# Principles of Reactivity: Energy and Chemical Reactions

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

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

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.

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

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

System: region of space where process occurs
Surroundings: region of space around system

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

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

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

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

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

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

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

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

Factors Determining Amount of Heat
Amount of material Temperature change Heat capacity

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

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

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?

Surroundings System Energy DE <0

Exothermic reactions release energy to the surroundings.
q < 0 (-) Exothermic

Heat Potential energy

Surroundings System Energy DE >0

Endothermic reactions absorb energy from the surroundings.
q > 0 (+) Endothermic

Heat Potential energy

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?

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?

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

Energy processes: Within a phase q=mHv or mHf Between phases q=mCpT

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?

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.

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.

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

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

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?

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

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

Bomb Calorimeter

Calorimeter Constant heat capacity that is constant over a
temperature range where q(cal) = CC x TC (CC = calorimeter constant)

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.

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?

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

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.

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?

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

Bomb Calorimeter thermometer stirrer full of water ignition wire
Steel bomb sample

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.

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

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

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?

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

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

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

Hess’ Law Reactants → Products
The enthalpy change is the same whether the reaction occurs in one step or in a series of steps

Molar heat capacity

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

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.

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

Example Given +1300. kJ + 394 kJ + 286 kJ
calculate DHº for this reaction

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

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

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

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

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

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 +

Example Now, You try it! Given DHº= +77.9kJ DHº= +495 kJ DHº= +435.9kJ
Calculate DHº for this reaction

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)

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.

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.

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.

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

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?

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

at constant pressure: E = qP + w or E = qP - PV

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

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

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

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

11. Combustion of Methane HC = ?

Write equations and find ΔHf0 for methane,
carbon dioxide, & water

HC = ?

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