* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 0 Quarter Three Assessment Review - SRHSchem
Survey
Document related concepts
Glass transition wikipedia , lookup
State of matter wikipedia , lookup
Temperature wikipedia , lookup
Degenerate matter wikipedia , lookup
Work (thermodynamics) wikipedia , lookup
Heat transfer physics wikipedia , lookup
Equation of state wikipedia , lookup
Countercurrent exchange wikipedia , lookup
Thermoregulation wikipedia , lookup
Heat transfer wikipedia , lookup
Electrolysis of water wikipedia , lookup
Transcript
Chemistry Quarter 3 Assessment Review www.srhschem.wikispaces.com Thermodynamics Determine the Specific Heat of a Metal 1. A group of students conducted an experiment to determine the specific heat of a metal. Their goal was to use the specific heat to identify the metal. Mass of metal Mass of water Initial temperature of metal Initial temperature of water Final temperature of water and metal βTwater βTmetal a. Add an arrow to the diagram that shows the direction of heat transfer when the hot metal is added to the water in the calorimeter. 20.0 g 100.0 g 100.0 °C 23.1 °C 26.3 °C Heat flows from warmer to cooler. Determine the Specific Heat of a Metal b. Calculate the heat gained by the water. The specific heat of water is 4.18 J/g°C. 1. A group of students conducted an experiment to determine the specific heat of a metal. Their goal was to use the specific heat to identify the metal. Mass of metal Mass of water Initial temperature of metal Initial temperature of water Final temperature of water and metal βTwater βTmetal 20.0 g 100.0 g 100.0 °C 23.1 °C 26.3 °C 3.2 °C π = π × π × βπ 4.18 π½ π = (100.0 π) × × (3.2 β) πββ βπ = π2 β π1 = 26.3 β β 23.1 β = 3.2 β π = 1337.6 π½ π = 1300 π½ Determine the Specific Heat of a Metal 1. A group of students conducted an experiment to determine the specific heat of a metal. Their goal was to use the specific heat to identify the metal. Mass of metal Mass of water Initial temperature of metal Initial temperature of water Final temperature of water and metal βTwater βTmetal 20.0 g 100.0 g 100.0 °C 23.1 °C 26.3 °C 3.2 °C c. How does this calculated heat relate to the metal? β What ever heat is gained by the water (endothermic) is exactly equal to the heat lost by the metal (exothermic). β This gives us a value for q to use when solving for the specific heat of the metal. Determine the Specific Heat of a Metal 1. A group of students conducted an experiment to determine the specific heat of a metal. Their goal was to use the specific heat to identify the metal. Mass of metal Mass of water Initial temperature of metal Initial temperature of water Final temperature of water and metal βTwater βTmetal 20.0 g 100.0 g 100.0 °C 23.1 °C 26.3 °C 3.2 °C - 73.7 °C d. Find the specific heat of the metal. π = π × π × βπ (π × βπ) (π × βπ) π π= π × βπ β1337.6 π½ π= (20.0 π) × (β73.7 β) βπ = π2 β π1 = 26.3 β β 100.0 β = β73.7 β 0.907 π½ π= πβ 0.91 π½ π= πβ Determine the Specific Heat of a Metal 1. A group of students conducted an experiment to determine the specific heat of a metal. Their goal was to use the specific heat to identify the metal. Mass of metal Mass of water Initial temperature of metal Initial temperature of water Final temperature of water and metal βTwater βTmetal 20.0 g 100.0 g 100.0 °C 23.1 °C 26.3 °C 3.2 °C -73.7 °C d. Which metal did the students have? β Based on the calculated value for specific heat (0.91 J/g°C), the sample was most likely aluminum. Specific Heat Values, J/g°C Al 0.900 Au 0.129 Cu 0.385 Enthalpy Diagrams 2. Consider the enthalpy diagram below. a. What is the heat of reaction, ΞH, for this reaction? β Difference between products and reactants β 50 kJ - 150 kJ = -100 kJ b. Make a claim about if the reaction is endothermic or exothermic. Support your claim with evidence. β Exothermic: the reactants are higher than the products, meaning heat must be given off βH= -100 kJ Enthalpy Diagrams 2. Consider the enthalpy diagram below. c. How much energy is required (activation energy Ea) to break reactant bonds? β Reactants to the top of the hill β 200 kJ β 150 kJ = 50 kJ d. Compare the stabilities of the products and the reactants. β Higher energy = less stable β Reactants are less stable than the products because they are higher energy. Ea= 50 kJ Specific Heat of a Substance 3. A sample that weighs 234 grams is heated by 18°C when 1460 calories of heat is added. What is the specific heat of the substance? π = π × π × βπ (π × βπ) (π × βπ) π π= π × βπ 1460 πππ π= (234 π) × (18 β) 0.346 πππ π= πβ 0.35 πππ π= πβ Endothermic and Exothermic Phase Change 4. Provide an example of a phase change that is endothermic and one that is exothermic. β Endothermic: any phase change that requires the substance to absorb heat β Melting β Boiling β Exothermic: any phase change that releases heat/ cools the substance β Freezing β Condensing Heat of Reaction (Stoichiometry) 5. Consider the reaction below. 67.8 ππ½ + π2 π + π2 π β 2ππ2 (π) a. What is the βH of the reaction? βπ» = +67.8 ππ½ b. Is the reaction endothermic or exothermic? How do you know? β The reaction is endothermic because heat is a reactant: heat must be absorbed for the reaction to occur. This means that βH is positive. c. How many grams of N2 are required to absorb 156 KJ of energy? 156 ππ½ 1 πππ π2 28.01 π π2 × × = 64.46 π π2 = 64.5 π π2 1 67.8 ππ½ 1 πππ π2 πππππ πππ π π2 = 2 × 14.007 = 28.014 π/πππ Entropy and Disorder 6. Entropy is a measure of the disorder of a system. For each of the following, determine if the disorder of the system is increasing or decreasing. a. Ice melts into a liquid β Increase b. Methane crystallizes into a solid β Decrease c. Solid iodine sublimes into a gas β Increase d. Water condenses on a cold can β Decrease Bond Breaking and Forming 7. In the reaction O2 + energy ο O + O, bonds are being (circle one: broken/ formed) and energy is (circle one: released/ absorbed). O O + energy ο O + O Standard Enthalpy of Formation 8. What is the standard enthalpy of formation of AgCl if 28.36 kJ of heat are given off when 32 g of AgCl are formed? β Enthalpy of formation has units of kJ/mol. 32 π π΄ππΆπ 1 πππ π΄ππΆπ × = 0.2233 πππ π΄ππΆπ = 0.22 πππ π΄ππΆπ 1 143.32 π π΄ππΆπ ππ½ 28.36 ππ½ ππ½ ππ½ βπ» = = = 127.01 = 127.0 πππ 0.2233 πππ πππ πππ Heat Changes During Solution Formation 9. A 2.0 g sample of a solid dissolves in 50 g of liquid water at 25.0°C. Once the solid is completely dissolve, the water temperature drops to 15.2°C. a. Is this process endothermic or exothermic? Explain. β The temperature of the water decreases, so the dissolving process must have absorbed that heat, making it endothermic. b. Calculate the amount of heat exchanged for this process. π = π × π × βπ 4.18 π½ π = (50 π) × × (β9.8 β) πββ βπ = π2 β π1 = 15.2 β β 25.0 β = β9.8 β π = 2048.2 π½ π = 2.0 × 103 π½ Heating and Cooling Curves 10. Using the word bank, label the heating curve below. Boiling Point/ Condensation Point Gas Liquid Melting Point/ Freezing Point Solid Word Bank ο· Solid ο· Liquid ο· Gas ο· Melting Point ο· Freezing Point ο· Boiling Point ο· Condensation Point ο· Phase Change Phase change occurs where graph is flat! Time (Min) Temperature (°C) 0 -10 1 -5 2 0 10. Heating Curve and Data. 3 5 4 5 a. Based on the data, what is the melting point? The boiling point? 5 5 6 10 β Melting Point: 1st flat part, 5°C 7 15 nd β Boiling Point: 2 flat part, 60°C 8 20 9 25 b. Which segment of the graph has the most kinetic energy? The least? 10 30 11 35 β Most: Gas; particles are moving the fastest and have the highest temperature 12 40 β Least: Solid; particles are vibrating in place and have the lowest temperature 13 45 14 50 c. Explain why the temperature remains constant during a phase 15 55 change. 16 60 17 60 β Heat is still being added, but the temperature remains constant. The added 18 60 heat is instead being used to break apart the intermolecular attractions that 19 60 are holding the substance together in that phase (i.e. as a solid) 20 60 21 65 22 70 Heating and Cooling Curves Behavior of Gases and Gas Laws Ideal vs. Real Gases 11. Under which conditions of temperature and pressure will a real gas behave most like an ideal gas? β Ideal gases experience no attraction between gas particles. β High temperature: gases are moving too fast (kinetic energy too high) to attract/ interact with each other β Low pressure: particles are diffuse (far apart), and unlikely to collide and experience attractions Partial Pressures 12. A sample of hydrogen gas is collected over water at a temperature of 29 °C. a. What two gases are present above the water? β Hydrogen gas β Water vapor Partial Pressures 12. A sample of hydrogen gas is collected over water at a temperature of 29 °C. b. If the total pressure is 300 mmHg, what is the pressure of the hydrogen gas alone? Total = Hydrogen Gas + Water Vapor PTotal = PHydrogen Gas + PWater Vapor 300 mmHg = PHydrogen Gas + 30 mmHg PHydrogen Gas = 270 mmHg Gas Pressure and Manometers 13. A sample of gas is collected in a manometer, as shown below. a. Which exerts more pressure, the gas or the atmosphere? Explain. β The atmosphere is exerting more pressure, as the mercury is forced up the tube towards the gas. b. If the pressure of the atmosphere is 755 mmHg, what pressure is exerted by the gas in the flask? πππ‘πππ πβπππ > ππππ Need to subtract the height difference (which is already in mmHg) 755 πππ»π β 23 πππ»π = 732 πππ»π Behavior of Gases 14. Consider the two cylinders below. a. How many moles of gas are present in each cylinder? β Both pistons have the same initial conditions of P, T, and V β Therefore as long as they are ideal gases, they have the same number of moles of each gas. Behavior of Gases 14. Consider the two cylinders below. b. If the conditions in cylinder A change to a pressure of 5 atm and a temperature of 300 K, what is the new volume of the gas? P2 = 5 atm V2 = ? L T2 = 300 K P1 = 2 atm V1 = 6.5 L T1 = 200 K π1 π1 π2 π2 = π1 π2 2 ππ‘π × 6.5 πΏ 5 ππ‘π × π2 = 200 πΎ 300 πΎ 2 atm × 6.5 πΏ × 300πΎ = (200πΎ × 5 ππ‘π × π2 ) (200 πΎ × 5 ππ‘π) (200 πΎ × 5 ππ‘π) π2 = 3.9 πΏ Gas Laws: Boyle, Charles, Gay-Lussac Gas Law Equation Practice 5.6 L of gas at a pressure of 1.5 atm is compressed to a volume of 4.8 L. What is the new pressure of the gas? π1 × π1 = π2 × π2 Graph π1 × π1 = π2 × π2 π1 × π1 = π2 π2 1.5 ππ‘π × 5.6 πΏ = π2 4.8 πΏ Pressure Boyle P1: 1.5 atm V1: 5.6 L P2: ? V2: 4.8 L Volume π2 = 1.75 ππ‘π Gas Laws: Boyle, Charles, Gay-Lussac Gas Law Equation π1 π2 = π1 π2 Graph Volume Charles Practice A balloon contains 45 L of helium at 25°C. If the temperature is increased to 55°C, what will the new volume of the balloon be? V1: 45 L T1: 25 °C + 273 = 298 K V2: ? T2: 55 °C + 273 = 328 K π1 π2 = π1 π2 π1 × π2 = π2 π1 45 πΏ × 328 πΎ = π2 298 πΎ Temperature π2 = 49.5 πΏ Gas Laws: Boyle, Charles, Gay-Lussac Gas Law Equation π1 π2 = π1 π2 Graph Pressure GayLussac Practice A container of gas is initially at 0.500 atm and 25 ΛC. What will the pressure be at 125 ΛC? P1: 0.500 atm T1: 25 °C + 273 = 298 K P2: ? T2: 125 °C + 273 = 398 K π1 π2 = π1 π2 π1 × π2 = π2 π1 0.500 ππ‘π × 398 πΎ = π2 298 πΎ Temperature π2 = 0.67 ππ‘π