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**Principles of Reactivity: Energy and Chemical Reactions**

AP Notes Chapter 5 Principles of Reactivity: Energy and Chemical Reactions

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**study of energy transfer or heat flow**

Thermodynamics study of energy transfer or heat flow Energy Kinetic Energy Thermal - Heat Mechanical Electrical Sound Potential Energy Chemical Gravitational Electrostatic

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**Energy is... The ability to do work. Conserved. made of heat and work.**

Work is a force acting over a distance. W=F x d Power is work done over time. P=W t Heat is energy transferred between objects because of temperature difference.

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**First Law of Thermodynamics**

Law of conservation of energy Total energy of the universe is constant Temperature and heat Heat is not temperature Thermal energy = particle motion Total thermal energy is sum of all a materials individual energies

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**The Universe is divided into two halves.**

the system and the surroundings. The system is the part you are concerned with. The surroundings are the rest. q into system = -q from surroundings

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**System: region of space where process occurs**

Surroundings: region of space around system

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**Thermodynamic Properties**

1. State functions -depend on nature of process NOT method -path independent -denoted by capital letters E → state function

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**Thermodynamic Properties**

2. Path functions -depends on how process is carried out -path dependent -denoted by lower case q & w → path functions

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Work →w w = - (PV) w = - P V - V P Constant P wP = - P V Constant V wV = - V P

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Work →w w = - P V assume ideal piston (i.e. no heat loss) w = E E = - P V in isothermics w = -q

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PV = nRT R=0.0821L atm/mol K In isothermic conditions -q = w = -nRT R=8.314 J/mol K

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DU = q + w In isothermics DU = 0 so q=-w DU = q - PDV In constant V or constant P w=0 so DU = q

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**Calorie - Amount of heat needed to raise the temperature of 1 gram H20, 1 degree centigrade**

Nutritional calorie [Calorie]= 1 Kcal SI unit → Joule 1 cal = J 1000cal = 1Kcal=4.184KJ

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**Direction of energy flow**

Every energy measurement has three parts. A unit ( Joules or calories). A number how many. and a sign to tell direction. negative - exothermic positive- endothermic No change in energy - isothermic

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**Factors Determining Amount of Heat**

Amount of material Temperature change Heat capacity

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Specific Heat Amount of heat needed to raise the temperature of 1 gram of material 1 degree centigrade Heat = q q = m Cp T where Cp = Heat capacity T = T2 – T1

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CP (J/mol.oC) H2O(s) = H2O(l) = H2O(g) =

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**What amount of heat is needed to just boil away 100**

What amount of heat is needed to just boil away 100. mL of water in a typical lab?

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Surroundings System Energy DE <0

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**Exothermic reactions release energy to the surroundings.**

q < 0 (-) Exothermic

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Heat Potential energy

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Surroundings System Energy DE >0

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**Endothermic reactions absorb energy from the surroundings.**

q > 0 (+) Endothermic

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Heat Potential energy

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**1. How much heat is absorbed when 10**

1. How much heat is absorbed when 10.0 g of H20 are heated from 200C to 250C?

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**2. How much heat is needed to completely vaporize 10**

2. How much heat is needed to completely vaporize 10.0 g of H2O at its boiling point?

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Heat of Vaporization Hvap(H2O) = kJ/mol Heat of Fusion Hfus(H2O) = 6.01 kJ/mol

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Energy processes: Within a phase q=mHv or mHf Between phases q=mCpT

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**3. How much total heat is used to raise the temperature of 100**

3. How much total heat is used to raise the temperature of g H2O from-15o to 125oC?

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**Some rules for heat and work**

Heat given off is negative. Heat absorbed is positive. Work done by system on surroundings is positive. Work done on system by surroundings is negative. Thermodynamics- The study of energy and the changes it undergoes.

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**First Law of Thermodynamics**

The energy of the universe is constant. Law of conservation of energy. q = heat w = work DE = q + w Take the systems point of view to decide signs.

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**What is work? Work is a force acting over a distance. w= F x Dd**

P = F/ area d = V/area w= (P x area) x D (V/area)= PDV Work can be calculated by multiplying pressure by the change in volume at constant pressure. units of liter - atm L-atm

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Work needs a sign If the volume of a gas increases, the system has done work on the surroundings. work is negative w = - PDV Expanding work is negative. Contracting, surroundings do work on the system w is positive. 1 L atm = J

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Examples What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? If J of heat are absorbed by the gas above. what is the change in energy? How much heat would it take to change the gas without changing the internal energy of the gas?

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**Calorimetry Measuring heat. Use a calorimeter. Two kinds**

Constant pressure calorimeter (called a coffee cup calorimeter) heat capacity for a material, C is calculated C= heat absorbed/ DT = DH/ DT specific heat capacity = C/mass

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**Calorimetry molar heat capacity = C/moles**

heat = specific heat x m x DT heat = molar heat x moles x DT Make the units work and you’ve done the problem right. A coffee cup calorimeter measures DH. An insulated cup, full of water. The specific heat of water is 1 cal/gºC Heat of reaction= DH = sh x mass x DT

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Bomb Calorimeter

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**Calorimeter Constant heat capacity that is constant over a**

temperature range where q(cal) = CC x TC (CC = calorimeter constant)

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**4. Determine the calorimeter**

constant for a bomb calorimeter by mixing 50.0 mL of water at 25oC with 50.0 mL of water at 60.0 oC. The final temp. attained is 40oC.

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**5. What is the molar heat of combustion of sulfur, if 2**

5. What is the molar heat of combustion of sulfur, if 2.56 grams of solid flowers of sulfur are burned in excess oxygen in the bomb calorimeter in #4?

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**6. Determine the molar heat of solution of LiOH**

6. Determine the molar heat of solution of LiOH. HINT: Find the calorimeter constant first.

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7. The heat was absorbed by 815 g of water in calorimeter which had an initial temp of C. After equilibrium was reached, the final temperature was C.

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Examples 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?

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**Calorimetry 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 DV = 0, PDV = 0, DE = q

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**Bomb Calorimeter thermometer stirrer full of water ignition wire**

Steel bomb sample

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Properties intensive properties not related to the amount of substance. density, specific heat, temperature. Extensive property - does depend on the amount of stuff. Heat capacity, mass, heat from a reaction.

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**Enthalpy abbreviated H H = E + PV (that’s the definition) PV = w**

at constant pressure. DH = DE + PDV the heat at constant pressure qp can be calculated from DE = qp + w = qp - PDV qp = DE + P DV = DH

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**Hess’s Law 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 DH Two rules If the reaction is reversed the sign of DH is changed If the reaction is multiplied, so is DH

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Enthalpy and DHf N2 + 2O2 2NO2 N2 + 2O2 + 68KJ/mol 2NO2 Which side is the heat written on? Why? Look on your Heat Sheets! Why is it only 34KJ/mol?

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H (kJ) 2NO2 68 kJ N2 + 2O2

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O2 + NO2 -112 kJ H (kJ) 180 kJ 2NO2 N2 + 2O2

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O2 + NO2 -112 kJ H (kJ) 180 kJ 2NO2 68 kJ N2 + 2O2

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**Hess’ Law Reactants → Products**

The enthalpy change is the same whether the reaction occurs in one step or in a series of steps

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Molar heat capacity

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8. Calculate the heat needed to decompose one mole of calcium carbonate.

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

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**Example Given calculate DHº for this reaction DHº= -1300. kJ**

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**Example Given +1300. kJ + 394 kJ + 286 kJ**

calculate DHº for this reaction

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**Example Given +1300. kJ + 394 kJ + 286 kJ**

1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g) + 394 kJ + 286 kJ calculate DHº for this reaction

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**Example Given + 2(394 kJ) + 286 kJ calculate DHº for this reaction**

1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g) 2 2 2 + 2(394 kJ) + 286 kJ calculate DHº for this reaction

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**Example Given + 788 kJ + 286 kJ calculate DHº for this reaction**

1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g) 2 4 2 + 788 kJ 2 + 286 kJ calculate DHº for this reaction

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**Example Given + 788 kJ + 286 kJ calculate DHº for this reaction**

1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g) 2 4 2 + 788 kJ 2 + 286 kJ calculate DHº for this reaction

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**Example Given calculate DHº for this reaction 1300.KJ C2H2 (g)**

2C(s) 788KJ H2(g) 286KJ 1300.KJ 788kJ + 286KJ 1300.KJ 1074KJ calculate DHº for this reaction

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**Example Given calculate DHº for this reaction 1300.KJ 1074KJ 226KJ **

DHf = 1300KJ – 1074KJ DHf = 226KJ (+sign shows Endothermic) calculate DHº for this reaction 226KJ +

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**Example Now, You try it! Given DHº= +77.9kJ DHº= +495 kJ DHº= +435.9kJ**

Calculate DHº for this reaction

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**Standard Enthalpies of Formation**

Hess’s Law is much more useful if you know lots of reactions. Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states. Standard states are 1 atm, 1M and 25ºC For an element it is 0 There is a table in Appendix 4 (pg A22)

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**Standard Enthalpies of Formation**

Need to be able to write the equations. What is the equation for the formation of NO2 ? ½N2 (g) + O2 (g) ® NO2 (g) Have to make one mole to meet the definition. Write the equation for the formation of methanol CH3OH.

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**Since we can manipulate the equations**

We can use heats of formation to figure out the heat of reaction. Lets do it with this equation. C2H5OH +3O2(g) ® 2CO2 + 3H2O which leads us to this rule.

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**Since we can manipulate the equations**

We can use heats of formation to figure out the heat of reaction. Lets do it with this equation. C2H5OH +3O2(g) ® 2CO2 + 3H2O which leads us to this rule.

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**9. Calculate Hf of SO3(g) from the data given. Forms of Equations:**

2SO2(g) + O2(g) 2SO3(g) H = kj or 2SO2(g) + O2(g) 2SO3(g) kj

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10. If the metabolism of glucose is combustion of C6H12O6(s), how much heat is produced by the metabolism of 1.00 g of glucose?

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**Additivity of Heats of Reaction**

H = Enthalpy H = E + PV DH = DE + DPDV

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at constant pressure: E = qP + w or E = qP - PV

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now, at constant P: VDP = 0 thus, H = E + PV & if E = qP - PV then H = qP

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**H = Hproducts - Hreactants**

H = Enthalpy a “state” function cannot be measured measure H H = Hproducts - Hreactants

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**Standard Enthalpy of Formation**

Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states

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**Thermodynamic Standard State**

T = 25oC and P = 1 atm Thermodynamic Data Appendix L of text p. A31 reference point- for ELEMENTS in standard state

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11. Combustion of Methane HC = ?

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**Write equations and find ΔHf0 for methane,**

carbon dioxide, & water

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HC = ?

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