Chemistry 203 Term Test 1 1 Lecture Slides Booklet Solutions - PowerPoint PPT Presentation

Chemistry 203 – Term Test 1 1  Lecture Slides  Booklet Solutions Friday  Test Solutions Intro  Extra notes and advice 3:30 5:00 to Stoichiometry and Gases  Advice on which exams and questions to write 5:00 to 5:30 Break  Corrections 5:30 to 6:45 Energy and Thermochemistry Available February 9 th at: 6:45 to 7:00 Break http://learnfaster.ca/blog/chem­203­term­test­1/ 7:00 to 8:20 Equilibrium 8:20 to 8:30 Break 8:30 to 9:30 Review Questions? Suggestions? E­mail me and I will try to find time to reply. psedach@learnfaster.ca (C) Pavel Sedach All the best! Learnfaster.ca

What is on the Exam? (Part 1) 2 Chem 203 is cumulative , however topics get expanded into Equilibrium, Kinetics and Electrochemistry. Some topics may not be tested directly in the future but are foundational to future concepts. 1 Graphs of P vs. V, P vs. T, etc. y = x or y = x Partial Pressure Working with ideal gases Percent yield Mole fraction P a = χ a P total PV = nRT Grams to atoms (Avogadro’s number) 𝑜 𝑛 𝑊 × 𝑁𝑋 = Stoichiometry Mass density 𝑊 Moles/liter from gas equation Solutions (moles/L and Liters) Stoichiometry through pictures Stoichiometry: Initial and Final P given constant V Gases Stoichiometry through gas laws P 1 V 1 n 1 R 1 T 1 P 2 V 2 = 2 sets of conditions: n 2 R 2 T 2 Redox Balancing (pre­requisite) Partial pressure and mass density through pictures Qualitative – if system expands, system DOES PV work If system contracts, work is done ON system At high pressures, molecular size counts, V real > V ideal Non­ideal gases At low temperatures, IMFs are felt and P real < P ideal (C) Pavel Sedach Learnfaster.ca

What is on the Exam? (Part 2) 3 Understand exothermic/endothermic diagrams Understand how rearranging an equation changes ΔH Kinetic Molecular Theory Average vs. Total Kinetic Energy Thermodynamics Hess’s Law: adding reactions adds enthalpy Molecules vs. Speed Graphs Define standard enthalpy of formation (compound made from elements) Molecules vs. Kinetic Energy Graphs Calculate enthalpy of reaction using heats of formation (P­R) How can a smaller container Math with Q = nΔH increase pressure? (more F/A) Qualitative treatment of heat capacity (based on Q = mcΔT ) Bond Dissociation Energy (R­P) Stoichiometry is a prerequisite. (C) Pavel Sedach Learnfaster.ca

Breakdown of old exams (C) Pavel Sedach 4 Learnfaster.ca Winter 2016 Fall 2015 Fall 2013 Spring 2013 Stoichiometry 7% 9% 43% 45% Kinetic Molecular Theory 11% 9% 3% 3% Gases & Partial Pressure 33% 23% 7% 33% Equilibrium 0% 23% 47% 6% Calorimetry and Enthalpy 33% 26% 9% Work and Energy 15% 11% 3% In recent years, for TT1, there has been a shift away from kinetics and more towards a broad overview of thermochemistry, work, heat and equilibrium. Gases and stoichiometry are foundational to this exam. The majority of the grades are from basic stoichiometry, gases and thermochemistry. Do your best on this exam and don’t fall behind – term test 2 and the final are progressively more difficult and are cumulative with the knowledge on this exam. IF you did not get a topic on TT1, you have a month to review before TT2. MAKE SURE you get it! (Cumulative Exams!)

What we offer 5 x2 Lecture Slides and Extra Questions 2 Copies Exam Package learnfaster.ca Full solutions Use the 2 nd copy to write closed book exams and grade yourself . Format: Cover every concept at least once through the lecture. Every question is based off exams If we finish a section early, we practice some exam questions. (C) Pavel Sedach Learnfaster.ca

Significant Figures (F2015 Q2) 6 Leading zeroes (red) don’t add information so they don’t count . 0.000011500 = 1.1500 × 10 −5 Zeroes on the right tell us precision. Ex. 50 means 50 ± 0.5 whereas 50.0 means 50.0 ± 0.05 Addition and Subtraction ­ Least number of decimals: 100 − 0.1547869 = 99.8452131 = 100 If two numbers use different scientific notation, convert to same notation: 17.2 × 10 3 − 5.5 × 10 2 = 172 × 10 2 − 5.5 × 10 2 = 166.5 = 167 Multiplication ­ Least number of sig. figs. 1230 × 16 = 19680 = 2.0 × 10 4 Evaluate sig figs. at every step but round only at the end. Practice: 24.71 − 24.0 = 24.71 × 24.0 = (C) Pavel Sedach Learnfaster.ca

Ion Concentrations (F2013 Q3) 7 If 50.0 mL of 0.100 M K 2 Cr 2 O 7 and 100 mL of 0.20 M K 3 PO 4 are combined with 50.0 mL of NaF , what is the concentration of potassium ions in solution? (C) Pavel Sedach Learnfaster.ca

Basic Stoichiometry 8 I have 6 wheels and 4 frames, given the following equation: 2 wheels + 1 frame  1 bicycle How many bicycles can I make? Given that 1 bicycle was broken during assembly, what was my percent yield? (C) Pavel Sedach Learnfaster.ca

Tips to Balancing Reactions (C) Pavel Sedach 9 H 2 Learnfaster.ca 1. Balance, 2. Convert to moles, 3. Use a molar ratio, 4. Convert moles to unit required N 2 O 2 F 2 If I react 15.0 g of Paraffin (mostly C 31 H 64 ) with 40.0 g of O 2 , what will my yield of CO 2 be? Cl 2 C 31 H 64 + O 2 → CO 2 + H 2 O Br 2 I 2 *When solving, we get 1.250 mol of O 2 and 0.0343 mol paraffin. After doing a mol:mol ratio, we get 0.825 mol CO 2 (36.3 g) from the oxygen and 1.065 mol of CO 2 from the paraffin. The oxygen is therefore the limiting reagent. If my actual yield of CO 2 g was 15.0 g, what was my percent yield? 41.3% How much of my excess reagent (in grams) was left? 3.38 g or 0.00774 mol of paraffin will be left over.

Stoichiometry with Pictures (F2013 Q4) 10 Whenever you encounter picture questions, ALWAYS convert the number of dots to moles and write them below the image: Given: 2 A + 3 2 B 2 → A 2 B 3 Which of the containers below will have the highest final yield of A 2 B 3 ? = A = B 2 (C) Pavel Sedach Learnfaster.ca

Review (W2016 Q2) 11 1. Balance, 2. Convert to moles, 3. Use a molar ratio, 4. Convert moles to unit required If we react 5.25 moles of sodium hydroxide with 6.75 moles of sulfuric acid, what is the final amount of all reagents? 2 NaOH s + H 2 SO 4 l → 2 H 2 O l +Na 2 SO 4 s all values in moles Initial 5.25 6.75 0 0 Change Final *Sometimes professors refer to “Final” as “Result” or “End” – the idea is the same – this is NOT an equilibrium ICE table, it’s an ICF table used for stoichiometry and one of the concentrations WILL go to zero! (C) Pavel Sedach *Remember this example – the above is how you evaluate the species present at ANY step in a titration OR an all gas system! Learnfaster.ca

PV = nRT Stoichiometry (W2016 Q1) 12 ) Pressure P × Volume V = moles n × ideal gas constant R × Temperature in Kelvin (T How many atoms of chlorine are present in are present in 3.45 g of PCl 5 g ? (C) Pavel Sedach Learnfaster.ca

Calculating the final pressure in a container (W2016 Q3) 13 For a gas system, PV = nRT and, if volume and temperature are constant we get: If I have a container with 1.50 atm of CH 4 g and 12.00 atm of O 2 g , what is the initial total pressure in the container? What is the total pressure AFTER reaction? (C) Pavel Sedach Learnfaster.ca

The Ideal Gas Law PV=nRT 14 (C) Pavel Sedach Learnfaster.ca If you have a mixture of CO 2 g (44.01 g/mol), N 2 g (28.02 g/mol) and C 12 H 22 O 11 g (MW=342 g/mol) they all behave the same! For instance, I have a container with 1 mol of CO 2 g , 1 mol of N 2 g and 2 mol of C 12 H 22 O 11 g (all at the same temperature) and my total pressure is 24 atmospheres, what portion of the pressure is each gas responsible for? moles pressure CO 2 g 1 mol N 2 g 1 mol C 12 H 22 O 11 g 2 mol total 24 atm This is because we assume all gases are ideal because they follow the “Kinetic Molecular Theory” (KMT). KMT says that regardless of the size of the gas molecule, they’re so small that the volume of a gas is zero. KMT also says that intermolecular forces (you know how water sticks to itself ?) are ZERO for any gas.

Kinetic Molecular Theory (C) Pavel Sedach 15 Learnfaster.ca Gas molecules ­can be treated like spheres of the same size. ­negligible volume compared to volume of space between gas molecules. ­not true at high pressures! (VOLUME is GREATER THAN IDEAL) ­still have mass (a more dense gas than another can settle to the bottom of a container) Gases expand to fill their container and collide with the container walls. This exerts pressure on the walls of the container: Pressure = Force Area = N m 2 Gases travel in constant, random, linear (straight­line) motion. gas molecule collisions are elastic (momentum and energy (E K ) are conserved). gases never lose energy (compressed gas can be used years later to run a barbecue despite constant collisions!) Molecules of gas experience no intermolecular forces, therefore each molecule moves at constant speed between collisions. ­ not true at low temperatures! (PRESSURE is LOWER THAN IDEAL)

Review (W2016 Q4/5/6) 16 At a given temperature, ALL gases (e.g. N 2 g vs. H 2 g ) have the same average E K . HOWEVER, E K = 1 2 𝑛𝑤 2 where m is Molecular Weight. We have two containers, 1 L of N 2 g at 25 ° C and 1L of H 2 g at 25 ° C, which gas would move faster? (C) Pavel Sedach Learnfaster.ca

Other Gas Laws (W2012 Q6) *assume other variables constant 17 (C) Pavel Sedach Learnfaster.ca 3 Scenarios: What does a Look like? y = x Direct Proportion V vs. T 𝑧 PV = nRT 𝑦 1 Inverse Proportion y = x OR xy = 1 𝑧 P vs. V Why? PV = nRT 𝑦 y = x 2 Geometric 𝑧 P vs. T Why? PV = nRT 𝑦