The following gas laws describes an ideal gas, where

Alief ISD Chemistry STAAR Review Reporting Category 4: Gases and Thermochemistry C.9.A Describe and calculate the relations between volume, pressure, number of moles, and temperature for an ideal gas as described by Boyles s law, Charles law, Avogadro s law, Dalton s law of partial pressure, and the ideal gas law. Gas Laws The following gas laws describes an ideal gas, where P = pressure of gas (atm), V = volume of gas (L), n = moles of gas, T = absolute temperature of gas (K), and R = the gas constant, 0.0821 L atm/mol K. The gas law that contains all four variables----p, V, T, and n--------is called the ideal gas law. Gas Law Equation Description Dalton s Law of Partial P T = P 1 + P 2 + P 3 + Pressures The total pressure of a mixture of gases equals the sum of the individual gas pressures Boyle s Law P 1 V 1 = P 2 V 2 P is inversely proportional to V at constant T (and at constant n) Charles Law V 1 = V 2 T 1 T 2 Avogadro s Law V 1 = V 2 n 1 n 2 V is directly proportional to T at constant P (and at constant n) V is directly proportional to n at constant T and P Ideal Gas Law PV = nrt P V is directly proportional to n T Standard Temperature and Pressure, STP: conditions of 0 o C (273 K) and 1 atm or 101.3 kpa (760 mm Hg); at STP, 1 mole of an ideal gas has a volume of 22.4L To solve problems using gas laws, convert each unit to those in R L atm mol K Sample Problem 1: The pressure on a balloon with a volume of 300 ml increases from 1.10 to 2.00 atmospheres (atm). Use Boyle s law to calculate the new volume of the balloon. 1. First, identify the variables. P 1 = 1.10 atm V 1 = 300 ml P 2 = 2.00 atm V 2 =? 2. Now, rearrange the equation algebraically to isolate the unknown. V 2 = P 1 x V 1 P 2

3. Insert the known values and solve. V 2 = 1.10 atm x 300mL 2.00 atm V 2 = 165 ml Sample Problem 2: The air in a balloon with a volume of 25.0 liters (L) is heated from 20 o C to 60 o C. If the pressure stays the same, what will be the new volume of the balloon? 1. First, identify the variables. V 1 = 25.0 L T 1 = 20 o C (convert o C to K by adding 273) = 293K T 2 = 60 o C = 333K V 2 =? 2. Rearrange the equation algebraically to isolate the unknown. V 2 = V 1 x T 2 T 1 3. Insert the known values and solve. V 2 = 25.0 L x 333 K 293 K V 2 = 28.4 L Sample Problem 3: A sample of gas in a 1.00 L flask at 1.50 atm contains 75.0 percent CO 2 and 25.0 percent H 2 O gas. Calculate the partial pressures of each gas. P CO2 = (0.750 x 1.50 atm) = 1.125 atm P H2O = (0.250 x 1.50 atm) = 0.375 atm Sample Problem 4: Calculate the temperature of 5.85 mol N 2 gas in a 12.0 L steel bottle under 10.0 atm of pressure. 1. First, identify the variables. P = 10.0 atm V = 12.0 L n = 5.85 mol R = 0.08206 atm/mol K T =? 2. Rearrange the equation algebraically to isolate the unknown. T = P x V n x R 3. Insert known values and solve. T = 10.0 atm x 12.0 L 5.85 mol x 0.08206 L atm/mol K T = 250 K = -23.0 o C

C.9.B Perform stoichiometric calculations, including determination of mass and volume relationships between reactants and products for reactions involving gases. Gas Stoichiometry Reactant and product relationships, including determination of mass and volume relationships, can be found for reactions involving gases. For example, suppose you want to determine the mass relationships between reactants and products for a reaction involving gases. You first convert the given mass to moles. Then you use the mole ratio from the balanced equation to calculate the number of moles of the wanted substance. Finally, you convert the moles of the wanted substance to mass. Sample Problem 1: Mass relationship between reactants and products for reactions involving gases What is the mass of oxygen gas produced when 29.2 g of water is decomposed by electrolysis according to this balanced equation? 2H 2 0 2H 2 + O 2 In order to solve this equation, you have to perform the following calculations: g H 2 O mole H 2 O mol O 2 g O 2 You need to find the mole ratio of H 2 O to O 2 using the balanced equation: 2H 2 0 2H 2 + O 2 Mole Ratio : 2 mol H 2 0 to 1 mol O 2 Can be written as: 2 mol H 2 0 OR 1 mol O 2 depending on problem 1 mol O 2 2 mol H 2 0 1. First, start with the given quantity and convert from mass to moles. 29.2 g H 2 0 x 1 mol H 2 0 18.0 g H 2 0 2. Then use your mole ratio from the balanced equation to convert moles of reactant to moles of product 29.2 g H 2 O x 1 mol H 2 0 x 1 mol O 2 18.0 g H 2 0 2 mol H 2 0 3. Finish by converting from moles of the product to mass of the product. 29.2 g H 2 O x 1 mol H 2 0 x 1 mol O 2 x 32.0 g O 2 = 26.0 g O 2 18.0 g H 2 0 2 mol H 2 0 1 mol O 2

Sample Problem 2: Volume relationships between reactants and products for reactions involving gases Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog. What volume (in L) of nitrogen dioxide gas is produced when 34 L of oxygen gas react with an excess of nitrogen monoxide? Assume conditions are at STP. 2NO + O 2 2NO 2 Recall that the following calculations need to be performed: L O 2 mol O 2 mol NO 2 L NO 2 1. First, start with the given quantity and convert from volume to moles by using the mole-volume ratio. 34 L O 2 x 1 mol O 2 22.4 L O 2 2. Then, convert from moles of reactant to moles of product by using the correct mole ratio from balance equation. 34 L O 2 x 1 mol O 2 x 2 mol NO 2 22.4 L O 2 1 mol O 2 3. Finish by converting from moles of the product to liters of the product. 34 L O 2 x 1 mol O 2 x 2 mol NO 2 x 22.4 L NO 2 = 68 L NO 2 22.4 L O 2 1 mol O 2 1 mol NO 2 C.9.C Describe the postulates of kinetic molecular theory Kinetic Molecular Theory Postulates of kinetic molecular theory perfectly describe an ideal gas: 1. Gases are made of molecules or atoms in constant, random motion 2. The volume of these molecules or atoms is extremely small. 3. Collisions among these molecules, atoms, and the container walls are elastic (no net energy loss or gain) 4. Attractive or repulsive forces among these molecules or atoms are extremely small 5. The average kinetic energy of these molecules or atoms is proportional to the absolute temperature of the gas in Kelvin (K).

C.11.A Understand energy and its forms, including kinetic, potential, chemical, and thermal energies Energy and Its Forms What is energy? Energy is the capacity to do work or produce heat. Energy does not have mass or volume. Units called joules (J) are used to measure energy. A common non-si unit of energy is the calorie. One calorie (cal) is the quantity of heat that raises the temperature of 1 g of water by 1 degree Celsius. The conversion relationships for joules and calories are: 1 J = 0.2390 cal 1 cal = 4.184 J Energy exists in different forms: Form Description Example Potential Stored energy Rock resting on a hill Kinetic Energy of motion Rock rolling down a hill Chemical Potential energy stored in Gasoline in car Thermal chemical bonds Kinetic energy of atoms or molecules motion A toaster transforms electrical energy to thermal energy Other forms of energy: light, heat, sound, mechanical motion, nuclear energy, gravitational potential energy, mechanical potential energy. Energy is always being transformed from one form to another. Some energy changes involve single transformations, while others involve many transformations. A toaster performs a single transformation as it transforms electrical energy to thermal energy. In a car engine, a series of transformations are required to make the car move. The chemical energy in the gasoline is ignited, producing light, heat, and sound in the engine. The ignition of the gasoline also pushes a piston, which makes the car move, resulting in mechanical motion. C.11.B Understand the law of conservation of energy and the processes of heat transfer Conservation of Energy and Heat Transfer Law of Conservation of Energy: Energy cannot be created or destroyed in an ordinary chemical reaction, but the form of energy can change. Example: A campfire transforms energy stored in the wood s chemical bonds (Potential energy) into thermal energy (kinetic energy), causing molecules in the air near the fire to move faster (temperature increases). Heat can be transferred 3 ways: 1. Conduction: objects in contact, like a warm hand melting ice 2. Convection: fluid motion or currents, like warm air rising 3. Radiation: electromagnetic waves, like those moving through empty space from the sun or like those made in a kitchen microwave.

C.11.C Use thermochemical equations to calculate energy changes that occur in chemical reactions and classify reactions as exothermic or endothermic. Energy Changes in Chemical Reactions Standard state: reference state for a substance (typically 1 atm, 25 o C or 298 K, and 1 mole); defining a property at standard state can allow it to be intensive (independent of size); indicated by a superscript of 0. Standard molar enthalpy of formation, H o f : difference in energy that occurs when one mole of a substance is formed from its pure elements (all at standard state conditions); H o f for any element = 0. means change or difference Enthalpy of reaction, H: energy released or absorbed during a chemical reaction. At standard state, H = H o f (products) H o f (reactants), where means sum of all (multiply each H o f by its coefficient. Reaction Enthalpy Change Description Exothermic H < 0 (negative) Heat is released Endothermic H > 0 (positive) Heat is absorbed Sample Problem 1: Calculate the change in energy for the following reaction at standard conditions. Is this reaction endothermic or exothermic? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) Substance H o f (kj/mol) CH 4-74.8 O 2 0 CO 2-393.5 H 2 O -241.8 Use the following formula to calculate the change in energy: H = H o f (products) H o f (reactants) H o f (products) = (1 mol)( -393.5 kj/mol) + (2 mol)( -241.8 kj/mol) = -877.1 kj H o f (reactants) = (1 mol)( -74.8 kj/mol) + (2 mol)(0 kj/mol) = -74.8 kj H = - 877.1 kj ( -74.8 kj) = - 802 kj (EXOTHERMIC, because H < 0)

Sample Problem 2: Calculate the change in energy for the following reaction at standard conditions. Is this reaction endothermic or exothermic? CaCO 3 (s) CaO (s) + CO 2 (g) Substance H o f (kj/mol) CaCO 3-1207 CaO -635.5 CO 2-393.5 Use the following formula to calculate the change in energy: H = H o f (products) H o f (reactants) H o f (products) = (1 mol)( -635.5 kj/mol) + (1 mol)( -393.5 kj/mol) = -1029 kj H o f (reactants) =(1 mol)( -1207 kj/mol) = -1207 kj H = -1029 kj (-1207 kj) = 178.0 kj (ENDOTHERMIC, because H > 0) C.11.D Perform calculations involving heat, mass, temperature change, and specific heat. Heat, Mass, Temperature Change, and Specific Heat Specific Heat: Specific heat is an intensive property. It does not vary with the amount of the substance. The specific heat of a substance is the amount of heat needed to raise the temperature of one gram of the substance by one degree Celsius. The unit of specific heat is joules per gram-degree Celsius (J/g o C). The equation for specific heat is: Q = mc p T, Q represents the heat input, m is for mass, c p is for specific heat, and T is for the change in temperature. Sample Problem 1: Given that c p of copper = 0.385 J/ g o C, calculate the heat absorbed by a 0.020 kg piece of copper metal that is heated from 25 o C to 125 o C. Note: Convert to common units so units will cancel (change kg to g). So, 0.020 kg = 20 g Q = mc p T = (20 g)( 0.385 J/ g o C)( 125 o C - 25 o C) = 770 J

Sample Problem 2: The specific heat of water is 4.179 J/ g o C. A 1,200 g water sample at 19 o C loses 10kJ of heat. What is the final temperature? Note: Convert to common units so units will cancel (change kj to J). So, 10kJ = 10,000 J T = Q = -10,000 J = -2 o C mc p (1,200 g)( 4.179 J/ g o C) (Q is negative since energy flows out of system) T = T final - T initial, so T final = T + T initial = -2 o C + 19 o C = 17 o C C.11.E Use calorimetry to calculate the heat of a chemical process Calorimetry The Science of Measuring Heat Flow Calorimeter: tool used to measure heat of a chemical process ( H); heat flows into a substance with known c p (specific heat) and the temperature change is recorded: H = -Q ; where Q = mc p T A calorimeter usually contains a carefully measured mass of a substance of known specific heat, such as water. As the reaction absorbs or releases energy, the temperature of the water will change. From that temperature change, the energy that the water absorbed or released can be calculated. Energy released by the reaction = Energy absorbed by the solution Sample Problem 1: A student is using a calorimeter to measure the molar heat of salvation of calcium chloride (CaCl 2 ). The calorimeter contains 100.0 g of water at 25.1 o C. After 10.0 g of CaCl 2 is fully dissolved, the temperature of water rises to 42.7 o C. (Hint: Remember that the specific heat of water is 4.18 J/g o C. 1. First, write the process as a chemical equation: CaCl 2 (s) Ca 2+ (aq) + 2Cl - (aq) H = H solution 2. Then calculate the temperature change of the water. T = 42.7 o C - 25.1 o C = 17.6 o C 3. Next, calculate the energy absorbed by the water using the following equation. Q water = mass water x c pwater x T Q water = (100.0 g) (4.18 J/ g o C) (17.6 o C) Q water = 7360 J Q water is positive, indicating that the water absorbs energy from the reaction. In a perfect calorimeter, Q water equals the amount of energy released by the dissolving of calcium chloride

However, Q water is the energy released when 10.0 g CaCl 2 dissolves, not 1 mol CaCl 2. To calculate the heat per mole, first divide 10.0 g CaCl 2 by the molar mass of CaCl 2, which is 110.98 g/mol. 10.0 g CaCl 2 x 1 mol CaCl 2 = 0.0901 mol CaCl 2 110.98 g CaCl 2 Then divide the heat change of the water by 0.0901 mol CaCl 2. Notice that because the heat change of the water is positive, the heat change of forming the solution is negative. H solution = -Q water / number of mol CaCl 2 H solution = -7360 J/0.0901 mol CaCl 2 H solution = -81700 J/mol CaCl 2 H solution = -87.7 kj/mol CaCl 2