 · Application of IGL  godaddy.IE Chem 1211/1203 Lecture Notes Unit 4 dc 394 KC 2 𝑚𝑜𝑙 KClO3 122.6 𝑔 KClO3 𝑔 KClO3 = (4.81 mg 2 03 𝑚𝑚𝑜𝑙 O ) ( 3 𝑚𝑜𝑙 O2 ) (KC103 0.394g 1 𝑚𝑜𝑙 KClO3 ) = 394 𝑚𝑔 KClO3 = 0.394 𝑔 KClO3 Q.6 Nitrogen gas is collected in an inverted−buret setup similar to the pneumatic trough (and to Experiment E). The collected N2 is generated by the given reaction. If 0.12 g of each of NaNO2 (MM = 69.00 g·mol‒1) and v NaNO2 + HSOv HSO3NH2 (MM = 97.1 g·mol‒ 3NH2 1 ) are mixed in 25°C water and NaHSO + N (g) + H O 4 2 2 the atmospheric pressure is 689 mm Hg, what is the maximum amount of N2 that can be collected? As mentioned on the footnote on p. 4‒10, determination of the limiting reagent is a likely component of this problem. It’s easy with a 1:1 stoichiometry: 1 𝑚𝑜𝑙 𝑚𝑜𝑙 HSO3 NH2 = (0.12 𝑔) ( ) = 1.23 𝑚𝑚𝑜𝑙 97.1 𝑔 1 𝑚𝑜𝑙 𝑚𝑜𝑙 NaNO2 = (0.12 𝑔) ( ) = 1.74 𝑚𝑚𝑜𝑙 69.0 𝑔 ℎ𝑎𝑣𝑒 1.23 𝑚𝑚𝑜𝑙 HSO3 NH2 ⇒ = 1.23 𝑛𝑒𝑒𝑑 1 𝑚𝑚𝑜𝑙 HSO3 NH2 · 4 – 17 · Application of IGL  Chem 1211/1203 Lecture Notes Unit 4 ℎ𝑎𝑣𝑒 1.74 𝑚𝑚𝑜𝑙 NaNO2 ⇒ = 1.74 𝑛𝑒𝑒𝑑 1 𝑚𝑚𝑜𝑙 NaNO2  HSO3NH2 is the L. R. Use that to determine mol N2: 1 𝑚𝑜𝑙 N2 1.23mmol molNNa 𝑚𝑜𝑙 ( 4.23 2 = 1.23 mmol 𝑚𝑚𝑜𝑙 ) HSO3 NH2 ( ) = 1.23 𝑚𝑚𝑜𝑙 nq 1 𝑚𝑜𝑙 HSO 3 NH2 Not all the gas collected is N2 though…the other gas is: H2O 𝑃H2O (25 °C) = 23.8 𝑚𝑚 𝐻𝑔 Ptotal PN 𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃 Pro 689 N2 + 𝑃H2 O = 689 𝑚𝑚 𝐻𝑔 mmHg 𝑃PN2 N2 = 𝑃𝑡𝑜𝑡𝑎𝑙 − 𝑃H2O = 689 − 23.8 𝑚𝑚 𝐻𝑔 = 665 𝑚𝑚 𝐻𝑔 Ptotal Ppe 689 23.8 mmHg 665 untly Must use PV = nRT to determine V: 𝑛𝑅𝑇 𝑉= 𝑃 nRI E−1 298 5 23 mmol 0082062 atm mot−1 (1.23 𝑚𝑚𝑜𝑙 )(0.08206 𝐿 ⋅ 𝑎𝑡𝑚 ⋅ 𝑚𝑜𝑙 ⋅ 𝐾 )(298 𝐾 ) = (665 𝑚𝑚665 𝐻𝑔mmHg )( 1 𝑎𝑡𝑚 ) 760 𝑚𝑚 𝐻𝑔 g =34.4m α 34.4 𝑚𝐿 Your lab experiment will differ slightly, however.3 3 In the lab, you’ll take hydrostatic pressure into account, and you will determine R, not the volume of gas (you’ll read that off the buret). · 4 – 18 · Application of IGL Chem 1211/1203 Lecture Notes Unit 4 Supplemental Questions Unit Conversion Practice: 1. Convert the following pressure units: (a) 4 atm into torr (b) 788 mm Hg into kPa (c) 722 torr into mm Hg 5 (d) 1.56×10 Pa into atm (e) 1.22 atm into mm Hg Simple Gas Laws: 2. A 35.8 L cylinder of Ar(g) is connected to an evacuated 1875 L tank. If the temperature is held constant and the final pressure of the gas is 721 mm Hg, what was the original pressure of the gas, in atm? 3. A fixed quantity of gas at 23 °C exhibits a pressure of 735 torr and occupies a volume of 5.22 L. (i) Use Boyle’s Law to calculate the volume of the gas if the pressure is increased to 1.88 atm while T is held constant (ii) Use Charles’ Law to calculate the volume of the gas if the temperature is increased to 165 °C while P is held constant 4. A sample of sulfur hexafluoride (SF6) gas occupies a volume of 5.10 L at 193 °C. If P remains constant, what temperature (in °C) is needed to reduce the volume to 2.50 L? 5. What is the effect of the following on the volume of 1 mol of an ideal gas? (i) The temperature is decreased from 700 K to 350 K (P constant) (ii) The temperature is increased from 350 °C to 700 °C (P constant) (iii)The pressure is increased from 2 atm to 8 atm (T constant) 6. A 93 L sample of dry air is cooled from 145 °C to – 22 °C while the pressure is maintained at 2.85 atm. What is the final volume? 7. SO2 is a major culprit in the formation of acid rain. The pressure of a 0.57 L sample is measured to be 1.5×104 Pa. If the volume was changed to 1.53 L, what is the new pressure of the gas? 8. A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 21 °C 9. A gas-filled weather balloon with a volume of 55.0 L is released at sea-level conditions of 755 mm Hg and 23 °C. The balloon can expand to a maximum volume of 835 L before exploding. When the balloon rises to an altitude at which the temperature is – 5°C and the pressure is 0.066 atm, will the balloon explode? Ideal Gas Law: 10. Tennis balls are usually filled with air or N2(g) to a pressure above atmospheric to increase their bounce. If a particular tennis ball has a volume of 0.144 L and contains 0.33 g of N2, what is the pressure inside the ball, in both kPa and atm? HINT: what assumption(s) do you have to make? 11. The Hindenberg was an H2(g) filled dirigible that exploded in 1937 (Oh the Humanity!). If the balloon held 2×108 L of gas at 23 °C and 1.0 atm, what mass of H2 is in the balloon? 12. A compressed gas cylinder contains 106.0 g of acetylene, C2H2, at a pressure of 51.0 atm. Some of the acetylene is used in a chemical reaction, and the pressure in the cylinder drops to 42.0 atm, how many grams of acetylene are left in the cylinder? Molar Mass, Dumas Bulb: 13. Calculate the molar mass of a gas at 388 torr and 45 °C if 206 ng occupies 0.206 L. ·1· 4: Supplement Chem 1211/1203 Lecture Notes Unit 4 Supplemental Questions 14. In the Dumas–bulb technique for determining the molar mass of an unknown liquid, a sample is vapourized at 100 °C in a boiling water bath and determining the mass of vapour required to fill the bulb (see figure on the right). An organic chemist isolates a colourless liquid from a petroleum sample. The following data is obtained while studying the properties of the vapour of the liquid using the Dumas method. Calculate the liquid’s molar mass: Volume of flask: 213 mL Mass of Flask and Gas: 78.416 g P: 754 torr Mass of Flask: 77.834 g 15. The following data was obtained in another Dumas-bulb molar mass determination: Volume of flask: 354 mL Mass of Flask and Gas: 78.846 g P: 0.976 atm Mass of flask: 77.834 g Calculate the liquid’s molar mass. Gas Stoichiometry: 16. A compound of silicon and fluorine consists of 33.0% Si and 67.0% F by mass. If 2.38 g of the compound occupies a volume of 0.210 L at 38 °C and 1.70 atm. Determine the compound’s molecular formula. 17. Calculate the volume of O2 at STP required for the complete combustion of 100 g of octane (C8H18). 18. A compound of silicon and fluorine consists of 33.0% Si and 67.0% F by mass. If 2.38 g of the compound occupies a volume of 0.210 L at 38 °C and 1.70 atm. Determine the compound’s molecular formula. 19. Calcium carbonate, CaCO3(s), decomposes upon heating to give CaO(s) and CO2(g). A sample of CaCO3 is decomposed and the CO2 is collected in a 250 mL flask. After the decomposition is complete, their pressure in the flask is 131.7 kPa at a temperature of 31 °C. (a) How many moles of CO2 were generated? (b) if the reaction was only 62% efficient, how much CaCO3(s) will be needed to produce the amount of CO2(g) in part (a)? 20. Airbags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes nitroguanidine (CH4N4O2, MM = 104.1 g·mol‒1) to decompose explosively according to the reaction shown below. What mass of nitroguanidine is needed to inflate a 70.0 L airbag at SATP (25°C, 1 atm)? CH4 N4 O2 (𝑠) → 2 H2 O(𝑔) + 2 N2 (𝑔) + C(𝑠) Partial Pressures, Dalton’s Law, i.e. Pneumatic Trough: 21. The evolved hydrogen gas from a reaction of 50.0 mL of 0.250 M sulfuric acid and 0.178 g of aluminum metal was collected over water in a pneumatic trough setup similar to that seen on page 4−15. The collected gas occupied 253 mL at 93.8 kPa barometric pressure and 20 °C: (a) Write the balanced chemical reaction (b) The % yield of hydrogen (c) The volume the collected hydrogen gas would occupy if it were dried at STP. 22. In a pneumatic trough setup, 185 mL of oxygen gas was collected over water at 25°C and 600 mm Hg atmospheric pressure. (a) How many moles of oxygen gas were collected? (b) if the water was removed, how much volume would the dry oxygen occupy at STP? ·2· 4: Supplement Chem 1211/1203 Lecture Notes Unit 4 Supplemental Questions Challenging, Comprehensive Problems: 23. A 2.115 g sample of cobalt metal reacts with 25.0 mL of 6.0 M HNO3(aq) and produces 2.56 L of NO2 gas collected over water (like the pneumatic trough set up at the end of the Unit 4 Lecture Notes) at 25°C and 50 kPa. Balance the reaction, give the complete molecular equation and calculate the percent yield: 𝑎𝑐𝑖𝑑𝑖𝑐 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠 Co(𝑠) + NO− 3 (𝑎𝑞) → NO2 (𝑔) + Co2+ (𝑎𝑞) 24. A 27.2 mg sample of the anti−malarial drug quinine (contains C, H, O, and N) exerted a pressure of 9.48 torr in a 250.0 mL container at 453 K in a Dumas−bulb type set up discussed in class. When 17.2 mg of quinine was completely burned in excess oxygen, 46.7 mg of carbon dioxide, 11.5 mg of water and 1.49 mg of nitrogen gas was produced. Determine the molecular formula of quinine. 25. A mixture of methane (CH4) and ethane (C2H6) at a total pressure of 380 torr is completely burned in excess O2 forming CO2 and H2O(l). The water and excess oxygen are removed and the remaining CO2 has a partial pressure of 700 torr at the same volume and temperature as the original methane/ethane mixture. What are the partial pressures and mole fractions of CH4 and C2H6 in the original mixture? HINT: you’ll need to have balanced combustion reactions! ·3· 4: Supplement