Mols, Stoich, and Thermo! PDF

Warm - Up 11/11 - 11/12 Grab a calculator and get out data booklet every day from here on out!!! Does the structure to the right (which is sucrose btw) represent a covalent or ionic molecule? Would you expect it to be soluble in water? Why or why not? What is the molecular formula of sucrose? Mols, Stoich and Thermo Units S1.4, R2.1, R1.1-1.3 Oh boy From now ‘till 1. Review over mols, 2025 molecular/empirical formulae, etc (S1.4) 2. Stoichiometry (R2.1) 3. Thermochemistry (R1.1 - R1.3) S1.4 - Counting Particles - The Mole Mols Review of the good ol’ mole - The mole (mol) is a unit (haven’t you heard?) - the SI unit of amount of substance. - A mol of an entity contains 6.02 x 1023 of that entity - 6.02 x 1023 - For every 1 mole of donuts, I have 6.02 x 1023 donuts - For every 1 mole of water, I have 6.02 x 1023 molecules of water - For every 1 mole of electrons, I have 6.02 x 1023 e- - Etcetcetcetc Mols - Example problem - If I have 2 mols of water molecules… - how many mols of oxygen atoms do I have? - How many mols of hydrogen atoms do I have? Ex Water is H2O For every 1 mol of water, I have 2 moles of hydrogen atoms, 1 mol of oxygen atoms So 2 mols of water molecules →2 moles oxygen atoms 2 mols of water molecules → 4 moles hydrogen atoms Why moles? - Moles matter because reactions occur PROPORTIONALLY - Ex: - 2H2 + O2 → 2H2O - What do those coeﬃcient “2s” actually mean? - The proportional ratio of hydrogen molecules to oxygen molecules to water! - 2 molecules of hydrogen + 1 molecule of oxygen → 2 molecules of water - We can scale that up using the mole concept An aside - atomic masses - The relative atomic mass (Ar) is the weighted mean of the naturally occurring isotopes relative to 1/12th of a carbon-12 atom - Relative atomic mass has no units, as it is a ratio between masses and masses (think about your units!) - Actual atomic masses are ratios with units of grams per mole - Relative atomic masses can be found in the data booklet Molar masses - The mass of one mole of a substance is called the molar mass. Molar mass is given the symbol M and has units of grams per mol. - Ex: For every 1 mol of helium atoms I have 4.00 grams **OF THE UTMOST IMPORTANCE - UNDERSTAND THAT MOLAR MASS IS A RATIO - UNDERSTAND IT HAS UNITS** Example problem I have a mole of water. - How many grams of oxygen atoms are present? - How many grams of hydrogen atoms are present? Molar masses Recall dimensional analysis… YOU *MUST* SHOW WORK WITH UNITS TO GET FULL MARKS IN THIS UNIT Dimensional analysis time USING DIMENSIONAL ANALYSIS, determine how many moles are in 10 grams of water. - Step 1: molar mass of H2O = (2 * 1.01) + 16.00 = 18.02 g H2O / mol H2O - Step 2: use molar mass to convert 10 g -> moles Practice Practice solns Empirical formula tips Emp. example prob Molecular formula Answers More molecular formula practice “Mol airlines” - 10 min to ﬁnish the identiﬁcation of the passengers and chemicals (top priority) and the “report” (1 sentence per bullet point is really all you need), can do on back of paper) Mols - An outlandish claim in that video: - “There are more atoms in a handful of sugar than stars in the universe” - How can we put that to the test? - Assume there are 200 sextillion - aka 2 * 1023 - stars. Mol of sugar vs EVERY STAR IN THE SKY - Molecular formula of sugar: - C12H22O11 - Number of stars in the universe: - 2 x 1023 (200 sextillion) - Already we can see we have more sugar molecules than stars in the universe! - BUT how many more atoms do I have? “All the stars in the sky are our enemy” 45 mols of atoms in one mol of sucrose 45 mols * (6.02 * 1023 atoms / 1 mol) = 2.709 * 1025 atoms 2.709 * 1025 atoms in a mol of sugar / 2 * 1023 stars in the universe = ~135 atoms in a mol of sugar for every star in the universe Ways to interpret this info: - There are 135 times more atoms in a mole of sugar than stars in the sky. - But that wasn’t the claim in the video….claim in the video was about HANDFUL of sugar 11 / 15 R2.1 - Stoichiometry - Stoichiometry is ultimately the study of ratios. - Chemical reactions occur on a mol basis. For the reaction below: 2Ba + O2 → 2BaO 2 moles of barium react with 1 mole of oxygen gas to create 2 moles of barium oxide. Given a mass or molar quantity of a product, we can use our knowledge of ratios to predict the theoretical yield of a product. The theoretical yield is the maximum possible yield, assuming all reactants are consumed. Stoich Problems For any stoichiometry problem, you must know what unit you are given and what unit you are looking for. Step one is to ALWAYS convert to moles unless you are already in moles. Step two is to multiply by the mole ratio with the unknown on top. Step three is to convert to the unit you are looking for Stoich Problems - Mol to Mol Given 2Ba + O2 → 2BaO: If we have 2.5 moles of oxygen gas and excess barium, how many mols of BaO will we create? (Note that what we are calculating is the theoretical yield of BaO) *Can use dimensional analysis or RICE tables - note that Griesel will be using dimensional analysis* Stoich Problems - Mass to Mol Given 2Ba + O2 → 2BaO: If we have 2.5 grams of barium, how many mols of BaO will we create? Assume we have excess O2. (Note that what we are calculating is the theoretical yield of BaO) *Can use dimensional analysis or RICE tables - note that Griesel will be using dimensional analysis* Stoich Problems - Mol to Mass Given 2B + 3F2 → 2BF3 If we ended up with a theoretical yield of 5 mols of boron triﬂuoride, how many grams of ﬂurorine did we use? Assume we have excess boron. *Can use dimensional analysis or RICE tables - note that Griesel will be using dimensional analysis* Stoich Problems: Mass to Mass Assuming we are synthesizing phosphorous trichloride from white phosphorus (P4) and chlorine gas: P4 + 6Cl2 → 4PCl3 If I have 10 grams of P4 and excess chlroine, how many grams of PCl3 can I create? Limiting Reagents - Burger… - Imagine we are trying to make a massive amount of cheeseburgers to feed a crowd. - Each cheeseburger is a single. It needs one patty, one slice of cheese, and 2 pieces of bread for buns. - If we have 102 pieces of bread, 100 pieces of cheese, and 61 patties: - How many burgers can I make? - What did I run out of ﬁrst? - How much of each supply do I have leftover? Burger? - I could only make 51 burgers. - I ran out of buns ﬁrst. - I had 49 pieces of cheese leftover. - I had 10 patties leftover. Limiting Reagent / Limiting Reactant - The chemical we run out of ﬁrst in a chemical reaction is called the limiting reagent (or limiting reactant). - In the previous example, we ran out of buns ﬁrst. The buns were our limiting reactant. - Even though my amount of buns was larger, due to the ratio of buns to patties to cheese to completed burgers (2:1:1:1), I still ran out of buns ﬁrst. - Similarly, in chemistry problems, we can have a larger quantity of the limiting reagent - it’s all about what runs out ﬁrst - Tip - you need to worry about limiting reagent only when you are given masses/quantities for multiple reactants. Limiting Reagent Problems P4 + 6Cl2 → 4PCl3 If I have 147 g of P4 and 615g of Cl2, how much PCl3 can I create? 11/20 Warm-Up on whiteboard Theoretical vs Experimental Yield Theoretical Yield – the maximum amount of product that can be produced from a given amount of reactant (found by using stoichiometry). Experimental Yield – the measured amount of product obtained from a reaction (what you got in the lab, AKA actual yield) Example Assume get a yield of 19.8 g when 10 g of sodium reacts with 14 g of chlorine gas. Calculate the percent yield of the reaction. (Think - what do you need to know?) Atom Economy - Atom economy is a measure of how “eﬃcient” a process is from a green-chemistry perspective - The calculation is given in your data booklet! Atom economy can be increased if a use can be found for byproducts in a reaction (waste not, want not!!!) https://www.youtube.com/watch?v=Zuyk4hfbjSA Concentration M = unit of molarity A 1.0 M solution is deﬁned as 1 mole of solute dissolved in 1 L (or 1 dm3) of solvent to create 1 L (or 1 dm3) of solution. Putting brackets around a chemical indicate that you are taking the concentration of that chemical in mol/L (or mol/dm3) Ex: [BaO(aq)] means “concentration of barium oxide in mol/L” Ionic Equations - Remember that in aqueous solutions, ions dissociate. - Consider a double replacement reaction: - AgNO3 (aq) + NaCl (aq) → NaNO3 (aq) + AgCl (s) - The AgCl(s) is not dissociated into its ions - it is solid. - We could rewrite that equation as thus: Ag+ (aq) + NO3- (aq) + Na+ (aq)+ Cl- (aq)→ Na+ (aq)+ NO3- (aq) + AgCl (s) Ionic Equations (Cont) Ag+ (aq) + NO3- (aq) + Na+ (aq)+ Cl- (aq)→ Na+ (aq)+ NO3- (aq) + AgCl (s) ^ this is called the overall ionic equation Notice how the sodium and nitrate ions don’t do anything in the equation. They are dubbed spectator ions because they just watch and don’t do a thing. The overall reaction can be simpliﬁed as: - Ag+ (aq) + Cl- (aq) → AgCl (s) ^this is called the net ionic equation Ionic Equations - Write the ionic equation and identify the spectator ions for the following reaction: 11/22 Warm Up 2 laptops per table: - Go to schoology - Do the assignment labeled “The Cranberry Problem” Half Reactions Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s) Write this as a overall ionic equation. What do you notice as a key difference compared to the previous problems? Half Reactions Zn (s) + Cu2+ (aq) + SO42- (aq) → Zn2+ (aq) + SO42- (aq) + Cu (s) Zinc is losing electrons as it gains charge. Zn (s) → Zn2+ (aq) + 2e- Copper is gaining electrons as it loses its charge. Cu2+ (aq) + 2e- → Cu (s) We can express these processes as half reactions. Half Reactions Zn (s) → Zn2+ (aq) + 2e- Cu2+ (aq) + 2e- → Cu (s) We would say zinc is getting oxidized, as it is losing electrons Copper is getting reduced, as it is gaining electrons The two half-reactions here are more useful to see what’s actually going on compared to the overall reaction. OILRIG - oxidation is loss, reduction is gain Half Reactions Single-replacement are not the only type of oxidation/reduction reactions! Consider this: 2Mg (s) + O2 (g) → 2MgO (s) Write the two half reactions. Hint: remember that coeﬃcients =/= subscripts. Standard Temperature and Pressure, along with gas stuff - Standard Temperature and Pressure (STP) is assumed to be 273.15 K (aka 0 degrees C) and 100 kPa (aka 1 atm). - *important* - at STP, 1 mole of gas is said to occult 22.7 L of space. - THIS IS IN YOUR DATA BOOKLET - sections 2 and 4! - Remember 1000 L = 1 m3, and that mL = cm3, L = dm3, thus 1000 L = m3 Post Thanksgiving Break Warm-Up If 55.5 g of cobalt (II) carbonate reacts with 5 mol of oxygen gas, what mass of cobalt (II,III) oxide is produced (note: do not worry about the weird formula of Co3O4). Reminder of the ideal gas law - While molarity was useful for operations involving solutions, gas properties are useful for stoichiometry involving gasses. - Recall that PV = nRT This equation, as well as the value for the ideal gas constant R, are present in your data booklet Complete Combustion Complete combustion is the reaction of a hydrocarbon with oxygen gas to produce CO2 gas and H2O gas. Ex: Ex: Complete Combustion Write the balanced chemical equation for the complete combustions of: 1. But-1-ene 2. Cyclobutane 3. But-1-yne Multiple Choice Work Time :) SKIP 3, 4, 6, 10, 12, 17, 18 12/5 - 12/6 Warm Up - Get out notes - Work on MC from last class for 10 minutes. - SKIP 3, 4, 6, 10, 12, 17, 18 Write the half reactions and the net ionic equation for the a reaction that has silver ions in solution and aluminum metal as reactants and solid silver metal and aluminum ions in solution as products. Identify what species are being oxidized and reduced. Also get out your notes :) Ag++e−→Ag LEOGER - loss of electrons is oxidation, gain of electrons is reduction. Silver is being reduced, aluminum getting oxidized A Note on Chemical Reactions - Chemical reactions involve bond breaking and bond formation - This means, for a chemical reaction to occur, products and reactants have to be different chemicals - different arrangement of atoms - If BONDS are not broken a chemical reaction does not occur - Phase change is not truly a chemical reaction, as no bonds are broken - When phase change occurs, IMFs are overcome -> bonds are NOT BROKEN Energy All chemical reactions are accompanied by energy changes. Energy is a measure of the ability to do work, that is to move an object against an opposing force. Examples: heat, light, sound, electricity and chemical energy which is the energy released or absorbed during chemical reactions. Energy and Bonds - KEY POINT - Breaking chemical bonds requires energy. - MAKING chemical bonds releases an equal amount of energy…more on this later Energy vs Temperature Heat is a form of energy. It is often measured in kJ or J (kilojoules or joules) Remember kinetic molecular theory! Temperature is the average kinetic energy of the particles of a substance. Conservation of Energy - Just like mass, in a chemical reaction energy is neither created nor destroyed - An energy input is needed to break bonds. Energy is output when bonds are formed. Exo vs Endothermic - If the energy output from bond formation is greater than the energy input needed to break bonds, then the leftover energy is given out to the surroundings as heat - This is an exothermic reaction - If the energy input to break bonds is greater than the energy output by bond formation, then energy needs to come from the surroundings - This is an endothermic reaction Needed Deﬁnition System – area of interest (Example: beaker and its contents) Surroundings – everything else in the universe An “open” system can exchange matter and energy with the surroundings. A “closed” system can only exchange energy, not matter, with the surroundings. Exothermic Endothermic Heat given out to surroundings Heat input required from the surroundings Speciﬁc Heat Capacity - Intrinsic property of a material - Pure water has a speciﬁc heat capacity of 4.18 kJ kg-1 K-1 - This means that it requires 4.18 J to heat 1 g of water by 1°C or K - Similarly, when 1 g of water cools by 1°C, it gives out 4.18 J of energy means the standard enthalpy change of reaction Standard conditions are deﬁned as the normal, most pure and stable state of a substance at 100 kPa/1 atm and 298 K. The little circle thing means “standard conditions”. I don’t have the unicode shortcut memorized hence the manifold copy-pasting. Enthalpy of Formation / Enthalpy of Combustion The enthalpy change of formation, ΔHfᶱ, is the enthalpy change when one mole of a substance in its standard state is formed from its elements in their standard states at a pressure of 100 kPa and at a temperature of 298 K The enthalpy change of combustion, ΔHcᶱ, is the enthalpy change when one mole of a substance in its standard state is completely combusted in oxygen under standard conditions of 100 kPa and at a temperature of 298 K Warm-Up 1/7-1/8 Welcome Back! - Using section 12 in your data booklet, state the enthalpy of these bond making/breaking processes, respectively (in kJ/mol): - 2H (g) → H2 (g) - H2 (g) → 2H (g) - Hint: The “trick” is for you to think about the sign on the enthalpy change! [2 min timer] Changes and Expectations - The year is going to become faster paced. - Expect nightly reading or homework. Bond Enthalpy Refresher Energy is required to break a chemical bond. Purely bond breaking is endothermic (requires energy) Ex: H2 (g) → 2H (g) Energy is released when bonds are formed. Purely bond making is exothermic. Ex: 2H (g) → H2 (g) Bond Enthalpies We tend to refer to average bond enthalpies. Some bond enthalpies are absolute as there is only one compound where that bond can exist (ex: H-F, O=O). However, think about all the different C-H bonds. The bond enthalpies will be inﬂuenced ever so slightly by the nature of the other atoms present. So we use the average. Average bond enthalpy is the average energy to break one mole of the bond in similar compounds, all in the gaseous state. Using Bond Enthalpies What if we have both bond breaking and formation? If all species are gaseous, we can use the following equation to determine the enthalpy change of a reaction: The ﬁrst term is positive because breaking bonds is endothermic. The second term is negative because forming bonds is exothermic. Bond Enthalpy Ex Consider the formation of nitrogen triﬂuoride: N2 (g) + 3F2 (g) → 2NF3 (g) All species are gaseous. We can use bond enthalpy to ﬁnd enthalpy change. HINT: how many and what types of bonds are being broken? How many and what bonds are being formed? Enthalpy of formation For the previous reaction, we are creating 2 mols of NF3. What if we were to ﬁnd the enthalpy of formation for NF3? RECALL: The enthalpy change of formation, ΔHfᶱ, is the enthalpy change when ONE MOLE of a substance in its standard state is formed from its elements in their standard states at a pressure of 100 kPa and at a temperature of 298 K. Did we have 1 mol of NF3? Answer Bond Enthalpies Applied We can tend to predict if a reaction will be exothermic or not due to the fact two factors inﬂuence bond enthalpy: - (1) the relative strengths of the bonds as measured by the bond enthalpies - (2) the relative number of bonds broken and formed An exothermic reaction corresponds to the formation of more bonds, stronger bonds, or both! Why Gas Only For Bond Enthalpies? - What is going on during a phase change? What is being broken (or overcome) when something, for example, goes from liquid to gas? Why Gas Only For Bond Enthalpies? - What is going on during a phase change? What is being broken (or overcome) when something, for example, goes from liquid to gas? - IMFs! - Consider: C(s) + 2H2(g) → CH4(g) - What is really happening? Think of what *every step* is during the formation of methane. Why Gas Only For Bond Enthalpies? - What is going on during a phase change? What is being broken (or overcome) when something, for example, goes from liquid to gas? - IMFs! - Consider: C(s) + 2H2(g) → CH4(g) - What is really happening? Think of what *every step* is during the formation of methane. - The carbon needs to change phases in addition to all the bond breaking and making. We cannot use avg. bond enthalpies to accurately describe this process, as phase changes must also be considered. Hess’ Law - Since enthalpies require gaseous state, we require other methods to determine enthalpy changes for most reactions - Hess’ Law - the total energy change in a reaction depends only on the ﬁnal and initial states and is independent of reaction pathway Hess’ Law Consider formation of methane: - C(s) + 2H2(g) → CH4(g) Due to the fact that we can form inﬁnitely many hydrocarbons AND have a solid, can’t use bond enthalpies. However, we know that we can ﬁnd the enthalpies of combustion for carbon, hydrogen, and methane (section 14 in data booklet)! Combustion Review The enthalpy change of combustion, ΔHcᶱ, is the enthalpy change when one mole of a substance in its standard state is completely combusted in oxygen under standard conditions of 100 kPa and at a temperature of 298 K We have enthalpy of combustion in our data booklets - section 14! Find the enthalpy of combustion for H2, C, and CH4 Enthalpy of formation of methane is equal to the [combustion of one mole of carbon + 2 moles of hydrogen] + [reverse (Bottom part here of combustion of represents combustion) methane!] Hess’ Law - Enthalpy of Formation The enthalpy change of formation, ΔHfᶱ, is the enthalpy change when one mole of a substance in its standard state is formed from its elements in their standard states at a pressure of 100 kPa and at a temperature of 298 K. So if we know enthalpies of formation (section 13 of data booklet), we can use Hess’ Law, too. Ex: 2Al (s) + [3/2] O2 (g) → Al2O3 (s) ΔHfᶱ = - 1676 kJ/mol Note the fractional coeﬃcient - since enthalpy of formation is deﬁned by the formation of one mole of something Hess’ Law - Summation of Equations - The simpler way of thinking about Hess’ Law (IMO) is as a system of equations. Consider: What if we know more information about other rxns? C(s) + O2 (g) → CO2 (g) ΔH = -393.5 kJ 2H2 (g) + O2 (g)→ 2H2O (l) ΔH = -571.6 kJ 2C2H2 (g) + 5O2 (g) → 4CO2 (g) + 2H2O (l) ΔH = -2602.2 kJ Manipulate “known” rxns to get to target rxn. I have to perform rxn 1 TWICE (look at coeﬃcient of C in target) I have to do rxn 2 “HALF A TIME” (look at coeﬃcient of H2 in target) I have to do rxn 3 “HALF A TIME” and IN REVERSE (I want ethyne as a product, and only 1!) End of class reminder - Reading homework due next class - begin thinking about IA for SL test-takers - Stoich/mol quiz next class 1/9 - 1/10 - Exit ticket review Get out data booklet, get a calculator and divider (don’t put it up yet tho). Going to pass out exit tickets when class starts. Biggest exit ticket mistakes: - Q1 - Using Section 11 instead of Section 12… - Incorrect number of O-F bonds being broken - Using O-O value instead of O=O value - Q2 - Saying that A→C was equal to A→B + C→B (will illustrate w cycle and system) Quiz 26.5 min Notes Check / MCQ Practice Get out R1.3 notes Work on the multiple choice practice problems for R1.2 (for IB pacing, should be done in less than 18 minutes). R1.3 - Intro - Why do we care about this unit? - The production of energy (realistically, as it relates to industry) - There are millions of moles of hydrocarbons combusting across the world as we speak as people drive their cars and heat up their houses - Chemical reactions require energy inputs and/or release heat energy to the environment - We can use thermodynamics to discuss fuel and energy. Combustion of metals When metals react with oxygen, they form metal oxides (duh?). Ex: 2Sr (s) + O2 (g) → 2SrO (s) 4K (s) + O2 (g) → 2K2O (s) (note - if we were to ﬁnd enthalpies of combustion for metals, we’d want the metal to have a coeﬃcient of 1) Remember that metal oxides are basic in solution. These combustion products, if they get into the water, can radically alter the pH. Combustion of non-metals Sulfur and nitrogen combust to form various oxides. Ex: S (s) + O2 (g) → SO2 (g) Sulfur is notable as it can react further with oxygen and water in the atmosphere to produce acid rain. 2SO2 + O2 (g) → 2SO3 (g) SO3 (g) + H2O (l) → H2SO4 (aq) Nitrogen’s combustion produces nitrogen oxides, which behave similarly. Combustion of non-metals Combustion of hydrogen Formation of water! 2H2 + O2 → 2H2O Keep this in mind for the part of this unit re: hydrogen fuel cells Complete Combustion of Hydrocarbons/Alcohols - We are mainly focused on the combustion of alkanes and alcohols - The complete combustion of any hydrocarbon or alcohol produces CO2 and water. - Complete combustion occurs w/ excess oxygen. - Most common fuels (propane, octane, etc) are alkanes. - Flip to section 14 of data booklet. Look at how much energy combusting those alkanes can release! PRACTICE - Write the equation for the complete combustion of: - Butane - butan-1-ol Incomplete combustion of hydrocarbons Incomplete combustion means Possible incomplete combustion of propane oxygen is limiting. Incomplete combustion both is less exothermic than complete combustion AND produces pollutants (CO, soot). It is very diﬃcult to determine exactly WHAT incomplete combustion reactions are occurring. The products of incomplete combustion can be CO or C, along w/ water. PRACTICE Write equations for the incomplete combustion of octane: - Where CO is a product - Where C is a product Why is oil still the dominant fuel? - While this is a massive can of worms one such reason is its high SPECIFIC ENERGY - the energy released per kilogram of fuel combusted - Use more frequently than enthalpies of combustion due to logistical constraints Speciﬁc Energy - We can convert from enthalpy of combustion to speciﬁc energy (dimensional analysis using molar mass!) - For octane: So the speciﬁc energy of octane is 47860 kJ Find the speciﬁc energy of ethanol Speciﬁc Energy Implication The speciﬁc energy of octane = ~47860 kJ/kg For ethanol, = ~29666 kJ/kg So burning an equal mass of octane releases approx 1.6 times more energy than ethanol. Energy density - Enthalpy of combustion is how much energy is released per mole combusted - Speciﬁc energy is how much energy is released per mass combusted - Energy density is how much energy is released per volume combusted Fossil Fuels The primary fossil fuels are coal, crude oil, and natural gas - Natural gas is a mixture, mostly methane - Crude oil is a very complicated mixture of hydrocarbons (mostly alkanes) - Different types of coal contain different % carbon by mass, and have different speciﬁc energies Intro to Fuel Cells 1/15 - 1/16 Warm-Ups 1. Write the equation associated with the incomplete combustion of propane: a. Where CO and water are products b. Where C and water are products 2. The following question is about methoxymethane (also called dimethyl ether, or DME) a. Using section 12 in your data booklet, calculate the enthalpy for the below reaction: C2H6O (g) + 3O2 (g) → 3H2O (g) + 2CO2 (g) b. The literature value for the enthalpy of combustion for DME is found to be -1406.4 kJ/mol. Calculate your percent error. Suggest a reason for the discrepancy. i. (Hint: the standard enthalpy of combustion assumes that all products and reactants are in their standard states at 298 K) Brief, Post-Break, R1.1 Review Standard Enthalpy Changes The enthalpy changes of reactions cna be measured but absolute enthalpies cannot be measured ΔHo - standard enthalpy change. Enthalpy change for standard conditions, where all compounds are in their standard states. Tends to correspond to 298 K and 100 kPa. Calorimetry - When heat is given out via exothermic reactions, or taken in by endothermic reactions, the temperature of the surroundings will change. - The magnitude and nature of the change depends on how much is being reacted and the speciﬁc heat capacity of the materials. - Re: speciﬁc heats - recall that pure water has a speciﬁc heat capacity of 4.18 kJ kg-1 K-1 - This means that it requires 4.18 kJ to heat 1 kg of water by 1°C or K - Implications: - It takes more energy to heat 100 kg of water 1 K than 10 kg of water for the same temperature change - The same energy input will increase the temperature of 10 kg more than it will 100 kg. Speciﬁc Heat Practice 1. Calculate the amount of energy needed to heat up 2 L of water from 298 K to 308 K. - Speciﬁc heat of water is 4.18 kJ kg-1 K-1, density is 1 g cm-3. 2. I have an equal mass of iron compared to the mass of water in the question above. I use the same amount of energy in the previous problem. If my iron starts at 298 K, what will its ﬁnal temperature be in oC? - The speciﬁc heat of iron is 0.499 kJ kg-1 K-1. Calorimetry - “Coffee Cup” Style - The temperature change is measured. - Using the speciﬁc heat of solution and the mass of reactants, we can determine the heat released and therefore enthalpy change of the reaction. - How would we do so? Turn and talk. Calorimetry - “Coffee Cup” Style - If we know the temperature change, mass of reactants, and speciﬁc heat of soln (assume same as water), we can calculate the heat of the reaction (think “mcat”)! - The enthalpy change is just the heat for your reaction divided by number of moles (think units) with the opposite sign. - Potential energy from the system is converted to heat. - This heat raises the temperature of water. - Since potential energy from the system was LOST, that’s why for coffee cup calorimetry, the positive temperature increase = exothermic reaction Calorimetry - “Coffee Cup” Style - We know that a positive temp change means the chemical potential energy is being converted to heat, which we are measuring via the thermometer. - BUT in reality some heat may be lost to the surroundings - This is why insulation matters - lowers the heat loss Calorimetry - “Bomb” Style 1.1 MC Practice Hint for #9 - the reaction is HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l). 1/17 - 1/22 Warm Up Using section 14 of your data booklet: - 300 cm3 of propane is ignited in a bomb calorimeter. Determine the temperature change (in degrees celsius) if 444 mL of water is heated by said combustion. - Assume the density of propane is 0.493 g/cm3

Mols, Stoich, and Thermo! PDF

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