(^) The standard enthalpy of formation
of a compound is the enthalpy change that
accompanies a reaction in which one mole
of pure compound in its standard state is
formed from its elements in their standard
states.
The formation of one mole of CH 4 in its
standard state from the elements carbon and
hydrogen in their standard states is represented
by
C(graphite)+ 2H 2 (g) CH 4 (g), ∆rH^0 = -74.8 kJ
or ∆fH^0 (CH 4 ) = -74.8 kJ mol-1
= [c ∆fH^0 (C) + d ∆fH^0 (D)] -
[a ∆fH^0 (A) + b ∆fH^0 (B)]
= ∑ ∆fH^0 (products) - ∑ ∆fH^0 (reactants)
(4.30)
4.10.7 Standard enthalpy of combustion
(∆cH^0 )
Consider the reaction
C 2 H 2 (g) +^5
2
O 2 (g) 2 CO 2 (g) + H 2 O(l),
∆rH^0 = -1300 kJ
In the above reaction, the standard enthalpy
change of the oxidation reaction, -1300 kJ
is the standard enthalpy of combustion of
C 2 H 2 (g).
The standard enthalpy of combustion of a
substnce is the standard enthalpy change
accompanying a reaction in which one mole
of the substance in its standard state is
completely oxidised.
Do you know?
(^) The reaction to form a
substance from its constituent
elements is hypothetical. It is not possible
to combine C and H 2 in the laboratory to
prepare CH 4. The enthalpy of reaction
for the formation of CH 4 can be obtained
indirectly by knowing the standard enthalpy
change for system. The value -74.8 kJ mol-1
corresponds to the hypothetical reaction.
4.10.6 Standard enthalpy of reaction from
standard enthalpies of formation
The standard enthalpies of formation
of compounds are used to determine standard
enthalpies of reactions.
Calculations of ∆rH^0 from ∆fH^0 of
compounds are based on the following.
i. Standard enthalpies of formation of an
element is zero.
∆fH^0 (H 2 ) = ∆fH^0 (Cl 2 ) = ∆fH^0 (C) = 0
ii. Standard enthalpy of formation of a
compound is equal to its standard enthalpy
∆fH^0 (compound) = H^0 (compound)
Consider the reaction
aA + bB cC + dD
Standard enthalpy of the reaction is given by
∆fH^0 = (cH^0 C+ dH^0 D) - (aH^0 A + bH^0 B)
Problem 4.10
Calculate standard enthalpy of reaction,
2C 2 H 6 (g) +7O 2 (g) 4 CO 2 (g) + 6 H 2 O(l)
Given that
∆fH^0 (CO 2 )= -393.5 kJ mol-1,
∆fH^0 (H 2 O)= -285.8 kJ mol-1 and
∆fH^0 (C 2 H 6 ) = -84.9 kJ mol-1
Solution -
∆rH^0 = ∑ ∆fH^0 (products) - ∑ ∆fH^0 (reactants)
= [4 ∆fH^0 (CO 2 ) + 6 ∆fH^0 (H 2 O)]
- [2 ∆fH^0 (C 2 H 6 ) + 7 ∆fH^0 (O 2 )]
= [4 mol × (-393.5 kJ mol-1) + 6 mol ×
(-285.8 kJ mol-1)] - [2 mol × (-84.9 kJ mol-1) + 0]
= -1574 kJ - 1714.8 kJ + 169.8 kJ
= -3119 kJ