Chemical Energetics , Thermodynamics Standard State
Download as PPTX, PDF0 likes44 views
This document defines standard state as conditions of 1 atmosphere pressure and 298.15K temperature, where the standard molar enthalpy of each element in its most stable state is defined as zero. It explains that standard enthalpy of formation (Hof) is the heat change when 1 mole of a substance is formed from its elements in their standard states. This value can be used to calculate the standard enthalpy change (Horeaction) of a chemical reaction from the standard enthalpies of formation of the reactants and products. Two example problems are included to demonstrate calculating Horeaction using given standard enthalpies of formation.
![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) ]](https://image.slidesharecdn.com/chemicalenergetics02-240317032304-f3a29fe3/85/Chemical-Energetics-Thermodynamics-Standard-State-5-320.jpg)
![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)](https://image.slidesharecdn.com/chemicalenergetics02-240317032304-f3a29fe3/85/Chemical-Energetics-Thermodynamics-Standard-State-6-320.jpg)
More Related Content
Similar to Chemical Energetics , Thermodynamics Standard State (20)
More from Arts Commerce & Science College, Tryambakeshwar, Nashik (7)
Recently uploaded (20)
Chemical Energetics , Thermodynamics Standard State
- 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