AP Chemistry Study Guide 6 v Evaporation vs. condensation Ø Vaporization and condensation are opposite processes Ø In an open container the vapor

Transcription

1 AP Chemistry Study Guide 6 v Evaporation vs. condensation Ø Vaporization and condensation are opposite processes Ø In an open container the vapor molecules generally spread out faster than they can condense Ø The net result is that the rate of vaporization is greater than the rate of condensation and there is a net loss of liquid Ø However, in an enclosed container the vapor is not allowed to spread out infinitely Ø The net result in a closed container is that at some time the rates of vaporization and condensation will be equal v Effect of intermolecular attraction on vaporization and condensation Ø The weaker the attractive forces between molecules the less energy they will need to vaporize Ø Also weaker attractive forcer means that more energy will need to be remover from the vapor molecules before then can condense Ø The net result will be more molecules in the vapor phase and a liquid that evaporates faster the weaker the attractive forces the faster the rate of evaporation Ø Liquids that evaporate easily are said to be volatile Ø Liquids that do not evaporate easily are said to be nonvolatile v Energetics of vaporization Ø When the high energy molecules that are lost form the liquid it lowers the average kinetic energy Ø If the energy is not drawn back into the liquid its temperature will decrease, therefore vaporization is an endothermic process Condensation is an exothermic process Ø Vaporization requires input of energy to overcome the attraction between the molecules v Heat of vaporization Ø The amount of heat energy is required to vaporize one mole of the liquid is called the heat of vaporization, change in Hvap Sometime called the enthalpy of vaporization Ø It is always endothermic therefore change in Hvap is positive Ø It is somewhat tempaerature dependent Ø Change in Hcondensation = - change in Hvap v Heating curve of a liquid Ø As you heat a liquid its temperature increases linearly until it reaches the boiling point Q = mass * Cs * change in T Ø Once the temperature reaches the boiling point all the added heat goes into boiling the liquid, the temperature stays constant Ø Once all the liquid had been turned into gas, the temperature can once again start to rise

2 v Melting = fusion Ø As a solid is heated its temperature rises and the molecules vibrate more vigorously Ø Once the temperature reaches the melting point the molecules have sufficient energy to overcome some of the attractions that hold them in position and the solid melts (or fuses) Ø The opposite of melting is freezing v Heating curve of a solid Ø As you heat a solid its temperature increases linearly until it reaches the melting point Ø Once the temperature reaches the melting point all the added heat goes into melting that solid Temperature stays constant Ø Once all the solid has been turned into liquid the temperature can again start to rise Ice/water always have a temperature of 0 degrees Celsius at 1 atm v Energies of melting Ø When the high energy molecules are lost form the solid it lowers the average kinetic energy Ø If energy is not drawn back into the solid its temperature will decrease therefore melting is an endothermic process and freezing is an exothermic process Ø Melting requires an input of energy to overcome the attractions among molecules v Heat of fusion Ø The amount of heat energy required to melt on moel fo the solid is called the heat of fusion, change Hfus

3 Ø It is always exothermic therefore change in Hfus is positive Ø Change in Hcrystalization = - change in Hfusion Ø Generally much less than change in Hvap Ø Change in Hsublimination = change in Hfusion + change Hvaporization v First law of thermodynamics Ø Energy cannot be created or destroyed Total energy in the universe cannot change You can transfer form one place to another Ø Change Euniverse = 0 = change Esystem + change in Esurroundings Ø Conservation of energy Ø Fro an exothermic reaction lost heat from the system foes into the surroundings Ø There are two ways that energy is lost from the system Converted to heat, q Used to do work w Ø Energy conservation requires that the energy change in the system is equal to the heat released plus work done Change in E = q + w Change in E = change in H + P * change in V Ø Change in E is a state function Internal energy change independent of how done v The energy tax Ø To recharge a battery with 100 kj of useful energy will require more than 100 kj because of the second law of thermodynamics Ø Every energy transition results in a loss of energy An energy tax demanded by nature Conversion of energy to heat which is lost by heating up the surroundings Ø Fewer steps generally results in a lower total heat tax v Thermodynamics and spontaneity Ø Thermodynamics predicts whether a process will occur under the given conditions Process that will occur are called spontaneous Nonspontaneous processes require energy input to go Ø Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction If the system after reaction has less potential energy than before the reaction the reaction is thermodynamically favorable Ø If a process is spontaneous THAT DOES NOT MEAN IT WILL OCCUR QUICKLY v Reversibility of a process Ø Any spontaneous process is irreversible because there is a net release of energy when it proceeds in that direction It will proceed in only one direction Ø A reversible process will proceed back and forth between the two end conditions Any reversible process is at equilibrium

4 This results in no change in free energy Ø If a process is spontaneous in one direction it must be nonspontaneous in the opposite direction Ø The direction of spontaneity can be determined by comparing the potential energy of the system at the start and at the end v Spontaneous process Ø Spontaneous processes occur because they release energy from the system Ø Most spontaneous processes proceed from a system of higher potential energy to a system at lower potential energy Exothermic Ø But there are some spontaneous processes that proceed from a system of lower potential energy to a system at higher potential energy Endothermic v Melting ice Ø When a solid melts the particles have more freedom of movement Ø More freedom of motion increases the randomness of the system when systems become more random energy is released we call this energy entropy Ø Melting is an endothermic process yet ice will spontaneously melt above 0 degrees Celsius v Factors affecting whether a reaction is spontaneous Ø There are two factors that determine whether a reaction is spontaneous they are the enthalpy change and the entropy change of the system Ø The enthalpy change, change in H is the difference in the sum of the internal energy and the PV work energy of the reactants to the products Ø The entropy change, change in S is the difference in randomness of the reactants compared to the products v Enthalpy change Ø Change in h is generally measured in kj/mol Ø Stronger bonds = more stable molecules Ø A reaction is generally exothermic if the bonds in the products are stronger than the bonds in the reactants Exothermic = energy released, change in H is negative Ø A reaction is generally endothermic if the bonds in the products are weaker than the bonds in the reactants Endothermic = energy absorbed change in H is positive Ø The enthalpy change is favorable for exothermic reactions and unfavorable for endothermic reactions v Entropy Ø The thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S S generally J/mol Ø S = k ln W K = Boltzmann constant = 1.38*10-23 J/K W is the number of energetically equivalent ways a system can exist Unit- less Random systems require less energy than ordered systems

5 v W Ø These are energetically equivalent states fot he expansion of a gas Ø It doesn t matter in terms of potential energy whether the molecules are all in one flask of evenly distributed Ø But one of these states is more probable than the other two v Macro states à microstates Ø These macro- state can be achieved through several different arrangements of the particles Ø The macro- state with the highest entropy also has the greatest dispersal of energy v Changes in Entropy, change in S Ø Change in S = Sfinal Sinitial Ø Entropy change is a favorable when the result is a more random system Change in S is positive Ø Some changes that increase the entropy are as follows: Reactions whose products are in a more random state

6 Solid more ordered than liquid, liquid more ordered than gas Reactions that have larger numbers of product molecules than reactant molecules Increase in temperature Solids dissociating into ions upon dissolving v Change in S Ø For a process where the final condition is more random than the initial condition change in Ssystem is positive and the entropy change is favorable for the process to be favorable for the process to be spontaneous Ø For a process where the final condition is more orderly than the initial condition, change in Ssystem is negative and the change is unfavorable for the process to be spontaneous Change Ssystem = change Sreaction = summation moles(s O products) summation moles(s o reactants) v Entropy change in state change Ø When materials change state the number of macro- states can have changes as well The more degrees of freedom the microstates have, the more macro- states are possible Solids have fewer macro- states than liquids which have fewer macro- states than gases Ø The second law of thermodynamics The second law of thermodynamics says that the total entropy change of the universe must be positive for a process to be spontaneous For reversible process change in Srnx = 0 For irreversible (spontaneous) process change in Srn > 0 Change in Suniverse = change in Ssystem + change in Ssuroundings The entropy of the system decreases then the entropy of the surroundings must increase by a large amount When change in Ssystem is negative change in Ssurroundings must be positive and big for a spontaneous process v Heat flow, energy and the second law Ø When ice is placed in water heat flows from the water into the ice Ø According to the second law heat must flow from water to ice because it results in more dispersal of heat the entropy of the universe increases v Heat transfer and changes in entropy of the surroundings Ø The second law demands that the entropy of the universe increases for a spontaneous process Ø Yet processes like water vapor condescending are spontaneous even though the water vapor is more random than the liquid water Ø If the process is spontaneous yet the entropy change of the process is unfavorable there must have been a large increase in the entropy of the surroundings Ø The entropy increase must come form heat released by the system the process must be exothermic

7 v Entropy change in the system and in the surroundings Ø When the entropy change in a system is unfavorable (negative) the entropy change in the surroundings must be favorable (positive) and the large in order to allow the process to be spontaneous v Heat change and change in Ssurroundings Ø When a system process is exothermic it adds heat to the surroundings increasing the entropy of the surroundings Ø When a system process is endothermic it takes heat from the surroundings decreasing the entropy of the surroundings Ø The amount of entropy of the surroundings changes depends on the original temperature The higher the original temperature the less effect addition or removal of heat has v Temperature dependence of change of Ssurroundings Ø When heat is added to surroundings that are cool is has more of an effect on the entropy than it would have if the surroundings were already hot Ø Water freezes spontaneous blow 0 degrees C because the heat released on freezing increases the entropy of the surroundings enough to make S positive Above 0 degrees Celsius the increase in entropy of the surroundings is insufficient to make S positive v Qualifying entropy change in surroundings Ø The entopry change in the surroundings is proportional in the amount of heat gained or lost qsurroundings = - qsystem Ø The entropy change in the surroundings is also inversely proportional to the temperature Ø At constant pressure and temperature the overall relationship is Change in Ssurroundings = (- qsystem/t) = - change in Hsystem/T v Gibbs free energy and spontaneity Ø It can be shown that T*change in Suuniverse = change in Hsys * T *change in Ssys Ø The Gibbs free energy G, is the maximum amount of work energy that can be released to the surroundings by a system for a sonstant temperarure and pressure system Gibbs free energy is often called the chemical potential because it is ananlogous to the storping of energy in a mechanical system Ø Change in Gsys = change in Hsys*T*change in Ssys Ø Because change in Suniverse determines if a process is spontaneous change in G also determines spontaneity Change in Suniverse is positive when spontaneous so change in G is negative Ø A process will be spontaneous if change in G is negative Change in G will be negative when Change in H is positive and change in S is positive Exothermic and more random Change in H is negative and large and change in S is negative and small

8 Change in H is positive but small and change is S is negative and large Or high temperature Change in G will be positive when change in H is positive and change in S is negative Never spontaneous at any temperature When change in G is = 0 the reaction is at equilibrium v Standard conditions Ø The standard state is the state of a material at a defined set of conditions Ø Gas = pure gas at exactly 1 atm pressure Ø Solid or liquid = pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest Usually 25 degrees Celsius Ø Solution = substance in a solution with concentration of 1 M v The third law of thermodynamics Ø The absolute entropy of a substance is the amount of energy it has due to dispersion of energy through the particles Ø The third law states that for the perfect crystal at absolute zero the absolute entropy = 0 kj/mol * K Therefore every substance that is not a perfect crystal is absolute 0 has the same energy of the entropy Therefore the absolute entropy of substances is always positive v Standard absolute entropies Ø Extensive Ø Entropies for 1 mole of a substance at 298 K for a particular state a particular allotrope a particular molecular complexity a particular degree of dissolution v Relative standard entropies: states Ø The gas state has a larger entropy than the liquid state at a particular temperature Ø The liquid state has a larger entropy than the solid state at a particular temperature v Relative standard entropies: Molar Mass Ø The larger the molar mass the larger the entropy Ø Available energy states are more closely spaced allowing more dispersal of energy through the states v Relative standard entropies: allotrope Ø The less constrained the structure of an allotrope is, the larger the entropy Ø The fact that the layers of graphite are not bounded together makes it less constrained v Relative standard entropies: molecular complexity Ø Larger more complex molecules generally have larger entropy Ø More energy states are available allowing more dispersal of energy through the states v Relative standard entropies: entropies dissolution Ø Dissolved solids generally have larger entropy distributing particles throughout the mixture

9 v The standard entropy change, change in S Ø The standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions Ø Change in S o reaction = (summation of moles of S o products) (summation of moles S o reactants) Ø Remember although the standard enthalpy of formation change in H o of an element is 0 kj/mol the absolute entropy at 25 degrees Celsius change in S o is always positive v Standard free energies of formation Ø The free energy of formation (change in Gf o ) is the change in free energy when 1 mole of a compound forms from its constituent elements in their standard state Ø The free energy of formation of pure elements in their standard state is zero v Change in G relationships Ø If a reaction can be expressed as a series of reactions the s of the change in G values of the individual reaction if the change in G of the total reaction Change in G is a state function Ø If a reaction is reversed the sign of the change in G value reverses Ø If the amount of materials is multiplies by a factor the value of the change in G is multiplied by the same factor The value of change in G of a reaction is extensive v Free energy Ø The free energy is the maximum amount of energy released from a system that is available to do work in the surroundings Ø For many exothermic reactions some of the heat is released due to the enthalpy change goes into increasing the entropy of the surroundings so it is not available to do work Ø Even some of the free energy is generally lost to heating up the surroundings v Free energy and reversible reactions Ø The change in free energy is a theoretical limit as to the amount of work that can be done Ø If the reaction achieves its theoretical limit it is a reversible reaction v Real reactions Ø In a real reaction some (if not most) of the free energy is lost as heat Ø Therefore real reactions are irreversible v Change in G under nonstandard conditions Ø Change in G = change in G o only when the reactants arn products are in their standard states Their normal state at the temperature Partial pressure of gas = 1 atm Concentration = 1 molar Ø Under non- standard conditions Change in G = change in G o + RTlnQ Q is the reaction quotient At equilibrium change in G = 0

10 Change in G o = - RTlnK v Change in G o and K Ø Because change in Grxn = 0 at equilibrium then Change in G o = - RTlnK Ø When K < 1 change in G o is positive and the reaction is spontaneous in the reverse direction under standard conditions Nothing will happen if there are no products yet Ø When K > 1 change in G o is negative and the ration is spontaneous in the forward direction under standard conditions Ø When K = 1 change in G o = 0 and the reaction is at equilibrium under standard conditions v Why is the equilibrium constant temperature dependent Ø Combine Change in G o = change in H o *- T*change in S Change in G o = - RTlnK Ø Then Ln(K) = (change H o reaction/r)(1/t) + (change in Srxn/R) Ø This equation is the form of y = mx + b Ø The graph of ln(k) versus inverse T is a straight line with slope (change H o reaction/r) and y- intercept (change in Srxn/R)