Study Guide for Exam 3
Study Guide for Exam 3 CHEM 120
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This 4 page Study Guide was uploaded by Leslie Pike on Saturday October 31, 2015. The Study Guide belongs to CHEM 120 at Western Kentucky University taught by Dr. Darwin Dahl in Summer 2015. Since its upload, it has received 85 views. For similar materials see College Chemistry I in Chemistry at Western Kentucky University.
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Date Created: 10/31/15
Study Guide for Exam 3 to be administered November 4th 2015 Chapter 5 Gases Boyle39s law PiViPfo Charles law ViTiVfTf Note TEMPERATURE MUST BE IN KELVINS FOR THIS TO WORK Avogadro s law ViniVfnf Combined gas law PiViniTiPfonfo ldeal gas law PVnRT where R00821LatmmolK STP0C and 1 atm Volume of a gas at STP is 22414Lmol Modi ed ideal gas law PmmrhoRT where mmmoar mas gmol and rhodensity gL Partial pressures PiXiPt Example problem sample of natural gas at 137 atm Contains 824 mol methane 0421 mol ethane 0116 mol propane What is the partial pressure of the propane xpmpaneo116824o421o116o0132 Ppropane00132137atm00181atm Note the ideal gas law is PVnRT At normal temperatures and pressures ALL gases obey this law Gases deviate from this law at extremely high pressures and low temperatures Also this law no longer applies once a gas becomes a liquid We assume for all problems worked in class that a the gas does not deviate from this law and b the gas does not become a liquid Sample problem You have a 15 g impure sample of potassium chlorate You heat it and it decomposes into potassium chloride and oxygen You collect 30 mL of oxygen gas over water The temperature is 20 C The pressure is 730 mmHg Calculate the percentage of potassium chlorate in the sample Balanced reaction 2KCO3 D 2KC 302 First realize that this is a stoichiometry problem You need to nd the number of moles of oxygen gas that you collected so that you can nd the number of moles of potassium chlorate that you originally had Convert this to grams and divide by the grams of sample to get the percentage To nd the number of moles of oxygen use the ideal gas equation PVnRT Solving for n gives us nPVRT To determine the pressure of the oxygen remember that your container will have a combination of oxygen and water vapor because you collected over water use the law of partial pressures PtPoxygen Pwater Pwater can be found on the table that Dr Dahl will provide you with 730Poxygen Poxygen7125 Now we can plug and chug Remember that pressure must be in atm volume must be in L and temperature must be in K 7125 xomo quot2 0082127320 0001169 mol 02 The rest is stoichiometry 022m01 KCl 03 122g KClO3 3mol 02 000169 mol 0095g KCZO3 mol KCl 03 Lastly nd the mass percentage 00951563 Gas equations Equation for velocity of a gas v3RTM Here R8314 Mmoar mas in KILOGRAMS per mole NOT GRAMS Ttemperature in Kelvin Kinetic energy The kinetic energy equation is 05mv2 This equation itself is not important for Chem 120 From this is derived the equation used to compare the velocities of two gases 2 m2 V2 E1 The kinetic energy of a gas is directly proportional to its temperature KE1T1 KE2T2 Therefore to get the relationship the kinetic energy of gas 1 and the kinetic energy of gas 2 divide the temperature of gas 1 by the temperature of gas 2 The rateofeffusion equation that compares two different gases is similar to the velocitymass relationship Rate 1 molar mass 2 Rate 2 molar mass 1 For example say it takes 2 liters of hydrogen gas molar mass 2 gmol 5 minutes to escape from a container How long would it take 3 liters of helium gas molar mass 4 gmol to escape from the same container The hydrogen escapes at a rate of 2L5min or 04Lmin The helium is three liters the number of minutes is unknown Filling in the equation 04Lmin3LX minsqrt4gmol2gmol Three liters of helium would take 106 minutes to escape the container Ch 6 Thermochemistry The total heat change of the system is equal but opposite to the total change of the surroundings For example when a system gains 100 k the surroundings lose 100 k and vice versa The total energy change of a system is equal to the heat exchange plus the amount of work done Memorize this equation Work Force Distance In the case of an expanding gas Work Outside pressure change in volume Important Concept Chemical equations can be written to contain a heat change along with the moles of products and reactants Saying that a reaction requires 100 k of heat is NO DIFFERENT from saying that a reaction requires ve moles of oxygen gas or 2 moles of sodium chloride When doing stoichiometry problems the heat is treated just like the compounds are treated Additionally the heat can be written outside of the equation as a delta H at the end If delta H is negative this means that the heat belongs on the PRODUCTS side If delta H is positive the heat belongs on the REACTANTS side When a chemistry problem says that heat is quotevolvedquot this means that the heat is given off and therefore belongs on the REACTANTS side of the equation The speci c heat of a substance is the amount of heat required to raise one gram of that substance one degree Celsius or Kelvin it makes no difference The speci c heat of water is 4184 Jg K YOU MUST MEMORIZE THIS ONE Dr Dahl will provide all other speci c heats The heat capacity of a substance is the heat required to raise a given amount of a substance 1 degree C The standard heat of formation of a substance is the delta H when one mole of that substance is formed from its elements at 1 atm THE STANDARD HEAT OF FORMATION OF A SUBSTANCE IN ITS ELEMENTAL STATE IS ALWAYS ZERO This is because no heat change takes place the substance is already in the elemental state The standard enthalpy of a reaction is calculated as follows sum of delta H of productssum of delta H of reactants Adding equations Chemical equations can be added This is quite useful when determining the delta H of a reaction in which none of the reactants are in their elemental state In a typical problem you will be given two or three equations and a quotfinalquot equation Yourjob is to cancel terms in the equations so that they add to the quotfinalquot equation Equations can be multiplied by a constant in this case delta H is also multiplied by a constant Remember delta H is the SAME as the moles of a reactant or product Equations can be reversed For instance if heat was on the product side it would now be on the reactant side causing a sign change in delta H Once you have multiplied andor reversed the equations so that terms not included in the quotfinalquot equation cancel out the delta Hs will add to give you the nal delta H Dr Dahl has very nice examples in the powerpoints
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