Chem Presentation Enthalpy

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


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


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! :)

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