The study of energy and the changes it undergoes.

 the ability to do work.  conserved.  made of heat and work.  a state function.  independent of the path, or how you get from point A to B.

 is divided into two halves. 1. System - the part you are concerned with. 2. Surroundings - the rest of the universe.  Exothermic reactions release energy to the surroundings.  Endothermic reactions absorb energy from the surroundings.

Potential energy Heat

Potential energy Heat

 Every energy measurement has three parts. 1. A unit ( Joules of calories). 2. A number how many. 3. and a sign to tell direction of energy flow.  negative - exothermic +positive- endothermic

System Surroundings Energy  E <0

System Surroundings Energy  E >0

 Heat given off is negative.  Heat absorbed is positive.  Work done by system on surroundings is negative.  Work done on system by surroundings is positive.

 The energy of the universe is constant= the law of conservation of energy.  q = heat  w = work   E = q + w Take the system’s point of view to decide signs.

 Work is a force acting over a distance.  w= F x  d  P = F/ area  d = V/area  w= (P x area) x  (V/area)= P  V  Work can be calculated by multiplying pressure by the change in volume at constant pressure.

 If the volume of a gas increases, the system has done work on the surroundings.  work is negative  w = - P  V  units of liter - atm  1 L atm = Joules  Expanding work is negative.  Contracting, surroundings do work on the system w is positive.

 What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure?  If 2.36 J of heat are absorbed by the gas above, what is the change in energy?

 Symbol: H  H = E + PV (that’s the definition)  At constant pressure:  H =  E + P  V  The heat at constant pressure q p can be calculated q p + w = q p - P  V q p =  E + P  V =  H  So, q =  H at constant pressure.

 Measuring heat.  Use a calorimeter.  Two kinds 1. Constant pressure calorimeter (aka coffee cup calorimeter)  A coffee cup calorimeter measures  H.  An insulated cup, full of water.  The specific heat of water is 1 cal/gºC or J/g ºC.  Heat of reaction=  H = Cp x mass x  T

 heat capacity for a material, C is calculated C= heat absorbed/  T =  H/  T  molar heat capacity = C/moles  specific heat capacity is dependent on the mass of material: C p = C/mass Cp has units of J/gC or J/gK  heat = Cp x mass x  T OR heat = molar heat x moles x  T  Make the units work and you’ve done the problem right.

 The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.  A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

2. Constant volume calorimeter is called a bomb calorimeter.  Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water.  The heat capacity of the calorimeter is known and tested.  Since  V = 0, P  V = 0,  E = q

 thermometer  stirrer  full of water  ignition wire  Steel bomb  sample

 Enthalpy is a state function.  It is independent of the path.  We can add equations to come up with the desired final product, and add the  H  Two rules 1. If the reaction is reversed the sign o  H is changed 2. If the reaction is multiplied, so is  H

 The enthalpy change for a reaction at standard conditions: (25ºC, 1 atm, 1 M solutions)  Symbol  Hº  When using Hess’s Law, work by adding the equations up to make it look like the answer.  The other parts will cancel out.

 Hº= -394 kJ  Hº= -286 kJ  Hº= kJ

Given Calculate  Hº for this reaction  Hº= +77.9kJ  Hº= +495 kJ  Hº= kJ

 Table of standard heats of formation: Appendix 4  The amount of heat needed to form 1 mole of a compound from its elements in their standard states.  Standard states are 1 atm, 1M and 25ºC  For an element, the  H  f is = 0

 Need to be able to write the equations.  What is the equation for the formation of NO 2 ? ½N 2 (g) + O 2 (g)  NO 2 (g)  Have to make one mole to meet the definition.

 We can use heats of formation to figure out the heat of reaction. Ex. C 2 H 5 OH +3O 2 (g)  2CO 2 + 3H 2 O  Heat of reaction = The sum of the heats of formation of the products – sum of the heats of formation of the reactants.

Spontaneity, entropy and free energy

 A reaction that will occur without outside intervention.  We can’t determine how fast.  We need both thermodynamics and kinetics (next chapter) to describe a reaction completely.  Thermodynamics compares initial and final states.  Kinetics describes pathway between.

 1st Law- the energy of the universe is constant.  Keeps track of thermodynamics but doesn’t correctly predict spontaneity. WE MUST ALSO CONSIDER ENTROPY  Entropy (S) is disorder or randomness  2nd Law - the entropy of the universe always increases.

 Defined in terms of probability.  Substances take the arrangement that is most likely.  The most likely is the most random.  Calculate the number of arrangements for a system.

 2 possible arrangements  50 % chance of finding the left empty Positional entropy

l 4 possible arrangements l 25% chance of finding the left empty l 50 % chance of them being evenly dispersed

l 4 possible arrangements l 8% chance of finding the left empty l 50 % chance of them being evenly dispersed

 Gases completely fill their chamber because there are many more ways to do that than to leave half empty.  There are many more ways for the molecules to be arranged as a liquid than a solid.  Gases have a huge number of positions possible = highest entropy. S solid <S liquid <<S gas

 Solutions form because there are many more possible arrangements of dissolved particles than if they stay bound as a solid.  Reactions that form gases or liquids are favorable since positional entropy is increased.

2nd Law   S univ =  S sys +  S surr  If  S univ is positive the process is spontaneous.  If  S univ is negative the process is spontaneous in the opposite direction

 For exothermic processes  S surr is positive.  For endothermic processes  S surr is negative.  Consider this process H 2 O(l)  H 2 O(g)   S sys is positive   S surr is negative   S univ depends on temperature.

 Entropy changes in the surroundings are determined by the heat flow.  An exothermic process is favored because by giving up heat, the entropy of the surroundings increases.  The size of  S surr depends on temperature   S surr = -  H/T   S surr = + when exothermic

 S sys  S surr  S univ Spontaneous? ---Yes +++No, reverse +-?At high temp. -+?At low temp.

 The entropy of a pure crystal at 0 K is 0. Gives us a starting point.  All others must be >0.  Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed in the appendix.  Entropy is a state function.  Σ Sº Products – Summation Sº reactants to find  Sº  More complex molecules have higher Sº.

 Free energy, G, is that energy free to do work.  The maximum amount of work possible at a given temperature and pressure = w max  Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.

 Another way to determine spontaneity  G=H-TS  Never used this way.   G=  H-T  S sys at constant temperature Divide by -T  G/T = -  H/T-  S  G/T =  S surr +  S sys  G/T =  S univ  If  G is negative at constant T and P, the Process is spontaneous.

 For the reaction H 2 O(s)  H 2 O(l)   Sº = 22.1 J/K mol  Hº =6030 J/mol  Calculate  G at 10ºC and -10ºC  Look at the equation  G=  H-T  S  Spontaneity can be predicted from the sign of  H and  S.

HH SS Spontaneous? +- At all Temperatures ++ At high temperatures, “entropy driven” -- At low temperatures, “enthalpy driven” +- Not at any temperature, Reverse is spontaneous

  Gº = standard free energy change.  Free energy change that will occur if reactants in their standard state turn to products in their standard state.  Can’t be measured directly, can be calculated from other measurements.   Gº=  Hº-T  Sº  Use Hess’s Law with known reactions.

 There are tables of  Gº f.  It is a state function.  Summation  Gº f products- summation reactants  Gº f reactants.  The standard free energy of formation for any element in its standard state is 0.  Remember- Spontaneity tells us nothing about rate.

  G tells us spontaneity at current conditions. When will it the reaction stop?  It will go to the lowest possible free energy which may be an equilibrium – no net change between products and reactants.  At equilibrium  G = 0, Q = K   Gº = -RTlnK

delta GK =0=1 0 >0<0

  Gº= -RTlnK =  Hº - T  Sº  ln(K) =  Hº/R(1/T)+  Sº/R  A straight line of lnK vs 1/T