Inorganic chemistry
 Standard enthalpy of formation and reaction.
 Calculating the standard enthalpy of reaction
 Direct method of calculating the standard
enthalpy of formation.
Assistance Lecturer Amjad Ahmed Jumaa
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1
Standard enthalpy of formation and reaction:
The standard enthalpy of formation (ΔHf° ) is defined as the heat
change that results when one mole of a compound is formed from its
elements at a pressure of (1 atm
Calculating the standard enthalpy of reaction
From standard enthalpies of formation, we can calculate the
standard enthalpy of reaction (ΔH°rxn). Consider the following
hypothetical reaction.
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aA + bB → cC + dD
Where, a,b,c and d are stiochiometric coefficients.
The standard enthalpy of reaction is given by:
ΔH° rxn = [ c ΔH°f ( C) + d ΔH°f (D) ] – [ a ΔH°f (A) + b ΔH°f (B)]
Where a,b,c and d all have the unit mol.
The equation can be written in the general form:
ΔH° rxn = ( (Σn ΔH°f (products) –( Σm ΔH°f (reactnts) )
Where (m) and (n) denote the stiochiometric coefficients for
the reactants and products ,and ( Σ ) (sigma) means '' the sum
of''.
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Example:
A reaction used for rocket engines is
N2H4 (l) + 2H2O2 (l) → N2 (g) +4H2O (l)
What is the standard enthalpy of reaction in kilojoules ? the
standard enthalpies of formation are:
ΔH°f [ N2H4 (l)] = 95.1 kJ.
ΔH°f [ H2O2(l)] = - 187.8 kJ.
ΔH°f [ H2O(l) ] = - 285.8 kJ.
Solution :
ΔH°rxn = ( (Σn ΔH°f (products) –( Σm ΔH°f (reactnts) )
ΔH°rxn = ΔH°f [N2 (g) ] +4ΔH°f [H2O(l)]-{ ΔH°f [N2H4(l)]+2[H2O2(l)]}
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Remember , the standard enthalpy of formation of any element
in its most stable form is zero. Therefore,
ΔH°f [ N2(g)] = 0.
ΔH°rxn= [ 0 + 4(-285.8 kJ)] –[ 95.1 kJ+ 2(- 187.8 kJ)] = 862.7 kJ.
Direct method of calculating the standard enthalpy of
formation
This method of measuring (ΔH°f) applies to compounds that
can be readily synthesized from their elements. The best way to
describe this direct method is to look at the following example.
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Example:
The combustion of sulfur occurs according to the following thermo
chemical equation:
S (rhombic) + O2(g) → SO2(g)
ΔH°rxn = - 296 kJ.
What is the enthalpy of formation of (SO2(g)).
Solution:
Step(1): recall that the enthalpy of reaction carried out under
standard-state conditions is given by
ΔH°rxn = ( (Σn ΔH°f (products) –( Σm ΔH°f (reactants) )
ΔH°rxn = [ΔH°f (SO2)(g) ] – [ΔH°f (S) + ΔH°f (O2 ) (g) ]
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Step(2): recall that the standard enthalpy of formation of any
element in its most stable form is zero. Therefore, ΔH°f of (S) =0
and ΔH°f of (O2 ) (g) = 0.
ΔH°rxn = [ΔH°f (SO2)(g) ] – [ΔH°f (S) + ΔH°f (O2 ) (g) ]
-296 kJ = [ΔH°f (SO2)(g) ] –[ 0+0]
ΔH°f (SO2)(g) = -296 kJ/ mol (SO2).
Note: you should recognize that this chemical equation as
written meets the definition of a formation reaction. Thus,
ΔH°rxn is ΔH°f of (SO2) (g).
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Indirect method of calculating the standard enthalpy of
formation, (Hess's law):
Hess's law, states that when reactants are converted to
products, the change in enthalpy is the same wither the reaction
takes place in one step on in a series of steps.
This means that if we can break down the reaction of interest
into a series of reactions for which (ΔH°rxn) can be measured , we
can calculate (ΔH°rxn) for the overall reaction. Let's look at an
example:
Example:
For the following heats of combustion with fluorine, calculate
the enthalpy of formation of methane , (CH4):
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(a) CH4 (g) + 4F2 (g) → CF4 (g) + 4HF (g)
(b) C(graphite) + 2F2(g) → CF4(g)
( c ) H2 (g) + F2 (g) → 2 HF (g)
ΔH°rxn= - 1942 kJ.
ΔH°rxn= - 933 kJ.
ΔH°rxn= - 542 kJ.
Solution:
Step (1): the enthalpy of formation of methane can be
determined from the following equation.
C(graphite) + 2H2 (g) → CH4(g)
ΔH°rxn= ?
If we can calculate (ΔH°rxn) for this reaction, we can
calculate ΔH°f (CH4) because the enthalpies of formation of (
C ) and ( H2(g) ) are zero. Why ? Remember that the standard
enthalpy of formation of any element in its most stable form
is zero.
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Step (2): since we want to obtain one equation containing only
(C,H2 , and CH4), we need to eliminate ( F2, CF4 , and HF), from
the first three equation (a-c). first, we note that equation (a)
contains methane (CH4), on the reactant side. Let's reverse (a)
to get (CH4) on the product side.
CF4 (g) + 4HF (g) →CH4 (g) + 4F2 (g) ΔH°rxn= + 1942 kJ.
Note : ΔH°rxn changed sign when reversing the direction of the
reaction.
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CF4 (g) + 4HF (g) → CH4 (g) + 4F2 (g) ΔH°rxn= + 1942 kJ.
C(graphite) + 2F2(g) → CF4(g)
ΔH°rxn= - 933 kJ.
2H2 (g) + 2F2 (g) → 4HF (g)
ΔH°rxn= - 1084 kJ.
C(graphite) + 2H2 (g) → CH4(g)
ΔH°rxn= - 75 kJ.
Step (3): since the above equation represents the synthesis of (CH4)
from its elements, we have.
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ΔH°rxn = [ΔH°f (CH4)] – [ΔH°f (C) + 2ΔH°f (H2)]
-75 kJ = [ΔH°f (CH4)] – [0 + 0]
ΔH°f (CH4) = -75 kJ / mol (CH4).
The first law of thermodynamics:
Applying the first law of thermodynamics:
The first law of thermodynamics states that energy can be
converted form one form to another, but cannot be created or
destroyed.
ΔEsys + ΔEsurr = 0
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Where:
The subscripts '' sys'' and '' surr'' denote system and surroundings,
respectively.
ΔE = q + w
Where:
ΔE is the change in internal energy of the system.
q is the heat exchange between the system and
surroundings.
w is the work done on (or by ) the system.
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Using the sign convention for thermo chemical processes, (q
is positive) for an endothermic process and (q is negative) for
an exothermic process. For work, (w is positive ) for (work
done on the system by the surroundings) and (w is negative
for work done by the system on the surroundings).
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Example:
A system does (975 kJ) of work on its surroundings while at the
same time it absorb ( 625 kJ) of heat. What is the change in energy,
ΔE, for the system.
Solution:
To solve this problem, you must make sure to get the sign
convention correct. The system does work on the surroundings; this
is an energy-depleting process.
w = - 975kJ.
The system absorbs (625 kJ) of heat. Therefore, the internal energy
of the system would increase.
q = + 625kJ.
ΔE = q + w = 625kJ + ( - 975kJ) = - 350 kJ.
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