Thermochemistry

thermochemistry explained

1. Dr Gaur Assistant professor Chemistry THERMOCHEMISTRY B.Sc Part 2 Physical chemistry

2. TOPICS COVERED Definition of Thermochemistry Types of Thermochemical Reactions Kirchoff’s Equation Calorimetry Laws of Thermodynamics Bond Enthalpy

3. WHAT IS THERMOCHEMISTRY ? The study of heat absorbed or released during chemical reactions.

4. TYPES OF THERMOCHEMICAL REACTIONS Endothermic Reactions (Heat is absorbed) Exothermic Reactions (Heat is released)

5. HEAT OF REACTION Heat of reaction is the amount of heat absorbed or liberated when the reactants are converted into products by the balanced chemical equation. If the reaction takes place at constant pressure, heat of reaction is represented by ∆H and if the reaction takes place at constant volume, it is represented by ∆E H 2 (g) + Cl 2 (g) 2HCl (g) ∆H = -184.1 kilo joule Standard Heat of Reaction ( ∆H ) is the change in enthalpy when reaction takes place under standard conditions ie . at 25 o C and 1 atm pressure.

6. EFFECT OF TEMPERATURE ON HEAT OF REACTION OR KIRCHHOFF’S EQUATION If we consider a reaction at constant pressure Then, q p = ∆H and ∆H = H p – H R - = (C p ) p - (C P ) R = ∆C P  

7. Integrating this equation within proper limits, we get…. d∆H = ∆C p dT ∆H 2 - ∆H 1 = ∆Cp (T 2 - T 1 ) ∆H 2 - ∆H 1 ∫ ∆H1 ∆H2 ∫ T 1 T 2 T 2 -T 1 = ∆ Cp

8. ∆E T2 - ∆E T1 T 2 – T 1 ∆ C V = C V(products) - C v(Reactants) KIRCHHOFF’S EQUATION AT CONSTANT VOLUME = ∆ C v

9. HEAT OF FORMATION OR ENTHALPY OF FORMATION Heat of Formation of a compound is the change of enthalpy (heat absorbed or released) during the formation of 1 mole of the substance from its constituent elements. Standard Heat of Formation ( ∆ f H ) of a compound is the enthalpy change during the formation of 1 mole of substance when reaction takes place in standard conditions.

10. CALORIMETRY Process of measuring the amount of heat released or absorbed during a chemical reaction is called calorimetry.

11. Bomb Calorimeter CONSTANT VOLUME NO VOLUME CHANGE NO WORK ∆ E = q + w = q = q v The heat given out by a reaction is absorbed by water. Weighted reactants are placed inside the bomb and ignited. The energy is determined by measuring the increase in the temperature of the water and other parts.

12. Calculations Change in temperature before and after combustion = T 2 C - T 1 o C Therefore, Heat released = Q = C∆T = C (T 2 - T 1 ) Where, C = Total Heat Capacity Q = n C bomb ∆ T bomb + n C water ∆ T water = m S bomb ∆ T bomb + m S water ∆ T water Fuel taken = x gm n fuel = x/M mole = n mole n mole will release Q heat 1 mole will release Q/n =q v = ∆ E ∆ H = ∆ E + ∆ n g RT

13. Enthalpy of Combustion When 1 mole of substance is completely burnt, the heat evolved is known as heat of combustion CH 4 + 2O 2 CO 2 + 2H 2 O ( ∆ c H ) = - 890.3 Kilojoule

14. ENTHALPY OF NEUTRALIZATION H + + A - + B + + OH - (1 g Eq Strong acid) (1g Eq Strong base) B + A - + H 2 O (Salt) (Water) ∆H = - 13.7 kcal

15. If weak acid and strong base or strong acid and weak base or weak acid and weak base are mixed then observed heat of neutralization is less than 13.7 kcal. The reason for this lower value is that some of the heat released in the neutralization is also used in the ionization of weak acid or weak base or both.

16. HCl( aq ) + NaOH( aq ) NaCl( aq )+H 2 O(l) ∆H = - 13.7 kcal HNO 3 ( aq ) +NaOH( aq ) NaNO 3 ( aq ) +H 2 O(l) ∆H = - 13.7 kcal HCl( aq ) + NH 4 OH( aq ) NH 4 Cl( aq )+H 2 O(l) ∆H = - 12.3 kcal CH 3 COOH( aq )+NH 4 OH( aq ) CH 3 COONH 4 ( aq ) +H 2 O(l) ∆H = - 11.9 kcal

17. LAW OF THERMOCHEMISTRY (Hess’s Law of Constant Heat Summation) The law states that total enthalpy change during the complete course of chemical reaction is same weather the reaction takes place in one step or several steps. ∆H 1 +∆H 2 +∆H 3 = ∆H

18. Example of Hess’s Law Formation of Carbon dioxide by carbon can take place in two ways, but the enthalpy change is same from both the ways. In single step ii) In two steps

19. Single Step Carbon is directly converted to CO 2 (g) as…. C (s) + O 2 (g) CO 2 ∆H = -393.5 kJ

20. Two Steps C (s) + 1\2 O 2 CO (g) ∆H 1 = -110.5 kJ CO (g) + 1\2 O 2 CO 2 (g) ∆H 2 = -283.0 kJ ∆H = ∆H 1 + ∆H 2 = -393.5 kJ

21. Applications of Hess Law Resonance energy Observed heat of formation – Calculated heat of formation

22. Other application of Hess’s law of constant heat summation Determination of lattice energy (BORN-HABER CYCLE) A + (g) + B - (g) AB (1 mole) + Lattice Energy

23. BOND ENERGY AND BOND ENTHALPY The amount of energy required to break 1 mole of bond of a particular type between the atoms in the gaseous state under 1 atm pressure and the specified temperature is called bond dissociation energy.

24. EXAMPLES H H(g) 2H(g) ∆H = +433 kJ/mole H Cl(g) H(g) + Cl(g) ∆H = +431 kJ/mole Cl Cl(g) 2Cl(g) ∆H = +242.5 kJ/mole

25. The bond dissociation energy also depends on the type of bond and on the type of molecule in which the bond is present. Consider the dissociation of water molecules which consists of O-H bonds. The dissociation occurs in two steps. H 2 O(g) H(g)+OH(g) OH(g) O(g)+H(g)

26. The average of these two bond dissociation energies gives the value of bond energy of O-H Bond energy of O-H bond 497.8 + 428.5 2 = 463.15 kJ/mol

27. Applications of bond energies Heat of a reaction ∑Bond energies of reactants – ∑Bond energies of products Heat of resonance Experimental or observed heat of formation-calculated heat of formation

28. Thank You