CHEM 1125Q 1st EditionExam # 1 Study Guide Lectures: 1 - 9Lecture 1 (January 21)EnergyIn system 1, the change in enthalpy is 300kJ, while in system 2, the change in enthalpy is -250kJ. Are the system reactions endothermic or exothermic? How do you know?System 1:- Since the ΔH is positive, that means that energy was absorbed by the system, making it endothermic.System 2:- Since the ΔH is negative, that means that energy was released by the system, making it exothermic.ThermochemistryThe energy produced by a reaction is measured as 3450J. How many calories is this?-Since we know that there are 4.184J for every 1 calorie, we can solve this by calculating:3450J (1 calorie/4.184J) = 824.6 calories when rounded to 4 significant figuresThermodynamicsWhat are the 3 types of systems and how does each one work?Open- In an open system, both mass and energy are able to be exchanged between the system and its surroundings.Closed- Unlike the name implies, the closed system is not the one that is unable to exchange anything with its surroundings. Instead, a closed system is able to exchange energy but not mass with its surroundings.Isolated- An isolated system is completely cut off from its surroundings, and therefore does not exchange mass or energy with its surroundings.Work and HeatA certain system absorbs 218kJ of heat energy from its surroundings and does 126kJ of work. What is the change in internal energy for the system?- First of all, we will need to use the equation: Δ U = q + w. Since the system is absorbing 218kJ of heat, we know that q will be positive because it is endothermic. Since 126kJ of work is being done BY the system, we know that w will be negative because the system is releasing that energy. Then we solve:Δ U = 218kJ + (-126kJ) Δ U = 92kJLecture 2 (January 26) Pressure and VolumeA sample of gas expands from 805mL to 1204mL at a constant temperature and at a pressure of1.35 atm. How much work (in J) is being done? (Hint: 1L x atm = 101.3J)- We know that the equation for work is: w = –PΔV. First we will need to find the change in volume of the gas in L:1204mL – 805mL = 399mL399mL (1L/1000mL) = 0.399LNow we can solve for work:w = -1.35atm x 0.399Lw = -0.539L x atm-0.539L x atm (101.3J/1L x atm) = -54.6JCalorimetryHow much heat (in kJ) is released in a reaction where 500mL water is heated from 19°C to 89°C? (Remember that the specific heat of water is 4.184J/g x °C)- We will be using the equation: q = smΔT. First, we need to solve for the mass of water. Since we know that the density of water is 1g/mL, we can solve it like this:500mL (1g/mL) = 500gThen we need to solve for the temperature change:89°C - 19°C = 70°CNow we can solve for q:q = 4.184 x 500 x 70q = 146,440J146440J (1kJ/1000J) = 146.4kJLecture 3 (January 28)Constant Volume CalorimetryIn a calorimetry experiment, the q of the water is found to be 4.541kJ while the q of the reactionis measured to be -6.397kJ. Find q for the bomb.- We will be using the equation: qbomb + qwater + qreaction = 0. Since we are given the q of water and the reaction, all we have to do is find the q of the bomb:qbomb + 4.541kJ + -6.397kJ = 0qbomb = 1.856kJLecture 4 (February 2)GasesState 5 characteristics of gases.- 1: Gases take the shape and volume of their container. 2: A gas’s density is much lower than that of its corresponding phases (liquid and solid). 3: A gas’s density depends on its environment’s temperature and pressure. 4: Gases can easily be compressed because of their low densities. 5: A gas can easily form mixtures with other gases.Molecular SpeedWhat is the difference between diffusion and effusion?Diffusion- Diffusion is when gas molecules move from higher pressure to lower pressure to avoid collisions.Effusion- Effusion on the other hand is when gas molecules escape from a hole in a container into a vacuum.Graham’s LawAn unknown diatomic gas moves at a rate that is 1.17 times as fast as CO2. What is the unknown gas?- For this problem we must use the equation: (urms1 / urms2) = √(Μ2/Μ1). We know that the mass of is 44.01g, so all we have to do is solve:1.17 = √(44.01g/Μ1)Μ1 = 32.15gTherefore our gas is oxygen, O2Lecture 5 (February 4)Gas LawsName and provide an equation for each of the 4 gas laws.Boyle’s Law- At a constant temperature and moles, a gas’s pressure is inversely proportional to its volume.V ∝ 1/PCharles’s Law- At a constant pressure and moles, a gas’s volume is directly proportional to its temperature.V ∝ TAvogadro’s Law- Equal volumes of gas, when at the same temperature and pressure, have the same number of moles.V ∝ nCombined Gas Law- This law is used when any/all variables changeP1V1 /n1T1 = P2V2 /n2T2Ideal Gas Equation3.6 moles of a gas are heated in a 20L container at 45°C. What is the pressure of the gas? (Remember, the gas constant R = 0.08206 (L x atm)/K)- We will be using the ideal gas equation: PV=nRT. First we have to convert the 45°C to Kelvin:45°C + 273.15K = 318.15KSince we are given every value except for the pressure, all we have to do is solve:P= (3.6 x 0.08206 x 318.15)/20P= 4.7atmLecture 6 (February 9)Dalton’s LawDefine Dalton’s Law of partial pressure.- Partial pressure means that when 2 or more gases occupy the same space, each exerts its own unique pressure, independent of the other gases around it. When added together, these partial pressures make up the total pressure of the gaseous mixture.Mole FractionsA tank contains 21.4g O2 and 31.7g CO2 at 21°C. The total pressure of the mixture is 5.65atm. What are the partial pressures of each gas?- We know that Χi = ni / nTotal and Χi = Pi / PTotal . First we need to find the total number of moles ofthe mixture:21.4g O2 (1 mol/32g) = 0.669mol O231.7g CO2 (1 mol/44.01g) = 0.720mol O2 0.669 + 0.720 = 1.389mol = nTotal Then we need to solve for X of each gas:ΧO2 = 0.669/ 1.389 = 0.482ΧCO2 = 0.720 / 1.389 = 0.518Now we can substitute X in and solve for the partial pressures:PO2 = 0.482 x 5.65 = 2.72atmPCO2 = 0.518 x 5.65 = 2.93atmLecture 7 (February 11)Surface TensionDistinguish between cohesion and adhesion.Cohesion- Cohesion is an attraction between like molecules. For example, water molecules are polar and attract to each other.Adhesion- Adhesion is an attraction between unlike molecules. For example, when little drops of water cling to your glass after you dump the water out, the water molecules are attracted to the glass molecules.Clausius-Clapeyron EquationMethanol boils at