Enthalpy, Calorimetry, Hess's Law

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Enthalpy, Calorimetry, Hess's Law

  1. 1. ObjectivesTo know how to calculate enthalpiesTo know how to solve problems on calorimetryTo know how to solve problems on Hess’ Law
  2. 2. How to Calculate Enthalpies How?
  3. 3. Let’s redefine some terms• Enthalpy (H) is the sum of the internal energy of the system plus the product of the pressure of the gas in the system and its volume Hsys = Esys + PV If pressure is kept constant, we can arrive at: H sys = q (at constant pressure) Where: H = H final – H initial q --- is heat
  4. 4. Let’s redefine some terms• Enthalpy of Reaction ( H) is the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants H = nH products – mH reactants n where and m are the coefficients of the products and the reactants in the balanced equation
  5. 5. Remember• Enthalpies are usually computed at standard conditions (25oC = 298K)• Note that enthalpies of formation of elements at standard conditions are equal to zero
  6. 6. Let’s Calculate• Example 1: Calculate the enthalpy of the oxidation reaction of benzene (C6H6) given with the chemical equation: C6H6(l) + 4.5O2(g) = 6CO2(g) + 3H20(l) 1. Find the enthalpies of formation for all chemical components of the reaction using a Table of Heat Formation H C6H6(g) = 48.85 KJ/mol H O2(g) = 0 H CO2(g) = -393.509 KJ/mol H H2O (l) = -285.83 KJ/mol
  7. 7. C6H6(l) + 4.5O2(g) = 6CO2(g) + 3H20(l)2. Multiply each enthalpy value on the corresponding reaction coefficient and sum up the enthalpies of formation C6H6(l) + 4.5O2(g) H C6H6(g) = 48.85 KJ/mol H O2(g) = 0 H initial = 48.95 KJ/mol+ 4.5 x O H CO2(g) = -393.509 KJ/mol H H2O (l) = -285.83 KJ/mol H initial = 48.95 KJ/mol3. Do the same to the final reagents 6CO2(g) + 3H20(l) H final = 6 x (-393.509 KJ/mol) + 3 x (-285.83 KJ/mol) H final = -3218.544 KJ/mol
  8. 8. H initial = 48.95 KJ/mol H final = -3218.544 KJ/mol4. Subtract the enthalpy of formation of the initial reagents from the final reagent H reaction = H final – H initial H reaction = -3218.544 KJ/mol – 48.95 KJ/mol H reaction = -3267.494 KJ/mol
  9. 9. Some Important Enthalpy Changes1. Enthalpy Change of Combustion - the enthalpy change which occurs when one mole of the substance is completely burnt in oxygen under standard conditionsEg. C (graphite) + ½ O2 (g)  CO2 (g) C (graphite) + O2 (g)  CO2 (g) chemist’s shorthand: H c,m [(graphite)] = -393.5 kJ/mol
  10. 10. Some Important Enthalpy Changes2. Enthalpy Change of Formation Eg. The SECF of methane, CH4, refersone mole of - the enthalpy change when to the change: the compound is formed from its elements C (graphite) + 2 H2 (g) ---->CH4 (g) H = -74.8 kJ/mol under standard conditions H f,m [CH4(g)] = -74.8 kJ/mol - may also be called Heats of Formation.  3. Enthalpy Change of Atomisation Eg. C (graphite  C of an element is kJ/mol The SMECA (g) H = 716.7 the enthalpy change when one mole of its atoms in the gaseous 716.7is formed from H at,m [(graphite)] = state kJ/mol the element under standard conditions . * Atomisation is always endothermic.
  11. 11. Some Important Enthalpy Changes 4. Enthalpy Change of Fusion - The enthalpy change when 1 mole of solid is converted to one mole of liquid at its melting point at standard pressure H fus,m [(H20)] = 716.7 kJ/mol H = 6.01 kJ/mol
  12. 12. Some Important Enthalpy Changes5. Enthalpy Change of Vaporisation - The enthalpy change when 1 mole of liquid is converted to one mole of gas at its boiling point at standard pressure H vap,m [(H20)] = 716.7 kJ/mol H = 41.09 kJ/mol
  13. 13. Calorimetry
  14. 14. What is it? Calor (Latin) + metry (Greek) = Calorimetry Science of measuring “heat” + “to measure” = the amount of heatTwo types of calorimetry• 1. measurements based on constant pressure• 2. measurement based on constant volume
  15. 15. Other termsCalorimeter – the device used to measure heat of reactionHeat capacity – the amount of heat required to raise its temperature by a given amount – SI unit: J/K Formula: q= C T where: q - heat C - heat capacity T - change in temperature = Tf-Ti
  16. 16. Other terms• Specific heat capacity – gives the specific heat capacity per unit mass of a particular substance - SI unit: J/kgK Formula: q= mc T where: q - heat m - mass c - specific heat capacity T - change in temperature = Tf-Ti
  17. 17. Other terms• Molar enthalpy of a substance Formula: H= mc T n where: H - enthalpy change m - mass c - specific heat capacity T - change in temperature = Tf-Ti n - moles of substance
  18. 18. Hess’s Law of Heat SummationBy Germain Henri Hess
  19. 19. What is it?• Hess Law of Heat Summation states that the heat absorbed or released during a reaction is the same whether the reaction occurs in one or several steps• Rules1. Make sure to rearrange the given equations so that the reactants and products are on the appropriate sides of the arrows2. If you reverse equations, you must also reverse the sign of H3. If you multiply/divide equations to obtain a correct coefficient, you must also multiply/divide the H by this coefficient
  20. 20. Get ready now forsome brainexercises

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