Standard Enthalpies of Formation:
Enthalpy changes for reactions can be used to assign enthalpies values to
substances and to the ions of aqueous solutions.
Standard Enthalpies of Compounds from the Formation Reactions:
The standard enthalpy of formation of any substance is defined as the
enthalpy of that substance compared to the enthalpies of the elements from
which it is formed.
Thus, the enthalpy change for the reaction in which a compound formed
from its elements is, therefore, the standard enthalpy of formation of that
compound.
For e.g.
C (graphite) + O2 (g) ---à
CO2 (g)
The measured value for ∆H (i.e. the enthalpy change) at 250C is -393.51
kJ mol-1.
This value gives us the enthalpy of 1 mol of CO2 (g) compared with the
enthalpy of 1 mol of C (graphite) and 1 mol of O2 (g).
Thus, the standard enthalpy of formation of CO2 (g) at 250C is -393.51
mol-1.
kJ
The definition of the standard enthalpy of formation gives a value of zero for
the standard enthalpy of formation of each of the elements in its stable or
designated form.
i.e.
For e.g.
C (graphite)
+
O2 (g)
---à
CO2 (g)
∆H0f (C) = 0
∆H0f (O2) = 0
∆H0f (CO2)= -393.51 kJ mol-1
The standard enthalpy of formation is represented as ∆H f where the degree
sign denotes the standard conditions (i.e. Pressure = 1 bar, and Temperature
= 250C). The subscript f stands for formation.
0
Standard Enthalpies of Formation from Other Chemical Reactions:
Reactions other than those in which the elements give the compounds can be
used to deduce the standard enthalpy of formation of a compound.
Problem 1:
Calculate the standard enthalpy of formation of CH3OH (l) at 250C from the
∆H value of -128.12 kJmol-1 that would be obtained for the reaction in
which 1 mol of CH3OH is formed from hydrogen and carbon monoxide.
The standard enthalpy of formation of carbon monoxide is -110.52 kJmol-1.
2H2 (g)
+ CO (g) ---à
CH3OH (l)
∆H0f = 0
∆H0f = -110.52
∆H0f = ?
The change in the enthalpy for the above reaction is
∆H0 =
∆H0f (products) - ∆H0f (reactants)
but
∆H0 = -128.12 kJmol-1
therefore
∆H0 =
∆H0f (products) - ∆H0f (reactants)
-128.12 kJmol-1 = ∆H0f (CH3OH) - (-110.52 kJmol-1)
∆H0f (CH3OH) = -238.64 kJmol-1)
Problem 2:
A rocket engine, which produces benign products, depends on the oxidation
of hydrazine by nitrogen dioxide. The equation for the reaction is
2 N2H4 (g) + 2 NO2 (g) --à 3 N2 (g) + 4 H2O (g).
If the standard enthalpies of formation (∆H0f ) for N2H4, NO2, and H2O are
given as +95.35 kJmol-1, +33.10 kJmol-1, -241.83 kJmol-1 respectively.
Calculate the change in enthalpy for gaseous reactants.
∆H0
=?
2 N2H4 (g)
∆H0f = 2(+95.35)
+ 2 NO2 (g)
∆H0f = 2(+33.10)
à 3 N2 (g) +
∆H0f = 0
The change in the enthalpy for the above reaction is
∆H0 =
∆H0f (products) - ∆H0f (reactants)
but
∆H0f (products) = 0 + 4 (-241.83 kJmol-1)
= -967.32 kJmol-1
4 H2O (g)
∆H0f = 4 (-241.83)
∆H0f (reactants) = 2 (+95.35 kJmol-1) + 2 (+33.10 kJmol-1)
= 190.70 kJmol-1 + 66.20 kJmol-1
= 256.90 kJmol-1
therefore
∆H0 =
∆H0f (products) - ∆H0f (reactants)
= - 967.32 kJmol-1 - (+256.90 kJmol-1)
= - 1224.22 kJmol-1
Standard Enthalpies of Formation of Ions in Aqueous Solutions:
Many chemical reactions involve ions in aqueous solutions. Thus, the data
for standard enthalpies for the formation of pure substances can be used to
find the enthalpy change for any reaction involving pure substances or
aqueous ionic species.
For e.g
A chemical process involving hydrogen gas dissolved in a large amount of
water.
Since HCl is strong electrolyte, the solution contains hydrated H+ (i.e. H3O+)
and Cl- ions. No HCl molecules are present.
The reaction can be then represented as follows:
( H20)
HCl (g) ------à
+
-
H (aq) + Cl (aq)
The symbol (aq) implies that the ion is present in large amounts of water.
The process of dissolving hydrogen chloride in water is very exothermic. If
the reactions were carried out at a pressure of 1 bar and at 250C, the thermal
surroundings will gain 75.85 kJ of energy for each mole of HCl that is
dissolved.
Thus, the enthalpy change for the above reaction is
∆H0 = - 75.85 kJmol-1
The standard enthalpy of formation of HCl (g) is -92.31 kJmol-1.
The standard enthalpy of formation for the reaction can be written as:
( H20)
+
-
HCl (g)
------à
∆H0f = (-92.31)
H (aq)
+
∆H0f [H+ (aq)]
Cl (aq)
∆H0f [Cl- (aq)]
The change in the enthalpy for the above reaction is
∆H0 =
∆H0f (products) - ∆H0f (reactants)
but
∆H0f (products) = ∆H0f [H+ (aq)] + ∆H0f [Cl- (aq)]
and
∆H0f (reactant) = ∆H0f [HCl (g)] = -92.31 kJmol-1
therefore we can write
∆H0 =
∆H0f (products) - ∆H0f (reactants)
-75.85 kJmol-1 = ∆H0f [H+ (aq)] + ∆H0f [Cl- (aq)] - (-92.31
kJmol-1)
∆H0f [H+ (aq)] + ∆H0f [Cl- (aq)] = -167.16 kJmol-1
This gives a value of the standard enthalpy of formation of the pair of ions
H+ and Cl-, in aqueous solution.
The value for the standard enthalpy of formation of H+ ions in dilute solution
is set to be equal to zero.
Hence
∆H0f [H+ (aq)] = 0
Thus, we can now calculate the value for the standard enthalpy of formation
of Cl- ions.
∆H0f [H+ (aq)] + ∆H0f [Cl- (aq)] = -167.16 kJmol-1
∆H0f [Cl- (aq)] = -167.16 kJmol-1 - ∆H0f [H+ (aq)]
∆H0f [Cl- (aq)] = -167.16 kJmol-1 - 0
∆H0f [Cl- (aq)] = -167.16 kJmol-1
Problem 3:
Dissolving 1 mol of KCl in a large amount of water at 250C produces an
enthalpy change of + 17.23 kJmol-1. The standard enthalpy of formation of
KCl (s) is - 436.75 kJmol-1. The standard enthalpy of Cl- (aq) is -167,16
kJmol-1. Calculate the standard enthalpy of formation of the aqueous
potassium K+, ion in the aqueous solution.
Thus, the enthalpy change for the reaction is
∆H0 = + 17.23 kJmol-1
The standard enthalpy of formation for the reaction can be written as:
( H20)
+
-
KCl (s)
------à
∆H0f = (-436.75)
K (aq)
+
∆H0f [K+ (aq)]
Cl (aq)
∆H0f = -167.16
The change in the enthalpy for the above reaction is
∆H0 =
∆H0f (products) - ∆H0f (reactants)
but
∆H0f (products) = ∆H0f [K+ (aq)] + ∆H0f [Cl- (aq)]
= ∆H0f [K+ (aq)] + ( -167.16 kJmol-1)
and
∆H0f (reactant) = ∆H0f [KCl (s)] = -436.75 kJmol-1
therefore we can write
∆H0 =
∆H0f (products) - ∆H0f (reactants)
17.23 kJmol-1 = ∆H0f [K+ (aq)] -167.16 kJmol-1 - (-436.75
kJmol-1)
∆H0f [K+ (aq)] = -252.56 kJmol-1
Problem 4:
Calculate the enthalpy change for the reaction in which the addition of
carbon dioxide to an aqueous solution containing Ca2+ ions brings about the
precipitation of calcium carbonate. Use the standard enthalpy of formation
for Ca2+ (aq), CO2 (g), CaCO3 (s), and H20 (l) as -542.83 kJmol-1, -393.51
kJmol-1, -1206.92 kJmol-1, -285.83 kJmol-1 respectively.
The reaction can be written as:
Ca2+ (aq) + CO2 (g) + H20 (l) ------à CaCO3 (s) + H+ (aq)
∆H0f = -542.83
-393.51
-285.83 kJmol
-1206.92
0
∆H0 =
∆H0f (products) - ∆H0f (reactants)
= (-1206.92 kJmol-1) - (-1222.17 kJmol-1)
= + 15.25 kJmol-1