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  • 8/11/2019 Chem Presentation Enthalpy

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    Chapter 15 - Standard enthalpy

    change of a reaction

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    Standard Enthalpy ChangesHana Amir and Madeley

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    Definition

    Any reaction that depends on temperature,

    pressureandstate

    The enthalpy change happens when all

    reactants and products are in their standard

    state.

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    Some enthalpies

    Enthalpy of reaction

    Enthalpy of formation

    Enthalpy of neutralization

    Enthalpy of hydration

    Enthalpy of combustion

    Enthalpy of solution Enthalpy of atomization

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    Standard Enthalpy Changes of

    ReactionDefinition:

    The heat change when molar quantities of reactants asspecified by the chemical equation react to form productsat standard conditions

    It depends in the physical state of reactants and theproducts and the conditions under which the reactionoccurs.

    Standard conditions are: 298K (25oC) and 1.00*105Pa

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    Standard Enthalpy change of

    Formation

    Enthalpy change that occurs when one mole of substanceisformed from its elements in the standard state under

    standard conditions.

    Standard conditions

    Temperature: 298K (25o C)

    Pressure: 1.00*105 Pa

    2C(graphite) + 3H2(g) + 1/2O2(g)-------> C2H5OH (I)

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    All elements in their standard states (oxygengas, solid carbon in the form of graphite, etc.)

    have a standard enthalpy of formation of zero,

    as there is no change involved.

    Eg:

    O(g) + O(g) ---->O2(g)

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    Energy Cycle

    Hreaction

    Hf(products)Hf(reactants)

    ProductsReactants

    Elements

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    From the diagram we get:

    The chemical change elements to products

    can either occur directly or indirectly .

    The Total enthalpy change must be the same for both routes.

    Hf (Products)=Hf (Reactants) +Hreaction

    This gives the general expression of:

    Hreaction =Hf (Products)- Hf (Reactants)

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    Example

    Calculate the enthalpy change for the reaction:

    C3H8(g) + 5O2(g)-------- 3CO2(g) + 4H2O(g)

    Standard change of formation: Hf /kJ mol-1

    C3H8(g) : -105

    CO2(g): -394H2O(l): -286

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    Steps:

    Write down the equation with the corresponding

    enthalpies of formation underneath:

    C3H8(g) + 5O2(g)-------- 3CO2(g) + 4H2O(g)

    (-105) 0 3(-394) 4(-286)

    Note: The standard enthalpies of formation are given in

    per mole , hence, they should be multiplied by the

    numbers of moles in the balance equation .

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    Hreaction =Hf (Products)- Hf (Reactants)

    Hreaction =3(-394) + 4(-286) -(-105)

    = -2221 KJ mol-1

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    Thermochemical Equations

    Balanced chemical equation for a reaction including the enthalpy of the reaction

    shown immediately after the equation.In a thermochemical equation, the coefficients represent moles and can

    therefore be fractional. The following is an example

    IB Data booklet > -227 kj mol -1

    Ethanol (C2H5OH) is made from the elements (C (Graphite)) and hydrogen

    (H2(g)) and oxygen (O2(g))

    __C (graphite) + __H2(g) +___O2(g)-----------> C2H5OH(I) H=227kj mol -1

    Balance the C, H and 0

    2C (graphite) + 3H2(g) +1/2 O2(g)-------------- C2H5OH(I) H=227 kj mol -1

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    Questions!

    1 .Use the table of standard enthalpies of formation at 25C to calculate

    enthalpy change for the reaction

    4NH3(g) + 5O2(g) 6H2O(g) + 4NO(g)

    2 .Write the thermodynamically equation for the standard enthalpy of

    formation of propanone enthalpy change

    CH3COCH3

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    Answers!

    1.1031.76 kJ mol1

    2. 3C (Graphite) + 3H2(g) + 1/2O2(g)-----------CH3COCH3

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    Standard enthalpy change of

    Combustion

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    What is standard enthalpy change of

    combustion?

    The standard enthalpy of combustion is the

    enthalpy change that occurs when one mole of

    substance burns completely under the standard

    conditions of 25 and 1 atm.

    Eg: C6H14(l) + 9O2(g) 6CO2(g) + 7H2O(l)

    The standard enthalpy of combustion is always negative

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    Exercise!

    Write down the enthalpy of combustion equations forthe following reactions!

    Methane

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

    Ethanol

    C2H5OH(l) + 3O2(g) CO2(g) + 2H2O(l)

    Propane

    C3H8(g)+ 5O2(g) 3CO2(g) + 4H2O(l)

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    Calculating standard enthalpy change

    Hreaction

    Hc(products)Hc(reactants)

    ProductsReactants

    Combustion Products

    H

    c(reactants) = H

    (products) + H

    H = Hc(reactants) - H

    c(products)

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    Question!

    May 2010 Paper 1 TZ 2B

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    Question!

    Give an equation for the formation of glucose.6C(graphite)+ 6H2(g) + 3O2(g) C6H12O6(s)

    Calculate the enthalpy of formation of glucose

    C: H

    = -394 Kjmol-1 H

    2:H

    = -286 Kjmol-1

    C6H12O6: H

    = -2803 Kjmol-1

    H = Hc(reactants) - H

    c(products)

    H = ( 6(-394) + 6(-286) + 3(0) ) - (-2803)= -1277 Kjmol-1

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    Question!

    Calculate the enthalpy change for the following reaction!

    C(s, graphite) C(s, diamond)

    C(s, graphite) + O2(g) CO2(g) H

    = -393 Kjmol-1

    C(s, diamond) + O2(g) CO2(g) H

    = -395 Kjmol-1

    Solution:

    C(s, graphite) + O2(g) C(s, diamond) + O2(g)

    CO2(g)

    H

    -395 Kjmol-1-393 Kjmol-1

    -393 Kjmol-1= -395 Kjmol-1+ H

    H = +2 Kjmol-1

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    Question

    May 2008 Paper 1 TZ 1A

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    Question

    Nov 2007 Paper 1 D

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    Comparison

    Hreaction

    Hc(products)Hc(reactants)

    ProductsReactants

    Combustion Products

    Standard enthalpy of combustion Standard enthalpy of formation

    H = Hc(reactants) - H

    c(products)

    Hreaction

    Hf(products)

    Hf(reactants)

    ProductsReactants

    Elements

    H = Hf(products) - H

    f(reactants)

    Its not CPR in chem! Its CRP! Think First Price

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    Question!

    May 2010 Paper 1 TZ 2A

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    Questions!

    May 2010 Paper 1 TZ1C

    May 2010 Paper 1 TZ1A

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    Question!

    Nov 2009 Paper 1 TZ1C

    Answers to all MCQ questions is the last letterin the identification of the paper from which the

    question was taken! :)

Chem Presentation Enthalpy

Jun 02, 2018

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Documents

  • 8/11/2019 Chem Presentation Enthalpy

    1/28

    Chapter 15 - Standard enthalpy

    change of a reaction

  • 8/11/2019 Chem Presentation Enthalpy

    2/28

    Standard Enthalpy ChangesHana Amir and Madeley

  • 8/11/2019 Chem Presentation Enthalpy

    3/28

    Definition

    Any reaction that depends on temperature,

    pressureandstate

    The enthalpy change happens when all

    reactants and products are in their standard

    state.

  • 8/11/2019 Chem Presentation Enthalpy

    4/28

    Some enthalpies

    Enthalpy of reaction

    Enthalpy of formation

    Enthalpy of neutralization

    Enthalpy of hydration

    Enthalpy of combustion

    Enthalpy of solution Enthalpy of atomization

  • 8/11/2019 Chem Presentation Enthalpy

    5/28

    Standard Enthalpy Changes of

    ReactionDefinition:

    The heat change when molar quantities of reactants asspecified by the chemical equation react to form productsat standard conditions

    It depends in the physical state of reactants and theproducts and the conditions under which the reactionoccurs.

    Standard conditions are: 298K (25oC) and 1.00*105Pa

  • 8/11/2019 Chem Presentation Enthalpy

    6/28

    Standard Enthalpy change of

    Formation

    Enthalpy change that occurs when one mole of substanceisformed from its elements in the standard state under

    standard conditions.

    Standard conditions

    Temperature: 298K (25o C)

    Pressure: 1.00*105 Pa

    2C(graphite) + 3H2(g) + 1/2O2(g)-------> C2H5OH (I)

  • 8/11/2019 Chem Presentation Enthalpy

    7/28

    All elements in their standard states (oxygengas, solid carbon in the form of graphite, etc.)

    have a standard enthalpy of formation of zero,

    as there is no change involved.

    Eg:

    O(g) + O(g) ---->O2(g)

  • 8/11/2019 Chem Presentation Enthalpy

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    Energy Cycle

    Hreaction

    Hf(products)Hf(reactants)

    ProductsReactants

    Elements

  • 8/11/2019 Chem Presentation Enthalpy

    9/28

    From the diagram we get:

    The chemical change elements to products

    can either occur directly or indirectly .

    The Total enthalpy change must be the same for both routes.

    Hf (Products)=Hf (Reactants) +Hreaction

    This gives the general expression of:

    Hreaction =Hf (Products)- Hf (Reactants)

  • 8/11/2019 Chem Presentation Enthalpy

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    Example

    Calculate the enthalpy change for the reaction:

    C3H8(g) + 5O2(g)-------- 3CO2(g) + 4H2O(g)

    Standard change of formation: Hf /kJ mol-1

    C3H8(g) : -105

    CO2(g): -394H2O(l): -286

  • 8/11/2019 Chem Presentation Enthalpy

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    Steps:

    Write down the equation with the corresponding

    enthalpies of formation underneath:

    C3H8(g) + 5O2(g)-------- 3CO2(g) + 4H2O(g)

    (-105) 0 3(-394) 4(-286)

    Note: The standard enthalpies of formation are given in

    per mole , hence, they should be multiplied by the

    numbers of moles in the balance equation .

  • 8/11/2019 Chem Presentation Enthalpy

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    Hreaction =Hf (Products)- Hf (Reactants)

    Hreaction =3(-394) + 4(-286) -(-105)

    = -2221 KJ mol-1

  • 8/11/2019 Chem Presentation Enthalpy

    13/28

    Thermochemical Equations

    Balanced chemical equation for a reaction including the enthalpy of the reaction

    shown immediately after the equation.In a thermochemical equation, the coefficients represent moles and can

    therefore be fractional. The following is an example

    IB Data booklet > -227 kj mol -1

    Ethanol (C2H5OH) is made from the elements (C (Graphite)) and hydrogen

    (H2(g)) and oxygen (O2(g))

    __C (graphite) + __H2(g) +___O2(g)-----------> C2H5OH(I) H=227kj mol -1

    Balance the C, H and 0

    2C (graphite) + 3H2(g) +1/2 O2(g)-------------- C2H5OH(I) H=227 kj mol -1

  • 8/11/2019 Chem Presentation Enthalpy

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    Questions!

    1 .Use the table of standard enthalpies of formation at 25C to calculate

    enthalpy change for the reaction

    4NH3(g) + 5O2(g) 6H2O(g) + 4NO(g)

    2 .Write the thermodynamically equation for the standard enthalpy of

    formation of propanone enthalpy change

    CH3COCH3

  • 8/11/2019 Chem Presentation Enthalpy

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    Answers!

    1.1031.76 kJ mol1

    2. 3C (Graphite) + 3H2(g) + 1/2O2(g)-----------CH3COCH3

  • 8/11/2019 Chem Presentation Enthalpy

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    Standard enthalpy change of

    Combustion

  • 8/11/2019 Chem Presentation Enthalpy

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    What is standard enthalpy change of

    combustion?

    The standard enthalpy of combustion is the

    enthalpy change that occurs when one mole of

    substance burns completely under the standard

    conditions of 25 and 1 atm.

    Eg: C6H14(l) + 9O2(g) 6CO2(g) + 7H2O(l)

    The standard enthalpy of combustion is always negative

  • 8/11/2019 Chem Presentation Enthalpy

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    Exercise!

    Write down the enthalpy of combustion equations forthe following reactions!

    Methane

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

    Ethanol

    C2H5OH(l) + 3O2(g) CO2(g) + 2H2O(l)

    Propane

    C3H8(g)+ 5O2(g) 3CO2(g) + 4H2O(l)

  • 8/11/2019 Chem Presentation Enthalpy

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    Calculating standard enthalpy change

    Hreaction

    Hc(products)Hc(reactants)

    ProductsReactants

    Combustion Products

    H

    c(reactants) = H

    (products) + H

    H = Hc(reactants) - H

    c(products)

  • 8/11/2019 Chem Presentation Enthalpy

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    Question!

    May 2010 Paper 1 TZ 2B

  • 8/11/2019 Chem Presentation Enthalpy

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    Question!

    Give an equation for the formation of glucose.6C(graphite)+ 6H2(g) + 3O2(g) C6H12O6(s)

    Calculate the enthalpy of formation of glucose

    C: H

    = -394 Kjmol-1 H

    2:H

    = -286 Kjmol-1

    C6H12O6: H

    = -2803 Kjmol-1

    H = Hc(reactants) - H

    c(products)

    H = ( 6(-394) + 6(-286) + 3(0) ) - (-2803)= -1277 Kjmol-1

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    Question!

    Calculate the enthalpy change for the following reaction!

    C(s, graphite) C(s, diamond)

    C(s, graphite) + O2(g) CO2(g) H

    = -393 Kjmol-1

    C(s, diamond) + O2(g) CO2(g) H

    = -395 Kjmol-1

    Solution:

    C(s, graphite) + O2(g) C(s, diamond) + O2(g)

    CO2(g)

    H

    -395 Kjmol-1-393 Kjmol-1

    -393 Kjmol-1= -395 Kjmol-1+ H

    H = +2 Kjmol-1

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    Question

    May 2008 Paper 1 TZ 1A

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    Question

    Nov 2007 Paper 1 D

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    Comparison

    Hreaction

    Hc(products)Hc(reactants)

    ProductsReactants

    Combustion Products

    Standard enthalpy of combustion Standard enthalpy of formation

    H = Hc(reactants) - H

    c(products)

    Hreaction

    Hf(products)

    Hf(reactants)

    ProductsReactants

    Elements

    H = Hf(products) - H

    f(reactants)

    Its not CPR in chem! Its CRP! Think First Price

  • 8/11/2019 Chem Presentation Enthalpy

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    Question!

    May 2010 Paper 1 TZ 2A

  • 8/11/2019 Chem Presentation Enthalpy

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    Questions!

    May 2010 Paper 1 TZ1C

    May 2010 Paper 1 TZ1A

  • 8/11/2019 Chem Presentation Enthalpy

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    Question!

    Nov 2009 Paper 1 TZ1C

    Answers to all MCQ questions is the last letterin the identification of the paper from which the

    question was taken! :)