Chemistry
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Name:
§ 06.06.a1
45 min
Date:
Energy: Standard Enthalpy of Formation
Most Important Ideas:
(
1.
)
(
& Reaction
)
2. Example: for the equation aA + bB  cC + dD:
( )
( )
( )
( )
Objective
The purpose of this activity is to calculate the
of a substance using the direct method. This
method is used when the
values are known for the compounds involved in the reaction.
Reference
Chang and Overby. Chemistry: The Essential Concepts, 6th ed. pp.196-199.
Background
The standard enthalpy of formation (
) for a substance is the ‘sea level’ reference for
determining its enthalpy, and indicates the relative stability of a substance. The superscript “0”
represents the standard-state conditions. Not to be confused with STP (0oC and 1 atm),
standard-state conditions are 1 atm and 25oC.
Because we can only know the relative enthalpy of a substance (not its absolute enthalpy),
chemists have assigned
for elements in their most stable forms at standard-state
conditions against all other substances are measured. For example, hydrogen exists as H2 at
1.
(
)
(
)
standard-state conditions so
and
There are two methods used to determine the standard heat of formation (
) of a substance:
1. Direct Method:
values are known for the compounds involved, and
2. Indirect Method (a.k.a., Hess’s law): the overall reaction is broken down in to a series of
steps with know enthalpies.
The direct method uses the difference between
In equation form:
(
of the products and
)
(
of the reactants:
)
Where:
  = sum
n & m = the stoichiometric coefficients for the products & reactants, respectively.
1
(
)
F:\330\330_sections\330_06_Energy\330_06_06_Standard Enthalpy_Formation_Reaction\330_06_06a1_Standard_Enthalpy_DIRECT_HW_KEY.docx (3/19/2014 2:06:00 PM)
Chemistry
§06.06.a1 Energy: Heat of Formation - Direct Method
p. 2
N.B. State symbols are very important for these calculations. (E.g.,
different from that of
for gas (steam) or
for solid (ice).
for liquid water is
Model
Problem
Calculate the
for glucose from the combustion reaction:
1C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)
Given:
(
)
(
, and
= –2821.8 kJ/mol
)
.
(
Why is
)
Solution
(
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(
(
(
(
))
(
(
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) (
)) (
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))
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?
Chemistry
§06.06.a1 Energy: Heat of Formation - Direct Method
p. 3
Problems
1. Which of the following standard enthalpy of formation values is not zero at 25°C? Na(s),
Ne(g), CH4(g), S8(s), Hg(l), H(g).
(6.45)
ANS
CH4(g) and H(g). All the other choices are elements in their most stable form
( H f  0 ). The most stable form of hydrogen is H2(g).
2. The
values of the two allotropes of oxygen, O2 and O3, are 0 and 142.2 kJ/mol,
respectively, at 25°C. Which is the more stable form at this temperature?
ANS
(6.46)
The standard enthalpy of formation of any element in its most stable form is zero. Therefore,
since
H f (O2 )  0, O2 is the more stable form of the element oxygen at this temperature.
3. Which is the more negative quantity at 25°C:
ANS
for H2O(l) or
for H2O(g)?
(6.47)
H2O(l)  H2O(g) Endothermic
H rxn  Hf [H2O( g )]  Hf [H2O(l )]  0
Hf [H2O( l )] is more negative since H rxn  0 .
You could also solve the problem by realizing that H2O(l) is the stable form of water at
25C, and therefore will have the more negative H f value.
4.
Predict the value of
(greater than, less than, or equal to zero) for these elements at 25°C:
(a) Br2(g) and Br2(l), (b) I2(g) and I2(s).
ANS
(6.48)
(a) Br2(l) is the most stable form of bromine at 25C; therefore, Hf [Br2 (l )]  0. Since Br2(g) is
less stable than Br2(l), Hf [Br2 ( g )]  0.
(b) I2(s) is the most stable form of iodine at 25C; therefore, Hf [I2 (s)]  0. Since I2(g) is less
stable than I2(s), Hf [I2 ( g )]  0.
Chemistry
§06.06.a1 Energy: Heat of Formation - Direct Method
p. 4
5. In general, compounds with negative values are more stable than those with positive
values. H2O2(l) has a negative
(–187.6 kJ/mol). Why, then, does H2O2(l) have a tendency
to decompose to H2O(l) and O2(g)?
ANS
(6.49)
2H2O2(l)  2H2O(l)  O2(g)
H2O2(l) has a tendency to decompose because H2O(l) has a more negative H f than
H2O2(l).
6. Suggest ways (with appropriate equations) that would enable you to measure the
values of Ag2O(s) and CaCl2(s) from their elements. No calculations are necessary.
(6.50)
ANS
Strategy: What is the reaction for the formation of Ag2O from its elements? What is the H f
value for an element in its standard state?
Solution: The balanced equation showing the formation of Ag2O(s) from its elements is:
2Ag(s) 
1
2
 Ag2O(s)
O2(g) 
Knowing that the standard enthalpy of formation of any element in its most stable form is
zero, and using Equation (6.18) of the text, we write:
H rxn   nHf (products)   mHf (reactants)
H rxn  [Hf (Ag2O)]  [2Hf (Ag) 
1 H (O )]
f
2
2
H rxn  [Hf (Ag2O)]  [0  0]
Hf (Ag2O)  Hrxn
In a similar manner, you should be able to show that Hf (CaCl 2 )  Hrxn for the reaction
 CaCl2(s)
Ca(s)  Cl2(g) 
7. Calculate the heat of decomposition for this process at constant pressure and 25°C:
(6.51)
CaCO3(s)  CaO(s) + CO2(g)
(CaCO3, s) = –1206.9 kJ/mol;
ANS
(CaO, s) = –635.6 kJ/mol;
(CO2, g) = –393.5 kJ/mol
H   [Hf (CaO)  Hf (CO2 )]  Hf (CaCO3 )
H  [(1)(635.6 kJ/mol)  (1)(393.5 kJ/mol)]  (1)(1206.9 kJ/mol)  177.8 kJ/mol
Chemistry
§06.06.a1 Energy: Heat of Formation - Direct Method
p. 5
8. The standard enthalpies of formation of ions in aqueous solutions are obtained by
arbitrarily assigning a value of zero to H+ ions; that is,
a. For the following reaction:
calculate
for the
([H+(aq)] = 0.
HCl(s)  H+(aq) + Cl–(aq)
(6.52)
= –74.9 kJ/mol
Cl– ions.
ANS
Strategy: The enthalpy of a reaction is the difference between the sum of the enthalpies of the
products and the sum of the enthalpies of the reactants. The enthalpy of each species
(reactant or product) is given by the product of the stoichiometric coefficient and the
standard enthalpy of formation, Hf , of the species.
Solution: We use the H f values in Appendix 2 and Equation (6.18) of the text.
H rxn   nHf (products)   mHf (reactants)
HCl(g)  H(aq)  Cl(aq)
H rxn  Hf (H )  Hf (Cl )  Hf (HCl)
74.9 kJ/mol  0  H f (Cl)  (1)(92.3 kJ/mol)
Hf (Cl  )   167.2 kJ/mol
b. Given that
for OH– ions is –229.6 kJ/mol, calculate the enthalpy of neutralization
when 1 mole of a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong
base (such as KOH) at 25°C.
ANS
The neutralization reaction is:
H(aq)  OH(aq)  H2O(l)
and,
H rxn  Hf [H2O(l )]  [Hf (H )  Hf (OH )]
Hf [H2O(l )]   285.8 kJ/mol (See Appendix 2 of the text.)
H rxn  (1)(285.8 kJ/mol)  [(1)(0 kJ/mol)  (1)(229.6 kJ/mol)]  56.2 kJ/mol