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Enthalpy. Most chemical and physical changes occur under essentially constant pressure (reactors open to the Earth’s atmosphere) very small amounts of work are performed as system expands or contracts

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enthalpy
Enthalpy
  • Most chemical and physical changes occur under essentially constant pressure (reactors open to the Earth’s atmosphere)
    • very small amounts of work are performed as system expands or contracts
    • the change in internal energy occurs primarily, or exclusively, as heat that is gained or lost.
enthalpy1
Enthalpy
  • If a process occurs at constant pressure and the only work done is PV work, the heat flow is described by the enthalpy of the system.
  • Enthalpy (H):
    • a state function defined by the equation:

H = E + PV

(Question: Are P and V state functions?)

enthalpy2
Enthalpy
  • Although the enthalpy of a system cannot be measured, the change in enthalpy (D H) can.
    • heat gained or lost by a system when a process occurs at constant pressure

D H = Hfinal - Hinitial = qP

where qP = heat gained/lost at constant pressure

enthalpies of reaction
Enthalpies of Reaction
  • The change in enthalpy can be found using:

D H = Hfinal - Hinitial

  • For a chemical reaction:
    • Hfinal = H products
    • Hinitial = H reactants
  • The enthalpy change for a chemical reaction is:

D H = Hproducts - Hreactants

enthalpies of reaction1
Enthalpies of Reaction
  • The enthalpy change that accompanies a chemical reaction is called the enthalpy of reaction
    • D Hrxn
    • Also called heat of reaction
enthalpy3
Enthalpy
  • For a chemical reaction,

D H = Hproducts - Hreactants

  • D H < 0 (negative)
    • heat is lost by system
      • exothermic

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

enthalpy4
Enthalpy
  • For a chemical reaction,

DH = Hproducts - Hreactants

  • D H > 0 (positive)
    • heat gained/absorbed

by the system

      • endothermic

CO2 (g) + 2 H2O (l) CH4 (g) + 2 O2 (g)

enthalpies of reaction2
Enthalpies of Reaction

P

  • If D Hrxn = positive
    • endothermic
      • heat must be added
  • If D Hrxn = negative
    • exothermic
      • heat is given off

H

R

time

R

H

P

time

enthalpies of reaction3
Enthalpies of Reaction
  • DHrxn is associated with a specific chemical reaction.
    • extensive property
      • Depends on the amount of material
  • Thermochemical equationsare balanced chemical equations that show the associated enthalpy change
    • balanced equation
    • enthalpy change (DHrxn)
enthalpies of reaction4
Enthalpies of Reaction
  • An example of a thermochemical equation:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

  • The coefficients in the balanced equation show the # moles of reactants and products that produced the associated DH.
    • If the number of moles of reactant used or product formed changes, then the DH will change as well.
enthalpies of reaction5
Enthalpies of Reaction
  • For the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

-890. kJ -890. kJ

1 mol CH4 2 mol O2

-890. kJ -890. kJ

1 mol CO2 2 mol H2O

enthalpies of reaction6
Enthalpies of Reaction
  • Guidelines for Using Thermochemical Equations:
    • Enthalpy is an extensive property
      • The magnitude of DH is directly proportional to the amount of reactant consumed or product produced
enthalpies of reaction7
Enthalpies of Reaction
  • The thermochemical equation for burning 1 mole of CH4 (g):

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

    • When 1 mole of CH4 is burned, 890. kJ of heat are released.
    • When 2 moles of CH4 are burned, 1780. kJ of heat are released.
enthalpies of reaction8
Enthalpies of Reaction

Example: How much heat is gained or lost when 10.0 g of CH4 (g) are burned at constant pressure?

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

Given:10.0 g CH4 (g)

-890. kJ

1 mol CH4

Find: heat

enthalpies of reaction9
Enthalpies of Reaction

Heat = 10.0 g CH4 x1 mole CH4 x -890. kJ

16.0 g CH4 1 mol CH4

= -556 kJ

Is this an endothermic or exothermic process?

enthalpies of reaction10
Enthalpies of Reaction

Example: How much heat is gained or lost when 10.0 g of water are formed at constant pressure in the following reaction?

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

Given:10.0 g H2O (g)

-890. kJ

2 mol H2O

Find: heat

enthalpies of reaction11
Enthalpies of Reaction

Heat = 10.0 g H2O x1 mole H2O x -890. kJ

18.0 g H2O 2 mol H2O

= -247 kJ

The negative sign indicates that heat was released to the surroundings.

enthalpies of reaction12
Enthalpies of Reaction
  • Guidelines for Using Thermochemical Equations (cont).
    • The enthalpy change for a reaction is equal in magnitude but opposite in sign to the DH for the reverse reaction.

2 H2O2 (l) 2 H2O (l) + O2(g) DH = -196 kJ

2 H2O (l) + O2(g) 2 H2O2 (l) DH = +196 kJ

enthalpies of reaction13
Enthalpies of Reaction
  • Guidelines for Using Thermochemical Equations (cont):
    • The enthalpy change for a reaction depends on the physical state of the reactants and products.

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) DH = -802 kJ

enthalpies of reaction14
Enthalpies of Reaction
  • Why does the DHrxn depend on the physical state of the reactants and products?
    • Energy is either absorbed or released when chemicals change from one physical state to another.

H2O (l) H2O (g) DH = + 44 kJ

calorimetry
Calorimetry
  • The enthalpy change associated with a chemical reaction or process can be determined experimentally.
    • measure the heat gained or lost during a reaction (or process) at constant P
      • Measure the change in temperature
calorimetry1
Calorimetry
  • Calorimetry:
    • the experimental measurement of heat gained or lost during a chemical reaction or process
  • Calorimeter
    • an instrument used to measure the heat gained or lost during a chemical reaction or process.
calorimetry2
Calorimetry
  • Calorimetry is used to experimentally determine:
    • Heat capacity or specific heat
    • DHrxn
      • Enthaply change for a reaction
    • DHfusion
      • Enthalpy change when a substance goes from the liquid to the solid state
    • DHvaporization
      • Enthalpy change when a substance goes from the liquid to the gas state
calorimetry3
Calorimetry

If you leave your keys and your chemistry book sitting in the sun on a hot summer day, which one is hotter?

Why is there a difference in temperature between the two objects?

calorimetry4
Calorimetry
  • The temperature increase observed when an object absorbs a certain quantity of energy is determined by its heat capacity (C).
  • Amount of heat required to raise the temperature of an object 1oC (or 1 K)
  • As heat capacity increases, more heat must be added to produce a specific temperature increase.
calorimetry5
Calorimetry
  • For pure substances, heat capacity is usually given for a specified amount of substance:
    • Molar heat capacity:
      • amount of heat required to raise the temperature of 1 mole of a substance by 1oC
    • Specific heat:
      • amount of heat required to raise the temperature of 1 g of a substance by 1oC
calorimetry6
Calorimetry
  • Specific Heat = quantity of heat transferred

mass x temp change

= q

mass x DT

  • Molar Heat = quantity of heat transferred

Capacity moles x temp change

= q

mol x DT

calorimetry7

Specific heat = q

m x DT

Calorimetry

Example:If 418 J is required to increase the temperature of 50.0 g of water by 2.00 K, what is the specific heat of water?

Given:q = 418 J

m = 50.0 g

DT = 2.00K

Find: specific heat

calorimetry8

Common units for specific heat are:

    • J cal
    • g.K g.oC
Calorimetry

Specific heat = 418 J = 4.18 J

50.0 g x 2.00 K g.K

calorimetry9

Molar heat capacity = q

mol x DT

Calorimetry

Example:What is the molar heat capacity of aluminum if it takes 9.00 J to raise the temperature of 5.00 g of aluminum from 298.0 K to 300.0 K?

Given:q = 9.00 J

m = 5.00 g

DT = 2.0 K

Find: molar heat

capacity

calorimetry10

Common units for molar heat capacity are:

    • J cal
    • mol.K mol.oC
Calorimetry

Molar heat capacity =

9.00 J x 27.0 g Al = 24 J

5.00 g x 2.0 K mol Al mol.K

calorimetry11
Calorimetry

Example:If the specific heat of Al (s) is 0.90 J/g.K, how much heat is required to raise the temperature of 10.0 kg of Al from 25.0oCto 30.0oC?

Given:C = 0.90 J/g.K

m = 10.0 kg

DT = 5.0 K

Find: heat

C = q

m x DT

calorimetry12
Calorimetry

q = C x m x DT

q = 0.90 J x10.0 kg x 1000 g x 5.0 K

g.K 1 kg

= 45,000 J = 45 kJ

C = q

m x DT

This equation is one that you will use OFTEN in calorimetry.

calorimetry13

Specific heat = q

m x DT

Calorimetry

Example:If the specific heat of Fe(s) is 0.45 J/g.K, what change in temperature would be observed when 1.0 kJ of heat is added to 45 g of Fe(s)?

Given:C = 0.45 J/g.K

m = 45 g

q = 1.0 kJ

Find:DT

calorimetry14
Calorimetry

Specific heat = q

m x DT

DT =1.0 kJ x 1000 J x g.K

45 g 1 kJ 0.45 J

= 49 K

DT = q

m x C

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