Standard Enthalpy Changes of Reactions

standard enthalpy changes during reactions

1. Standard Enthalpy Changes XI FDC CHEMISTRY CHAPTER 11 SIDRA JAVED

2. Relationship between Internal Energy of a system and Thermal Energy at constant Temperature and Pressure

3. Enthalpy According to the first law of Thermodynamics: ΔE = q p + PΔV Where q p is the heat transferred to the system from the surrounding PΔV is work done by the surrounding on system at constant pressure

4. If the system does work on surrounding ΔE = q p - PΔV Rearranging the equation: q p = ΔE + PΔV q p is the heat absorbed or evolved by the reaction at constant pressure is also known as enthalpy of reaction ΔH

5. Enthalpy ΔH = q p = ΔE + PΔV Thermal energy for a reaction is equal to change in internal energy of the system plus pressure volume work done by the system at constant pressure This thermal energy is also called enthalpy change of reaction

6. Enthalpy of a reaction is defined as the system's internal energy plus product of its pressure and volume H = E + PV

7. Enthalpy at constant volume: In case of solids and liquids reactants, there is no appreciable change in volume. Therefore, ΔV = 0 ΔH = ΔE + PΔV ΔH = ΔE + P(0) ΔH = ΔE

8. Standard enthalpy change Standard enthalpy change is the enthalpy change at constant pressure (1 atm) and constant temperature (25 o C) and is denoted by ΔH o

9. Relation between Enthalpy Change and heat of reaction

10. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure ΔH = Heat given off or absorbed during a reaction at constant pressure.

11. Standard states and standard enthalpy changes ΔH varies with conditions so standard values are used. Standard values are calculated when all substances are in their standard states.

12. The standard conditions The standard conditions are: 1 atm pressure for gases 1 atm pressure and 25 o C temperature for any element or compound in its stable physical state 1M concentration for any aqueous solution

13. Standard enthalpy of reaction ΔH r o It is the enthalpy change in a chemical reaction when reactants and products are in their standard states and their molar quantities are same as shown by balanced chemical equation. 2H 2(g) + O 2(g) → 2H 2 O (l) ΔH r o = -571.6 KJ CH 4(g) + 2O 2(g) → CO 2(g) + H 2 O (l) ΔH r o = -890.0 KJ

14. Standard enthalpy of formation ΔH f o It is defined as the enthalpy change that accompanies the formation of one mole of a compound from its elements with all substance in their standard states. H 2(g) + ½ O 2(g) → H 2 O (l) ΔH f o = -258.8 KJ mol -1 S (s) + O 2(g) → SO 3(g) ΔH f o = -395.2 KJ mol -1

15. Standard enthalpy of combustion ΔH c o It is defined as the enthalpy change when one mole of a substance is completely burnt in excess of oxygen under standard conditions. C (s) + O 2(g) → CO 2(g) ΔH c o = -393.5 KJ mol -1 CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O (l) ΔH c o = -285.8 KJ mol -1

16. Standard enthalpy of atomization ΔH at o It is defined as the enthalpy change when one mole of gaseous atoms are formed from its elements under standard conditions. ½ H 2(g) → H (g) ΔH at o = +218 KJ mol -1 ½ Cl 2(g) → Cl (g) ΔH at o = +121 KJ mol -1

17. Standard enthalpy of neutralization ΔH n o It is defined as the enthalpy change when one mole of H + ions from an acid combine with one mole of OH - ions from a base to form one mole of water under standard conditions. NaOH (aq) + HCl (aq) → NaCl (aq) + H 2 O (l) ΔH n o = -57.4 KJ mol -1 H + (aq) + OH - (aq) → H 2 O (l) ΔH n o = -57.4 KJ mol -1

18. Standard enthalpy of solution ΔH sol o It is defined as the enthalpy change when one mole of a substance is dissolved in so much solvent that further dilution results in no detectable heat change under standard conditions. NH 4 Cl (s) NH 4 + (aq) + Cl - (aq) ΔH sol o = +15.1 KJ mol -1 HCl (g) H + (aq) + Cl - (aq) ΔH sol o = + 75 KJ mol -1 H 2 O H 2 O

19. Bond Dissociation Energy The difference between B.D.E and bond energy determines whether the reaction absorbs or releases energy overall ΔH r =Σ B.D.E reactants - Σ B.E products The amount of energy required to break one mole of a particular bond to form neutral atoms is called Bond Dissociation Energy (B.D.E)

20. 2H 2(g) + O 2(g) → 2H 2 O (l) ΔH o =? 2 mol H 2 = 2 H-H bonds 1 mol O 2 = 1 O-O bonds 2 mol H 2 O = 4 O-H bonds ΔH r o =Σ B.D.E reactants - Σ B.E products Σ B.D.E = 2B.D.E of H 2 + 1B.D.E of O 2 Σ B.D.E = 2(436KJ) + 493.6KJ = 1365.6 KJ Σ B.E = 4 B.E of O-H bonds Σ B.E = 4 (460KJ) = 1840 KJ ΔH r o =Σ B.D.E reactants - Σ B.E products ΔH r o =1365.6 - 1840 = -474.4 KJ

21. End of Lesson