CHEM 1441 1st EditionExam # 3 Study Guide Lectures: 20 - 26Lecture 20 (March 16) Kinetic Molecular Theory at a certain temperature, each gas has the same average (avg) kinetic energy Kinetic Energy (avg) = (1/2)mass(velocity(avg))2 ; KE=(1/2)mv2 Each gas has to have a different velocity in order to calculate the average velocity For 1 gas: low temp, slower High temp, faster KE(avg)=(3/2)RT ; R is a gas constant As KE increases, temperature increases {direct relationship} Graham’s Law of Effusion The lighter the gas, the faster the rate of effusion As the rate of effusion increases, the molar mass decreases {inverse relationship} As the time of effusion increases, the molar mass increases r1 = M2 r is rate, M is molar mass r2 √ M1 Thermodynamics/Thermochemistry Energy is the capacity to do work Heat energy is temperature difference Work energy is a force acting through a distance- Work=Pressure∆Volume Work is negative if it takes energy out of the system Work is positive if it adds energy to the system Energy is possed by everything and relates to both heat and work Heat (q) Energy (E) Work (w)  Chemical Energy Have to define the system and surroundings As reaction occurs, the temperature goes up. The thermometer ALWAYS measures the heat of the SURROUNDINGS.O2 gas is heavier and slowerHe gas is lighter and fasterVelocityNumber of molsthermometerReaction 1st Law of Thermodynamics The total energy of the universe is constant; it can be neither created or destroyed Can’t directly measure E, but can measure ∆E (Energy final – Energy initial) Anytime you see delta in front (∆Temperature, ∆Enthalpy, ∆Volume) it indicates a state function Units to memorize! 1 cal = 4.184 J 1 Cal = 1000 cal = 1kcal How heat, work, and energy are related E = q + w - E is total energy of the system- q is the heat that is transferred- w is the amount of work done on systemLecture 21 (March 18) Enthalpy (H) measues how much energy is in the system altogether- It is the sum of internal energy and the product of a system’s volume @ constant pressure q = mC∆T J = g(J/ g . oC)∆oC- q is heat- m is mass in g- C is specific heat capacity in J/ g . oC- ∆T is change in temp. in oC CONCEPT CHECK- How much heat is absorbed by a 1 gram penny (Cs (specific heat capacity) = 0.385J/ g . oC) to warm it from -8 oC to 37 oC? q = mC∆T = 1g (0.385 J/ g . oC) (37-(-8)) = 17.325 J q is positive because the object is absorbing the heat (it gained heat; endothermic) If q is transferred from:- System Surroundings q is exothermic; system loses heat- System Surroundings q is endothermic; system gains heat- Always determine q by looking at it from the systems perspective  CONCEPT CHECK- A system releases 622 kJ of heat and does 105 kJ of work on the surroundings. What is the change in internal energy for the system?Surroundings∆E = q + w heat is released, so q is negative = -622 kJ + (-105 kJ) work is done ON surrdngs(energy lost), so w is neg = -727 kJ Enthalpy (H)- ∆H is the change in amount of heat energy per mol- ∆H = q/mol- CONCEPT CHECK A student dissolved 1 gram NaCl in 50 grams of water at 25oC. The temperature drops to 23oC. What is the ∆H per mol of NaCl for this process?NaCl(s)  NaCl (aq) ∆H = q/mol ; q = mC∆T∆H = mC∆T/mol = [51g (4.184 J/g.oC)(23-25)oC ] / (1g NaCl/58.45g/mol) 51g comes from 1g NaCl + 50g water = 24944.5 J  24.9 kJ temp of surrounding decreased, rxn is positiveLecture 22 (March 20) CONCEPT CHECK heat transfer problem- A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g of water at 22.7oC. When the substances reach thermal equilibrium, the final temperature is 24.2oC. What is the mass of the copper block? Cs=0.385 J/g.oC -q lost = q gained -q lost by Cu block = q gained by H2O Cs water = 4.18 J/g.oC mC∆T = mC∆T m(0.385)(24.2-65.4) = 95.7 (4.18)(24.2-22.7) m Cu = 600.61/15.86 = 37.86 g Pressure-Volume Work W=P∆V USE EQUATION ONLY IN COSTANT PRESSURE Units: P – atm , V – liters, W – Joules  To convert atm, multiply it by (8.31446 J/0.0802574 L) or 101.3 J CONCEPT CHECK If you inflate a balloon from vol. of 0.100 L to 1.85 L against an external pressure of 1 atm (constant pressure), how much work is done? W = P∆V = (1 x 101.3)(1.85-1) = 86.1 J CONCEPT CHECK Mg metal reacts with Hydrochloric acid according to the following reation: Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)In an experiment to determine the enthalpy change ∆H, We combine 0.158g Mg with enough HCl to make a 100mL solution in a coffee-cup calorimeter. Temperature rises from 25.6 oC to 32.8oC. Use the density of the solution at 1.00g/mL and C=4.184 J/g.oC. Find the change in enthalpy per mol of Mg. ∆H=q/mol ; q= mC∆T q is negative because the temp went up, q = (100g)(4.184)(32.8-25.6) = 3012.48 J heat went out to surroundings; exothermic ∆H = -3012.48/((0.158g/24.3g/mol)) = -463311.79 J  -4.63x10^5 J/mol Mg Enthalpy ∆H = ∆E + ∆(PV) ; ∆H = ∆E + ∆PV + P∆V  Coffee-cup equation when pressure is constant: ∆H = ∆E + P∆V Bomb calorimeter when volume is constant: ∆H = ∆E + ∆PV ∆E = q = C(heat capacity of calorimeter)x∆T CONCEPT CHECK When 1.010g of sucrose (C12H22O11) undergoes combustion in a bomb calorimeter, the temperature rises from 24.92oC to 28.33oC. Find ∆E per mol of sucrose. C=4.9 kJ/oC (∆E)/mol