TWO Trends in Nature Order  Disorder   High energy  Low energy 

Law of Conservation of Energy
Energy can be converted from one form to another but can neither be created nor destroyed. (Euniverse is constant) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

The Two Types of Energy Potential: due to position or composition - can be converted to work Kinetic: due to motion of the object KE = 1/2 mv2 (m = mass, v = velocity) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

(a) In the initial positions, ball A has a higher potential energy than ball B. (b) After A has rolled down the hill, the potential energy lost by A has been converted to what other forms? Frictional heating Of the hill Increase PE of Ball B

Temperature v. Heat Temperature reflects random motions of particles, therefore related to kinetic energy of the system. Heat involves a transfer of energy between 2 objects due to a temperature difference Copyright©2000 by Houghton Mifflin Company. All rights reserved.

This infrared photo of a house shows where energy leaks occur
This infrared photo of a house shows where energy leaks occur. The more red the color, the more energy (heat) is leaving the house.

there are two ways to transfer energy work and heat
Work: force acting over a distance Ball A does work on Ball B Thus… there are two ways to transfer energy work and heat

Work and heat are pathway dependent
The way the energy transfer is divided between work and heat depends on .. Pathway If all energy goes to friction in the hill and Ball B does not move…. no work is done. Work and heat are pathway dependent

…….Regardless of the pathway the actual energy change remains constant. Total energy transfer is independent of the pathway Energy is a state function. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

State Function (state property)
Depends only on the present state of the system - not how it arrived there. It is independent of pathway. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

System and Surroundings
System: chemical reaction Surroundings: Everything else in the universe Universe = System + Surroundings Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Exo and Endothermic Heat exchange accompanies chemical reactions. Exothermic: Heat flows out of the system (to the surroundings). [product] Endothermic: Heat flows into the system (from the surroundings). [reactant] Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 6.2: The combustion of methane releases the quantity of energy change ∆(PE) to the surroundings via heat flow. This is an exothermic process.

Figure 6.3: The energy diagram for the reaction of nitrogen and oxygen to form nitric oxide. This is an endothermic process: Heat {equal in magnitude to ∆(PE)} flows into the system from the surroundings.

Concept Check Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. Your hand gets cold when you touch ice. The ice gets warmer when you touch it. Water boils in a kettle being heated on a stove. Water vapor condenses on a cold pipe. Ice cream melts. Exo Endo Exothermic (heat energy leaves your hand and moves to the ice) Endothermic (heat energy flows into the ice) Endothermic (heat energy flows into the water to boil it) Exothermic (heat energy leaves to condense the water from a gas to a liquid) Endothermic (heat energy flows into the ice cream to melt it)

Law of Conservation of Energy is also called:
First Law of Thermodynamics: The energy of the universe is constant. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Exothermic versus endothermic.
Endothermic * products have Higher PE Exothermic * reactants have Higher PE (Higher PE – weaker bonds) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

First Law cont. Sign is the direction of the flow. (system point of view) q = + endothermic system E increase q = - exothermic system E decrease Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Common type of work in chemistry
Done by a gas (through expansion) Done to a gas (through compression) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

How does a car engine work?
Combustion of a gas: expand in cylinder: pushes piston: motion of car: Endo or exothermic q (– or +) (∆E to surroundings) w (- or +) (work done to surr.) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 6.4: (a) The piston, moving a distance ∆h against a pressure P, does work on the surroundings. (b) Since the volume of a cylinder is the area of the base times its height, the change in volume of the gas is given by ∆h x A = V.

Work is a force applied over a distance.
Force per unit area Work is a force applied over a distance. Work= force x distance = F x Δh SINCE , F = P x A Work = P x A x Δh ΔV = final V – initial (piston moving)

Since V = (A x ∆h) (area of the piston x height)
∆V = Vfinal - Vinitial = A x ∆h substitute into work: Work = P x (A x Δ h) = PΔV For a gas expanding w is (-) due to work flowing out of the system. W = -P ΔV

PV Work Calculate the work associated with the expansion of a gas from 46.L to 64 L at a constant external pressure of 15atm. For a gas at constant P, w= -P∆V

Just a note… For an ideal gas, work can occur only when its V changes. Thus, if a gas is heated at constant volume, the pressure increases but no work occurs.

P = 15 atm And ∆V = 64-46 = 18 L W = -15atm x 18 L = -270 L∙atm

A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes form 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ∆E for the process (1L atm = J) 8 x 107 J

Enthalpy = H = E + PV (state funx.)
WHAT IS ENTHALPY? At constant pressure, the only work allowed is PV where qP = heat at constant pressure is calculated from E = qP + w = qP  PV qP = E + PV= H H = energy flow as heat (at constant pressure) **Heat of reaction and change in enthalpy are interchangeable**

DH = H (products) – H (reactants)
Enthalpy (H) is used to quantify the heat flow into (or out of) a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 exotherm DH > 0 endotherm 6.4

Heat Capacity (amount of heat needed to raise T)
C= q q=(∆H) ∆T Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Some Heat Exchange Terms
specific heat capacity (energy required to raise the T of 1g of substance 1°C) heat capacity per gram = J/°C∙ g or J/K ∙ g molar heat capacity ( raise T of 1mol, 1°C) heat capacity per mole = J/°C ∙ mol or J/K ∙ mol Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Problem #27 39.1 mol He V=876L, T=0.0º C, P=1.00atm T2 =38.0º C, V2 =998L, P=1.00atm Calculate q,w,∆E Molar heat capacity = 20.8 J/º C∙mol Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Problem One mole of H2O(g) at 1.00 atm and 100.ºC has a V= 30.6 L. 1mol H2O(g) condensed to 1mol H2O(l) At P=1.00 atm T= 100.ºC q=40.66kJ (released) If D=0.906g/cm3 …. Calculate ∆E for the condensation of one mole of water at 1.00 atm and 100. ºC Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Constant-pressure calorimetry
Measurement of heat using a simple calorimeter (determine heat associated w/ a chemical reaction) Used to determine ΔH (heat) of a reaction in solution. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 6.5: A coffee-cup calorimeter made of two Styrofoam cups.

Energy released= energy absorbed by by the reaction the solution = specific heat capacity x mass of solution x increase in T. = s x m x ΔT Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 6.6: A bomb calorimeter. Constant Volume
1.Material is put in a container with pure oxygen 2.Wires are used to start combustion 3. Container is put in a container of water 4. Heat capacity is known and tested Since ∆V=0, P∆V =0 And ΔE = q + w = q = qv .

Energy released by the reaction at constant volume
=Temp. increase x energy required to change the temp by 1°C. =ΔT x heat capacity of calorimeter Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Heat of the Reaction Extensive property-depends on the amount of the substance. (Energy, heat capacity, mass, heat from a reaction) Intensive property-does not depend on the amount (Temperature, density, specific heat) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Hess’s Law Enthalpy (state function)
Reactants  Products The change in enthalpy is the same whether the reaction takes place in one step or a series of steps.

Oxidation of N2 to produce NO2
N2(g) + 2O2(g) NO2(g) ΔH1=68kJ Or.. N2(g) + 2O2(g) NO (g) ΔH2=180kJ 2NO(g) + O2(g) NO2(g) ΔH3=-112kJ N2(g) + 2O2(g) NO2(g) ΔH=68kJ

Figure 6.7: The principle of Hess's law. The same change in enthalpy occurs when nitrogen and oxygen react to form nitrogen dioxide, regardless of whether the reaction occurs in one (red) or two (blue) steps. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Calculations via Hess’s Law
1. If a reaction is reversed, H is also reversed. N2(g) + O2(g)  2NO(g) H = 180 kJ 2NO(g)  N2(g) + O2(g) H = 180 kJ 2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer. 6NO(g)  3N2(g) + 3O2(g) H = 540 kJ

Two forms of carbon are graphite, the soft, black, slippery material used in “lead” pencils and as a lubricant for locks, and diamond, the brilliant, hard gemstone. Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396kJ/mol), calculate ∆H for the conversion of graphite to diamond: Cgraphite(s)  Cdiamond(s) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Cgraphite(s)  Cdiamond(s)
Cgraphite(s) +O2 (g)  CO2 (g) ∆H=-394 kJ Cdiamond(s) + O2 (g)  CO2 (g) ∆H=-396 kJ Cgraphite(s) +O2 (g)  CO2 (g) ∆H=-394 kJ CO2  Cdiamond(s) + O2 (g) ∆H=-(-396 kJ) Cgraphite(s)  Cdiamond(s) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Diborane (B2H6) is a highly reactive boron hydride, which was once considered as a possible rocket fuel for the US space program. Calculate ∆H for the synthesis of diborane from its elements, according to the equation 2B(s) + 3H2 (g)  B2H6(g) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Standard Enthalpies of Formation
ΔHf° change in enthalpy that accompanies the formation of 1 mol of compound from its elements w/all substances in their standard states

Standard States Compound Element
For a gas, pressure is exactly 1 atmosphere. For a solution, concentration is exactly 1 molar. Pure substance (liquid or solid), it is the pure liquid or solid. Element The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.

Formation of NO2 ½ N2(g) + O2(g)  NO2(g) H°f = 34 kJ/mol
..both elements are in standard states ..1mol of product is formed.

Pathway for combustion of methane.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) Reactions are first taken apart in (a) and (b) then used to assemble the products in reactions (c) and (d).

Figure 6.9: A schematic diagram of the energy changes for the reaction CH4(g) + 2O2(g)  CO2(g) + 2H2O(l). Reverse of the formation

is reverse of formation of CH4 ΔHºf = 75kJ/mol
(b) O2 is already an element ΔHºf = 0 (c) formation of CO ΔHºf = -394kJ/mol (d) formation of H2O ΔHºf = -286kJ/mol -572kJ/mol **must multiply by 2, 2 mols needed**

Change in Enthalpy Can be calculated by subtracting the enthalpies of formation of reactants from products. Hrxn° = npHf(products)  nrHf(reactants) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Using the standard enthalpies of formation listed in Table 6.2, Calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water. This is the first step in the manufacture of nitric acid. 4NH3(g) + 7O2 (g)  4NO2 (g) + 6H2O(l) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Methanol (CH3OH) is often used as a fuel in high-performance engines in race cars. Using the data in Table 6.2, compare the standard enthalpy of combustion per gram of methanol with that per gram of gasoline. Gasoline is actually a mixture of compounds, but assume for this problem that gasoline is pure liquid octane (C8H18). Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 6.12: Greenhouse effect.

Figure 6.14: Coal gasification.

TWO Trends in Nature Order  Disorder   High energy  Low energy 

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TWO Trends in Nature Order  Disorder   High energy  Low energy 

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