Fundamentals of Thermo-Chemistry

2,594 views

Published on

Energy and Enthalpy
First Law of ThermoDynamics
Hess's Law
Kirchoff's Relation

Published in: Education
  • Be the first to comment

Fundamentals of Thermo-Chemistry

  1. 1. Thermochemistry Dr. Ruchi S. Pandey
  2. 2. Energy Changes and Thermochemistry <ul><li>Energy: capacity to do work </li></ul><ul><ul><li>Thermal energy is the energy associated with the random motion of atoms and molecules </li></ul></ul><ul><ul><li>Chemical energy is the energy stored within the bonds of chemical substances </li></ul></ul><ul><ul><li>Nuclear energy is the energy stored within the collection of neutrons and protons in the atom </li></ul></ul><ul><ul><li>Electrical energy is the energy associated with the flow of electrons </li></ul></ul><ul><ul><li>Potential energy is the energy available by virtue of an object’s position </li></ul></ul>Dr. Ruchi S. Pandey <ul><li>Heat is the transfer of thermal energy between two bodies that are at different temperatures. </li></ul><ul><li>Thermochemistry is the study of heat change in chemical reactions. </li></ul>
  3. 3. Some definitions and Conventions <ul><li>System: part of the universe we are interested in. </li></ul><ul><li>Surrounding: the rest of the universe. </li></ul><ul><li>Boundary: between system & surrounding. </li></ul><ul><li>Exothermic: energy released by system to surrounding. </li></ul><ul><li>Endothermic: energy absorbed by system from surroundings </li></ul><ul><li>Work ( w ): product of force applied to an object over a distance. </li></ul><ul><li>Heat ( q ): transfer of energy between two objects </li></ul>Dr. Ruchi S. Pandey
  4. 4. Energy and First Law of Thermodynamics <ul><li>Energy cannot be created or destroyed. </li></ul><ul><li>Energy of (system + surroundings) is constant. </li></ul><ul><li>Internal Energy: total energy of a system. </li></ul><ul><li>Involves translational, rotational, vibrational motions. </li></ul><ul><li>Cannot measure absolute internal energy. </li></ul><ul><li>Change in internal energy, </li></ul>Dr. Ruchi S. Pandey <ul><li>Mathematically the first law says, </li></ul><ul><li>Internal Energy is a State function : depends only on the initial and final states of system, not on how the internal energy is used. </li></ul>
  5. 5. Enthalpy <ul><li>Most chemical reactions occur at constant P, so the need to define enthalpy, </li></ul><ul><li>Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. </li></ul>Dr. Ruchi S. Pandey <ul><li>Enthalpy is also a state function, and an extensive property </li></ul><ul><li>When the process occurs at constant pressure, </li></ul><ul><li>We know that, </li></ul><ul><li>So we can write, </li></ul>
  6. 6. Enthalpy change for Chemical reactions <ul><li>For a chemical reaction, </li></ul>Dr. Ruchi S. Pandey H products < H reactants  H < 0 H products > H reactants  H > 0
  7. 7. Thermochemical Equations Dr. Ruchi S. Pandey <ul><li>The stoichiometric coefficients always refer to the number of moles of a substance </li></ul><ul><li>If you reverse a reaction, the sign of  H changes </li></ul><ul><li>If you multiply both sides of the equation by a factor n, then  H must change by the same factor n. </li></ul><ul><li>Change in enthalpy depends on state </li></ul>H 2 O( g )  H 2 O(l)  H = -44 kJ <ul><li>The physical states of all reactants and products must be specified in thermochemical equations. </li></ul>H 2 O ( s ) H 2 O ( l )  H = 6.01 kJ H 2 O ( l ) H 2 O ( s )  H = - 6.01 kJ 2H 2 O ( s ) 2H 2 O ( l )  H = 2 x 6.01 = 12.0 kJ H 2 O ( s ) H 2 O ( l )  H = 6.01 kJ H 2 O ( l ) H 2 O ( g )  H = 44.0 kJ
  8. 8.  H for some typical chemical reactions Dr. Ruchi S. Pandey
  9. 9. Dr. Ruchi S. Pandey Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (  H 0 ) as a reference point for all enthalpy expressions. The standard enthalpy of formation of any element in its most stable form is zero. Standard enthalpy of formation (  H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f  H 0 (O 2 ) = 0 f  H 0 (O 3 ) = 142 kJ/mol f  H 0 (C, graphite) = 0 f  H 0 (C, diamond) = 1.90 kJ/mol f
  10. 10. Dr. Ruchi S. Pandey Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end. The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn a A + b B c C + d D  H 0 rxn d  H 0 (D) f c  H 0 (C) f = [ + ] - b  H 0 (B) f a  H 0 (A) f [ + ]  H 0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  -
  11. 11. Hess’s Law <ul><li>If a reaction is carried out in a number of steps,  H for the overall reaction is the sum of  H for each individual step. </li></ul><ul><li>When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps </li></ul><ul><li>For example: </li></ul><ul><li>CH 4 ( g ) + 2O 2 ( g )  CO 2 ( g ) + 2H 2 O( g )  H = -802 kJ </li></ul><ul><li>2H 2 O( g )  2H 2 O( l )  H = - 88 kJ </li></ul><ul><li>CH 4 ( g ) + 2O 2 ( g )  CO 2 ( g ) + 2H 2 O( l )  H = -890 kJ </li></ul>Dr. Ruchi S. Pandey
  12. 12. Hess’s Law contd.. <ul><li>The enthalpy of solution (  H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. </li></ul>Dr. Ruchi S. Pandey  H soln = H soln - H components  H soln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
  13. 13. Heat of Dilution <ul><li>Enthalpy change when a solution containing 1 mol of solute is diluted from one conc to another. </li></ul><ul><ul><li>Differential heat of dilution : enthalpy change when 1 mole of solvent is added to a large volume of the solution of known conc. so that no appreciable change in the conc of the solution occurs. </li></ul></ul><ul><ul><li>Example: </li></ul></ul><ul><ul><li> KCl(s) + 20 H 2 O  KCl(20H 2 O)  H 1 =15.9kJ </li></ul></ul><ul><ul><li>KCl(s) + 200 H 2 O  KCl(200H 2 O)  H 2 =18.58kJ </li></ul></ul><ul><ul><li>Heat of dilution=  H 1 -  H 2 </li></ul></ul>Dr. Ruchi S. Pandey Heat of Hydration <ul><li>Enthalpy change when an anhydrous or a partially hydrated salt combines with the requisite amount of water to form a new hydrated stable salt. </li></ul><ul><ul><li>Example: CuSO 4 (s) + 5 H 2 O(l)  CuSO 4 .5H 2 O(s) </li></ul></ul><ul><ul><li> H is almost always negative (why?) </li></ul></ul>
  14. 14. <ul><li>Enthalpy change when 1 mol of a compound is completely burnt in requisite amount of oxygen and the products are formed in their stable form. </li></ul><ul><ul><li>e.g. CH 4 (g) + 2O 2 (g)  CO 2 (g) +2H 2 O(l)  H= -890.36 kJ </li></ul></ul><ul><ul><li>Advantages: </li></ul></ul><ul><ul><ul><li>Is easier to measure and so can be used to measure the heat of formation of compounds which may be difficult otherwise. </li></ul></ul></ul><ul><ul><ul><li>To measure calorific values of foods and fuels </li></ul></ul></ul>Heat of Combustion Dr. Ruchi S. Pandey Heat of Neutralization <ul><li>Enthalpy change when 1gm equivalent of an acid is neutralized by a base in dilute solutions. </li></ul><ul><ul><li>H + (aq) + OH - (aq)  H 2 O(l)  H= -57.32,  H ∞ = -55.84 </li></ul></ul><ul><li>Strong acid-strong base </li></ul><ul><ul><li>HCl/HNO 3 with NaOH/KOH/LiOH  H≈ -57.32 </li></ul></ul><ul><li>Weak acid-strong base or strong acid weak base </li></ul><ul><ul><li>Always less than -57.32. Why? </li></ul></ul><ul><ul><li>Some energy goes in to the complete dissociation of the weak acid/base  enthalpy of ionization </li></ul></ul><ul><ul><li>Examples: acetic acid and NaOH and HCl and NH 4 OH </li></ul></ul>Dr. Ruchi S. Pandey
  15. 15. <ul><li>Not all rxns are carried out under standard conditions and so it is important to know how enthalpy of a rxn changes with T </li></ul><ul><li>aA + bB  cC + dD </li></ul><ul><li>  rxn H = (cH f,C + dH f,D ) - (aH f,A + bH f,B ) </li></ul><ul><li>On varying the temperature at constant pressure, </li></ul><ul><li>We know that, </li></ul><ul><li>Now, </li></ul><ul><li>Rearranging the above equation, </li></ul><ul><li>On integrating at constt. Pressure, </li></ul>Effect of Temperature on heat of reactions Dr. Ruchi S. Pandey Kirchhoff’s Relation
  16. 16. Kirchhoff’s Relation <ul><li>To solve Kirchhoff’s relation we need to know the temperature dependence of C P </li></ul><ul><li>Case I: When C P is independent of T, the relation then becomes, </li></ul><ul><li>i.e., if temp. goes from T 1 to T 2 </li></ul><ul><li>Case II: When C P is dependent on T, </li></ul><ul><li>Then variation of C P with T can be written as, </li></ul><ul><li>Where, </li></ul><ul><li>Substituting in the Kirchhoff’s relation, we get, </li></ul><ul><li>Which on integration gives, </li></ul>Dr. Ruchi S. Pandey
  17. 17. Relation between energy and enthalpy <ul><li>We have discussed earlier that </li></ul><ul><li>Now, P  V=0 for solids or liquids </li></ul><ul><li>But, for gaseous reactants and products, P  V=nRT(assuming the gas to be an ideal gas) </li></ul><ul><li>Then, for a chemical reaction, we can write the enthalpy change to be, </li></ul><ul><li>Or, </li></ul><ul><li>Or, </li></ul>Dr. Ruchi S. Pandey
  18. 18. Adiabatic Flame Temperature <ul><li>Temperature which the system attains if the changes in the system are carried out under adiabatic conditions. </li></ul><ul><ul><li>Adiabatic conditions: when there is no heat exchange </li></ul></ul><ul><ul><li>Example, if an exothermic reaction like combustion, occurs adiabatically then the heat released raises the temperature of the products of combustion, sufficiently to produce a flame. </li></ul></ul><ul><ul><ul><li>So, Stage I : reactants change to products at constant pressure </li></ul></ul></ul><ul><ul><ul><li>Stage II : temperature of products goes from T 0 to T f </li></ul></ul></ul><ul><ul><ul><li>Since there is no heat exchange,  H=0 therefore, -  H I =  H II , so now the Kirchhoff’s relation can be written as, </li></ul></ul></ul><ul><ul><ul><li>If C P is independent of T, then, </li></ul></ul></ul><ul><ul><ul><li>Or, </li></ul></ul></ul><ul><ul><ul><li>Flame temperature can then be defined by rearranging the above equation, </li></ul></ul></ul>Dr. Ruchi S. Pandey

×