Enthalpy: is the heat absorbed or released by a system at constant pressure.

It is impossible to measure enthalpy directly. Only changes in enthalpy at constant pressure can measured “ΔH”

ΔH = HfinalH - initial

It is impossible to measure enthalpy directly. Only changes in enthalpy at constant pressure can measured “ΔH”

ΔH = Hfinal - Hinitial

Units of heat energy. We will use joules (J) or kilojoules (kJ).Units of heat energy. We will use joules (J) or kilojoules (kJ).

qp is the heat at constant pressure

qp = ΔH( absorbed or realsed heat)

At constant pressure, the change in enthapy ΔH of the system is equall to the energy flow as heat .

This means that for a reaction studied at constant pressure, the flow of heat is a measure of the change in enthalpy for the system.

In general for a chemical reaction :

ΔH = Hproducts – Hreactants

Forexothermicreactions, Δ H is negative Forendothermicreactions, Δ H is positive

In general for a chemical reaction :

ΔH = Hproducts – Hreactants

For exothermic reactions, ΔH is negativeFor endothermic reactions, ΔH is positive

Example: ,When 1 mole of methane is burned at constant pressure

KJ of energy 890 is realesed as heat. Calculate ΔH for a process in which 5.8 g sample of methane is burned at?constant pressure

:Solution♦Convert 5.8 g of methane to mole

n= mass/ molar mass = 5.8/16=0.36 mol of methane

♦:At constant pressure ΔH = q pKJ/ mol CH 890 - =4

mol of CH 1 4 KJ 890 -

mol of CH 0.36 4 ?

ΔH = q p= 1 mol = - 320 KJ /))0.36 mol(*)KJ 890-((

Thus, when 5.8 g of CH 4is burned at constant pressure

ΔH = heat flow= - 320 KJ

Example: When 1 mole of methane is burned at constant pressure,

890 KJ of energy is realesed as heat. Calculate ΔH for a process in which 5.8 g sample of methane is burned at constant pressure?

Solution:♦Convert 5.8 g of methane to mole

n= mass/ molar mass = 5.8/16=0.36 mol of methane

♦At constant pressure: ΔH = qp = - 890 KJ/ mol CH4

1 mol of CH4 - 890 KJ

0.36 mol of CH4 ?

ΔH = qp = ))-890 KJ(*)0.36 mol((/ 1 mol = - 320 KJ

Thus, when 5.8 g of CH4 is burned at constant pressure

ΔH = heat flow= - 320 KJ

The following guidelines are helpful in writing and interpreting thermochemical equations:

1 -specify the physical states of all reactants and products, because they help determine the actual enthalpy changes

2-If we multiply both sides of a thermochemical equation by a factor n, then DH must also change by the same factor. Returning to the melting of ice

3-When we reverse an equation, we change the roles of reactants and products. Consequently, the magnitude of DH for the equation remains the same, but its sign changes.

SolutionSolution

Standard enthalpy of formation (ΔHfo)

the subscript “f” stands for formation

The superscript “°” represents standard-state 1 atm and 25°C

By convention, the standard enthalpy of formation of any element in its most stable form is zeroBy convention, the standard enthalpy of formation of any element in its most stable form is zero

the standard enthalpy of formation of a compound as the heat change that results when 1 mole of the compound is formed from its elements at a pressure of 1 atm.

the standard enthalpy of formation of a compound as the heat change that results when 1 mole of the compound is formed from its elements at a pressure of 1 atm.

Molecular oxygen (O 2 ) is more stable than the other allotropicform of oxygen, ozone (O 3 ), at 1 atm and 25°C. Thus,

DH°f (O2) = 0, but DH°f (O3) = 142.2 kJ/mol

graphite is a more stable allotropic form of carbon than diamond at 1 atm and 25°C, so we have ΔH°f (C, graphite) = 0 andΔH°f (C, diamond) = 1.90 kJ/mol

n denote the stoichiometric coefficients for the reactants and products,and ∑(sigma) means “the sum of ”.

Note that in calculations, the stoichiometric coefficients are just numbers without units

ΔH°f values can be determined by applying the direct method or the indirect method.

ΔH°f values can be determined by applying the direct method or the indirect method.

The Direct MethodThe Direct Method

graphite and O 2 are stable allotropic forms of the elements, it followsthat Δ H ° f (C, graphite) and Δ H ° f (O 2 , g ) are zero

Calculate ΔH0 for the following reaction:

2Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (l)

ΔHfo of Fe2O3, Al2O3 and Fe(l) = - 822.2, - 1669.8 and

12.40 kJ/mol

ΔHo = ∑ n ΔHfo (products) - ∑ n ΔHf

o (reactants)

ΔHo = [(ΔHfo (Al2O3))+ (2× ΔHf

o (Fe))]- [(2×ΔHfo (Al))+(ΔHf

o (Fe2O3))]

ΔHo = [ (-1669.8)+ (2×12.40)] – [2×(0)+(-822.2)] - = 822.8 kJ

SolutionSolution

The Indirect MethodThe Indirect Method

Δ H °f can be determined by an indirect approach, which is based on Hess’s law

♦ If the reaction proceeds too slowly, or side reactions produce substances other than the desired compound.

Hess’s law can be stated as follows: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

Hess’s law can be stated as follows: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

the standard enthalpy of formation of carbon monoxide (CO). We might represent the reaction as

burning graphite also produces some carbon dioxide (CO2 ), so we cannot measure the enthalpy change for CO directly as shown. So we must apply Hess’s law.

The alternative is to regard the reaction as taking place in two steps .

♦we reverse Equation (b) to get (c)

Notes:Notes:

♦If a reaction is reversed, the sign of ΔH is also reversed.A + B Z ΔH = +231 kJ

A + B Z ΔH = -231 kJ

♦Enthalpy change ( H) is proportional to the amount of reactants and products. If the coefficients in a balance equation are multiplied by an integer, the value of ΔH is multiplied by the same integer.

A + 1/2B 2Z ΔH = + 231 kJ 2A + B 4Z ΔH = + 462 kJ

Example :Calculate ΔH for the conversion of graphite to

diamond:Cgraphite (s) Cdiamond )s( ΔH =?

The combustion reaction are: Cgraphite )s( + O2)g( CO2)g( ΔH1 = - 394 kJ

Cdiamond )s( + O2)g( CO2)g( ΔH2 = - 396 kJ

Answer:By reversing reaction )2( and add to reaction )1( gives the required reaction:Cgraphite )s( + O2)g( CO2)g( ΔH1 = -394 kJ

CO2)g( Cdiamond )s( + O2)g( ΔH3 = +396 kJ

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Cgraphite )s( Cdiamond )s( ΔH = ΔH1 + ΔH3 = +2 kJ

Since ΔH is positive the reaction is endothermic.

Example: Determine ΔH for the following reaction: C 3graphite4H + (s) 2 (g) C3H8? = ΔH )g(

:The combustion reaction areC3H85 O + )g( 2 )g( 3CO 24 H + )g( 2O )l( ΔH= - 2220.1 KJ

CgraphiteO + )s( 2 )g( CO 2ΔH= - 393.5 KJ )g(

H2O ½ + )g( 2 )g( H 2O )l( ΔH= - 585.8 KJ

Solution ♦Reverse the first equation3CO2)g( + H 42O )l( C 3H8 + )g( O 52)g( ΔH= + 2220.1 KJ

♦Multiply the second equation by 3 3Cgraphite + )s( 3O2)g( CO 32ΔH= + 1180.5 KJ )g(

Multiply the third equation by 44H2 + )g( O 22)g( 4H2O )l(ΔH= + 2343.2 KJ

3Cgraphite4H + )s( 2 )g( C3H8)g(

Example: Determine ΔH for the following reaction:3 C graphite (s) + 4H2 (g) C3H8 )g( ΔH = ?

The combustion reaction are: C3H8 )g( + 5 O2 )g( 3CO2 )g( + 4 H2O )l( ΔH= - 2220.1 KJ

C graphite )s( + O2 )g( CO2 )g( ΔH= - 393.5 KJ

H2 )g( + ½ O2 )g( H2O )l( ΔH= - 585.8 KJ

Solution ♦ Reverse the first equation3CO2 )g( + 4 H2O )l( C3H8 )g( + 5 O2 )g( ΔH= + 2220.1 KJ

♦ Multiply the second equation by 33C graphite )s( + 3O2 )g( 3 CO2 )g( ΔH= + 1180.5 KJ

Multiply the third equation by 44H2 )g( + 2 O2 )g( 4H2O )l( ΔH= + 2343.2 KJ

3C graphite )s( + 4H2 )g( C3H8 )g(

Example: Use Hess’s law to determine ΔH for the reaction:C2H4 (g) + Cl2 (g) C2H4Cl2 )l( ΔH =?

The combustion reactions are4 HCl )g( + O2 )g( 2 Cl2 )g( + 2 H2O )l( ΔH = - 202.5 KJ

2 HCl )g( + C2H4 )g( + 1/2 O2 C2H4Cl2 )l( + H2O

ΔH = - 319.6 KJSolution

♦Reverse the first equation and divide by

Cl2 )g( + H2O )l( 2 HCl )g( + 1/2O2 )g ( ΔH = + 101.25 KJ

♦Sum the last equation and the second equation Cl2 )g( + H2O )l( 2 HCl )g( + 1/2O2 )g ( ΔH = + 101.25 KJ2 HCl )g( + C2H4 )g( + 1/2 O2 C2H4Cl2 )l( + H2O ΔH = - 319.6 KJ

C2H4 )g( + Cl2 )g( C2H4Cl2 )l( ΔH = - 218.3 KJ