Barnegat High School AP Chemistry Chapter 5 Thermodynamics

System System Surroundings Surroundings State property – depend only on the state of the system, not on the way the system reached the state State property – depend only on the state of the system, not on the way the system reached the state

System and Surroundings The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston).

volume – state property volume – state property heat flow – not a state property – value depends on how the system reached that state. heat flow – not a state property – value depends on how the system reached that state.

The universe is divided into two halves. is divided into two halves. the system and the surroundings. the system and the surroundings. The system is the part you are concerned with. The system is the part you are concerned with. The surroundings are the rest. The surroundings are the rest. Exothermic reactions release energy to the surroundings. Exothermic reactions release energy to the surroundings. Endothermic reactions absorb energy from the surroundings. Endothermic reactions absorb energy from the surroundings.

Energy The ability to do work or transfer heat. The ability to do work or transfer heat. Work: Energy used to cause an object that has mass to move. Work: Energy used to cause an object that has mass to move. Heat: Energy used to cause the temperature of an object to rise. Heat: Energy used to cause the temperature of an object to rise.

Potential Energy Energy an object possesses by virtue of its position or chemical composition.

Copyright © Houghton Mifflin Company. All rights reserved.6–86–8 Initial Position In the initial position, ball A has a higher potential energy than ball B.

Copyright © Houghton Mifflin Company. All rights reserved.6–96–9 Final Position After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heading) and to the increase in the potential energy of B. After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heading) and to the increase in the potential energy of B.

Kinetic Energy Energy an object possesses by virtue of its motion. 1 2 KE =  mv 2

Units of Energy The SI unit of energy is the joule (J). An older, non-SI unit is still in widespread use: The calorie (cal). 1 cal = J 1 J = 1  kg m 2 s2s2

Work Energy used to move an object over some distance. w = F  d, where w is work, F is the force, and d is the distance over which the force is exerted.

Work needs a sign If the volume of a gas increases, the system has done work on the surroundings. work is negative Expanding work is negative. Contracting surroundings do work on the system w is positive.

Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects.

First Law of Thermodynamics Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. Use Fig. 5.5

Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. Use Fig. 5.5

Internal Energy By definition, the change in internal energy,  E, is the final energy of the system minus the initial energy of the system:  E = E final − E initial Use Fig. 5.5

Changes in Internal Energy If  E > 0, E final > E initial – Therefore, the system absorbed energy from the surroundings. – This energy change is called endergonic.

Changes in Internal Energy If  E < 0, E final < E initial – Therefore, the system released energy to the surroundings. – This energy change is called exergonic.

Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is,  E = q + w.

 E, q, w, and Their Signs

Text Pg. 171 Sample Exercise 5.2

Copyright © Houghton Mifflin Company. All rights reserved. 6–23 Work vs. Energy Flow

Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic.

Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system to the surroundings, the process is exothermic.

State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

State Functions However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. – In the system below, the water could have reached room temperature from either direction.

State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. And so,  E depends only on E initial and E final.

State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its  E is the same. – But q and w are different in the two cases.

Work When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. w = −P  V

Enthalpy If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure- volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV

Enthalpy When the system changes at constant pressure, the change in enthalpy,  H, is  H =  (E + PV) This can be written  H =  E + P  V

Enthalpy Since  E = q + w and w = −P  V, we can substitute these into the enthalpy expression:  H =  E + P  V  H = (q+w) − w  H = q So, at constant pressure the change in enthalpy is the heat gained or lost.

Endothermicity and Exothermicity A process is endothermic, then, when  H is positive.

Endothermicity and Exothermicity A process is endothermic when  H is positive. A process is exothermic when  H is negative.

Pg. 176 Sample and practice 5.3 Pg. 176 Sample and practice 5.3

Enthalpies of Reaction The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants

Enthalpies of Reaction This quantity,  H, is called the enthalpy of reaction, or the heat of reaction.

The Truth about Enthalpy Pg Enthalpy is an extensive property. 2.  H for a reaction in the forward direction is equal in size, but opposite in sign, to  H for the reverse reaction. 3.  H for a reaction depends on the state of the products and the state of the reactants.

Pg. 179 Sample and practice (most important out of all the samples)

Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure  H through calorimetry, the measurement of heat flow.

Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1  C) is its heat capacity. We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

Heat Capacity and Specific Heat Specific heat, then, is Specific heat = heat transferred mass  temperature change C s = q m  Tm  T

Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Copyright © Houghton Mifflin Company. All rights reserved.6–46 A Coffee Cup Calorimeter Made of Two Styrofoam Cups

Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/mol-K), we can measure  H for the reaction with this equation: q = m  s   T

Pg. 181 sample Pg. 181 sample Pg. 182 sample Pg. 182 sample

Bomb Calorimetry Reactions can be carried out in a sealed “bomb,” such as this one, and measure the heat absorbed by the water.

Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy,  E, not  H. For most reactions, the difference is very small.

Bomb Calorimetry

Pg. 184 sample

Enthalpy Heat Heat Remember: Remember: q reaction at constant pressure = q reaction at constant pressure = ∆H = H product – H reactants ∆H = H product – H reactants -∆H exothermic -∆H exothermic +∆H endothermic +∆H endothermic Remember reaction diagrams Remember reaction diagrams

Thermochemical equations Specify ∆H in kilojoules Specify ∆H in kilojoules Workbook Handout – Pg. 158 Read through and do sample problems - up to pg. 163 Workbook Handout – Pg. 158 Read through and do sample problems - up to pg. 163 Rules of Thermochemistry Masterton Handout Pg Rules of Thermochemistry Masterton Handout Pg Hess’s Law Hess’s Law Summarized on page 206 Summarized on page 206

Rules of Thermochemistry 1. The magnitude of ∆H is directly proportional to the amount of reactant or product 2. ∆H for a reaction is equal in magnitude but opposite in sign to ∆H for the reverse reaction 3. The value of ∆H for a reaction is the same whether it occurs in one step or in a series of steps. Called Hess’s Law

∆H is directly proportional to amount of reactants or products. ∆H is directly proportional to amount of reactants or products. When one mole of ice melts, 6.00 kJ of heat is absorbed, ∆H = kJ When one mole of ice melts, 6.00 kJ of heat is absorbed, ∆H = kJ If one gram of ice melts, If one gram of ice melts, ∆H = 6.00kJ/18.02 = kJ. ∆H = 6.00kJ/18.02 = kJ.

In general, ∆H can be related to amount by the conversion factor approach H 2(g) +Cl 2(g)  2HCl (g) ∆H = -185kJ H 2(g) +Cl 2(g)  2HCl (g) ∆H = -185kJ When 1.00 g of Cl 2 reacts, When 1.00 g of Cl 2 reacts, ∆H = 1.00g Cl 2 x 1 molCl 2 x -185kJ = ∆H = 1.00g Cl 2 x 1 molCl 2 x -185kJ = 70.90g Cl 2 1 mol Cl g Cl 2 1 mol Cl kJ

∆H for a reaction is equal in magnitude but opposite in sign to ∆H for the reverse reaction H 2 O (s)  H 2 O (l) ∆H = +6.00kJ; 6.00kJ H 2 O (s)  H 2 O (l) ∆H = +6.00kJ; 6.00kJ absorbed absorbed H 2 O (l)  H 2 O (s) ∆H = -6.00kJ; 6.00kJ evolved evolved

Hess’s Law If Equation 1 + Equation 2 = Equation 3, then ∆H 3 = ∆H 1 + ∆H 2 If Equation 1 + Equation 2 = Equation 3, then ∆H 3 = ∆H 1 + ∆H 2 Often used to calculate ∆H for one step, knowing ∆H for all other steps and for the overall reaction Often used to calculate ∆H for one step, knowing ∆H for all other steps and for the overall reaction

C (s) + ½ O 2(g)  CO (g) ∆H 1 = ? C (s) + ½ O 2(g)  CO (g) ∆H 1 = ? CO (g) + 1/2 O 2(g)  CO 2(g) ∆H 2 = kJ CO (g) + 1/2 O 2(g)  CO 2(g) ∆H 2 = kJ C (s) + O 2(g)  CO 2(g) ∆H 3 = C (s) + O 2(g)  CO 2(g) ∆H 3 = ∆H 1 = kJ ∆H 1 = kJ

Examples Pg. 204 – , 8.5, 8.6 Examples Pg. 204 – , 8.5, 8.6 Go over homework for this topic Go over homework for this topic

Heat of Formation, Heats of Reaction The standard heat of formation (ΔH o f ) is the change of enthalpy for the reaction that forms a compound from its pure elements under standard conditions. The standard heat of formation (ΔH o f ) is the change of enthalpy for the reaction that forms a compound from its pure elements under standard conditions. The standard heat of formation (ΔH o f ) for pure elements at standard conditions is zero. The standard heat of formation (ΔH o f ) for pure elements at standard conditions is zero. Standard heats of formation can be used to estimate the ΔH o f of any reaction. Standard heats of formation can be used to estimate the ΔH o f of any reaction.

Put new slide here

Copyright © Houghton Mifflin Company. All rights reserved.6–64 Standard States Compound Compound For a gas, pressure is exactly 1 atmosphere. For a gas, pressure is exactly 1 atmosphere. For a solution, concentration is exactly 1 molar. For a solution, concentration is exactly 1 molar. Pure substance (liquid or solid) Pure substance (liquid or solid) Element Element The form [N 2 (g), K(s)] in which it exists at 1 atm and 25°C. The form [N 2 (g), K(s)] in which it exists at 1 atm and 25°C.

ΔH o f and heats of formation ΔH o f and heat of formation Δ H o f = Σ H o f ( products) - Σ H o f ( reactants)

Enthalpies of Formation Standard molar enthalpy of formation – equal to the enthalpy change when one mole of the compound is formed at a constant pressure of 1 atm and a fixed temp (usually 25 o C Standard molar enthalpy of formation – equal to the enthalpy change when one mole of the compound is formed at a constant pressure of 1 atm and a fixed temp (usually 25 o C ∆H of ∆H of Can calculate Standard Enthalpy values ∆H o from these values Can calculate Standard Enthalpy values ∆H o from these values

The standard enthalpy change, ∆H o, for a given thermochemical equation is equal to the sum of the standard enthalpies of formation of the product compounds minus the sum of the standard enthalpies of formation of the reactant compounds The standard enthalpy change, ∆H o, for a given thermochemical equation is equal to the sum of the standard enthalpies of formation of the product compounds minus the sum of the standard enthalpies of formation of the reactant compounds

∆H o = ∑∆H o f products - ∑∆H o f reactants ∆H o = ∑∆H o f products - ∑∆H o f reactants Elements in their standard states can be omitted because their heats of formation are zero Elements in their standard states can be omitted because their heats of formation are zero Use Reference Table on Pg. 207 Use Reference Table on Pg. 207

ΔH o f and heats of formation Example: Calculate the Δ H o f for the reaction, Example: Calculate the Δ H o f for the reaction, PbO 2 (s) + 2H 2 SO 4 (l) + Pb(s)  2PbSO 4 (s) + 2H 2 O (l) PbO 2 (s) + 2H 2 SO 4 (l) + Pb(s)  2PbSO 4 (s) + 2H 2 O (l) Given the following Δ H o f information Given the following Δ H o f information

Species Δ H o f (kJ) PbO 2 (s) H 2 SO PbSO H 2 O-68.3 Δ H o f = Σ H o f ( products) - Σ H o f ( reactants) Δ H o f = [(2 x ) + (2 x -68.3)] – [(66.3) + (2 x )] Δ H o f = kJ

Pg. 207 Example Pg. 207 Example Pg. 208 do examples 8.7 and 8.8 Pg. 208 do examples 8.7 and 8.8

Bond energies are the amount of energy given off when bonds are formed, or the amount of energy used when bonds are broken. Bond energies deal with reactants and products in their gaseous state under standard conditions. Breaking bonds is an exothermic process. Making bonds is an endothermic process. Heat of reaction can be estimated by finding the difference between the bonds made and the bond energies of the bonds broken. Bond energies used in this way to find heats of reaction is an example of Hess’s Law.

ΔH o and Bond Energies ΔH o reaction = ΣH o bond energies (bonds broken) - ΣH o bond energies (bonds made)

 Bond enthalpy always positive  Pg. 211 of Masterton handout – table of bond enthalpies – heat always absorbed when bonds are broken

 The bonds in the reactants are stronger than those in the products  And  There are more bonds in the reactants than in the products  Bond entalpy is larger for a multiple bond than for a single bond between the same two atoms  Bottom of pg. 211

 First Law of Thermodynamics ◦ The first law of thermodynamics is commonly referred to as the law of Conservation of Energy. ◦ More specifically, the first law states that the changes in internal energy is equal to the difference between the energy supplied to the system as heat and the energy removed from the system as work performed on the surroundings.

 The energy of the universe is constant.  Law of conservation of energy.  q = heat  w = work   E = q + w  Take the systems point of view to decide signs.  In any process, the total change in energy of a system,  E, is equal to the sum of the heat, q, and the work,w, transferred between the system and the surroundings

 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  Expanding work is negative.  Contracting, surroundings do work on the system w is positive.

 Pg. 213 Example 8.9