FYBSC, Pune university

1. Standard State
F. Y. B. Sc (CBCS Pattern 2019)
Sem.: I Paper : I
Mr. M. K. Jopale
Arts, Commerce & Science College, Tryambakeshwar (Nashik)
Email ID : manohar@mvptryambakcollege.ac.in

2. Standard State
• It is define as the standard molar enthalpy of every element at 1
atm pressure and 298.15 K in the most stable state is taken as
ZERO
• These condition of 1 atm pressure and 298.15 K are called
standard state condition
• It is not possible to determine absolute enthalpy of substances
• For examples H2 as gas, Br2 as liquid & I2as solid their standard enthalpy
are taken as zero at 1 atm pressure and 298.15 K temperature.
• In case of allotropes we consider most stable form under the
condition of standard state

3. Enthalpy of formation and standard enthalpy of
formation
• Heat of Formation(or enthalpy of formation {∆Hf} ) : The enthalpy of
formation of a substance is defined as the heat change that takes
place when 1 mole of substance is formed from elements under given
condition of T & P. it is represented by symbol ∆Hf
• Standard enthalpy of formation (∆H
o
f ) of a substance is defined as the heat
change accompanying the formation of 1 mole of the substancein the
standard state from its elements, also taken in the standard state (i.e.,
298 K & 1 atmospheric pressure) it is represented by symbol ∆H
o
f

4. • Examples:
C (s) + O2 (g) CO2, ∆Ho
f = -393.5 kJ
When 1 mole of CO2 (g) is formed from its elements, C (s) and O2 (g) in
its standard states, 393.5 kJ of heat is produced. Hence the heat of
formation of gases CO2 is 393.5 kJ mole-1
Importance of standard enthalpies of formation :
By knowing standard enthalpies of formation of the different compounds involved
in chemical reaction,
The standard enthalpy change of the given reaction can be obtain using the
formula
∆H
o
reaction = {Sum of the standard heats of formation of products} –
{Sum of the standard heats of formation of reactants }
∆H
o
reaction = ∑ ∆H
o
f (Products) - ∑ ∆H
o
f (Reactants) (1)

5. • Thus for general reaction
aA + bB cC + dD ….. (2)
• ∆H
o
reaction = [c ∆H
o
f (C) + d ∆H
o
f (d)] – [a ∆H
o
f (A) + b ∆H
o
f (B) ]
Examples : 1.
For reaction,
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
Than,
∆H
o
reaction = [∆H
o
f (CO2) + 2 ∆H
o
f (H2O)] – [ ∆H
o
f (CH4) + 2 ∆H
o
f (O2) ]

6. Problems
• Calculate ∆H
o
for reaction
CO2 (g) + H2 (g) CO (g) + H2O (l)
• Given that ∆H
o
f for CO2 (g), H2 (g),CO (g), and H2O (l) are -393.5, 0.0, -111.3
And -241.8 kJ mol-1 respectively.
Solution : Reaction
Step 1 : CO2 (g) + H2 (g) CO (g) + H2O (l)
Step 2:
∆H
o
reaction = [∆H
o
f (CO2) + ∆H
o
f (H2)] – [ ∆H
o
f (CO) + 2 ∆H
o
f (H2O) ]
= [(-111.3)+ (-241.8)] – [(-393.5) + (0)] = -353.1 + 393.5
∆H
o
reaction = + 40.4 kJ ( Enthalpies for every element in standard state
is assumed as ZERO)

7. Problem 2.
Calculate ∆H
o
for formation of
C2H5OH (g) + 3O2 (g) 2CO2 (g) + 3H2O (l)
Given that ∆H
o
f for CO2 (g), C2H5OH (g), and H2O (l) are -393.3,
-277.0, & -285.8 kJ mole-1 respectively.
Answer : -1367.4 kJ
Problems