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THERMOCHEMISTRY- A

THERMOCHEMISTRY- A. By Dr. Hisham E Abdellatef. 2011-2012. THERMOCHEMISTRY. The study of heat released or required by chemical reactions or heat changes caused by chemical reactions. SURROUNDINGS. Surrounding: the part of the universe which surround the system.

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THERMOCHEMISTRY- A

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  1. THERMOCHEMISTRY- A By Dr. Hisham E Abdellatef 2011-2012

  2. THERMOCHEMISTRY The study of heat released or required by chemical reactions or heat changes caused by chemical reactions

  3. SURROUNDINGS • Surrounding: the part of the universe which surround the system. System: part of universe selected for the thermodynamic study. HEAT HEAT HEAT HEAT SYSTEM SYSTEM EXOTHERMIC ENDOTHERMIC

  4. Heat of Reaction • Heat of reaction: the change in energy which accompanies a chemical reaction.  H • Endothermic reaction: the reaction which is supplied by heat (absorb heat and  H is + ve). • Exothermic reaction: the reaction which is accompanied with evolution of heat (evolve heat and  H is - ve).

  5. System and Surroundings In chemical reactions, heat is often transferred from the “system” to its “surroundings,” or vice versa. • The substance or mixture of substances under study in which a change occurs is called thethermodynamic system (or simply thesystem.) • Thesurroundingsare everything outside of the thermodynamic system.

  6. Burning fossil fuels is an exothermic reaction Exothermic If E < 0, Efinal < Einitial cellular respiration of glucose

  7. Endothermic If E > 0, Efinal > Einitial Photosynthesis is an endothermic reaction (requires energy input from sun)

  8. Internal Energy E Kinetic energy (EK) Potential energy (EP) Energy due to motion Energy due to position (stored energy) What is Energy?

  9. Specific HEAT J°g-1C-1 • The amount of heat which required to raise the temperature of 1gof substance by 1°C. Molar heat capacity • The amount of heat which required to raise the temperature of one mole of substance by 1°C molar heat capacity = specific heat x M.wt.

  10. Heat of reaction: • The change in energy which accompanies a chemical reaction. • Heat of combustion: • The change in enthalpy (heat evolved) when one mole of the substance is completely burned in presence of excess oxygen. • Heat of formation: • The change in enthalpy (heat evolved) when one mole of the substance is formed from its elements their standard state. (t = 25 °C and P = 1 atm).

  11. Which has the larger heat capacity? HEAT CAPACITY • J°C-1Specific Heat x mass • The amount of heat which required to raise the temperature of the substance by 1°C.

  12. How do we relate change in temp. to the energy transferred? Heat capacity (J/oC) = heat supplied (J) temperature (oC) Heat Capacity = heat required to raise temp. of an object by 1oC

  13. Specific Heat Capacity (Cs) J / oC / g Heat capacity J / oC = = g Mass Specific heat capacity is the quantity of energy required to change the temperature of a 1g sample of something by 1oC

  14. Specific Heat Capacity How much energy is transferred due to T difference? The heat (q) “lost” or “gained” is related to a) sample mass b) change in T and c) specific heat capacity

  15. Specific Heat Capacity Substance Spec. Heat (J/g•K) H2O 4.184 Ethylene glycol 2.39 Al 0.897 glass 0.84 Aluminum

  16. Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?

  17. heat gain/lose = q = (sp. ht.)(mass)(∆T) Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? where ∆T = Tfinal - Tinitial q = (0.897 J/g•K)(25.0 g)(37 - 310)K q = - 6120 J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.

  18. James Joule 1818-1889 UNITS OF ENERGY 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. 1000 cal = 1 kilocalorie = 1 kcal But we use the unit called the JOULE 1 cal = 4.184 joules

  19. heat energy transferred work done by the system energy change FIRST LAW OF THERMODYNAMICS ∆E = q + w Energy is conserved! مَحْفُوظة

  20. heat transfer in (endothermic), +q heat transfer out (exothermic), -q w transfer in (+w) w transfer out (-w) SYSTEM ∆E = q + w

  21. 1st Law of Thermodynamics: • Energy can neither created not destroyed, only transformed from one form to another

  22. Exothermic Endothermic 6.2

  23. Enthalpy Diagrams • Values of DH are measured experimentally. • Negative values indicate exothermic reactions. • Positive values indicate endothermic reactions. An increase in enthalpy during the reaction; DH is positive. A decrease in enthalpy during the reaction; DH is negative.

  24. ExothermicExamples • Oxidation – wooden splint burning (giving off light, heat, CO2, H2O • Burning H2 in air, • body reactions, • dissolving metals in acid, • mixing acid and water, • sugar dehydration

  25. Endothermic Examples • Electrolysis (breaking water down into H2 and O2 by running electricity in it) • Photosynthesis, pasteurization, canning vegetables • 2 H2 + O2 2H2O + energy • 4 g + 32 g  36 g 136 600 cal • 2H2O + energy  2H2 + O2 • 36 g 136 600 cal  4g + 32 g

  26. Changes in Internal Energy • If E > 0, Efinal > Einitial • Therefore, the system absorbed energy from the surroundings. • This energy change is called endergonic.

  27. Changes in Internal Energy • If E < 0, Efinal < Einitial • Therefore, the system released energy to the surroundings. • This energy change is called exergonic.

  28. Thermochemical equations • factors which affects the quantity of heat evolved or absorbed during a physical or a chemical transformation. • Amount of the reactants and products • Physical state of the reactants and products • Temperature • Pressure

  29. Thermochemical equations It must essentially: • be balanced; • give the value of ΔE or ΔH corresponding to the quantities of substances given by the equation; • mention the physical states of the reactants and products . The physical states are represented by the symbols (s), (L), (g) and (aq) for solid, liquid, gas and aqueous states respectively.

  30. Example of thermochemical equation H2 + ½ O2 → H2O ΔH = -68.32 Kcal 1 mole of hydrogen reacts with 0.5 mole of oxygen, one mole of water is formed and 68.32 Kcal of heats evolved at constant pressure. But, not specify whether water is in the form of steam or liquid H2 (g) + ½ O2 (g)→ H2O (L) ΔH = -68.32 Kcal H2 (g) + ½ O2 (g)→ H2O (g) ΔH = -57.80 Kcal. Effect of temperature ?????

  31. Standard enthalpy change ΔH°. • The heat change at • 298 K and • one atmosphere pressure is called the standard heat change or standard enthalpy change. It is denoted by ΔH°.

  32. Enthalpy (H) of the reaction(Comes from Greek for “heat inside”) • the sum of internal energy and the product of this pressure and volume. H = E + PV • E is the internal energy, • P is the pressure and • V is the volume of the system. • It is also called heat content. • ΔH = H product – H reactants = Hp – Hr

  33. Calculation of ΔH from ΔE • When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) H = E + PV At constant pressure and temperature The enthalpy of a chemical is measured in kilojoules per mole (kJmol-1).

  34. At constant volume H = E + PV For solid and liquid H = E Liquids and solids are neglected in calculation of n n = moles of gaseous products - moles of gaseous reactants.

  35. In case of gases H = E + PV (I) ΔV =Δn x V • Δn = no of moles of products - no of moles of reactants PxΔV = PVxΔn (II) • But PV = RT (for one mole of gas) • Putting RT in place of PV in equation (II) we get PΔV = RTΔn • Substituting the value of P AV in equation (I) we get ΔH = ΔE + Δn RT

  36. R = 1.987 cal. (=2 cal.) or = 8.314 joules

  37. Example 1: • Calculate  E for the following reaction: 2 CO(g) + O2(g) → 2 CO2(g) Where  H = - 135272 cal. At 25°C Solution:  n = 2 -3 = -1H = E + nRT T(K) = 25 + 273 - 135272 cal. = E + (- 1x2x298) E = - 135272 + 596 = - 134676 cal.

  38. Example 2: Calculate E and H for vaporization of water at 100°C and 1 atm., the specific heat of vaporization of water under these conditions is 540 cal/g. Solution: H2O(l) ↔ H2O(vap) Sp. heat of vaporization = 540 cal/g. 1 mole H2O = 18g. H = molar heat of vaporization = 540 x 18 at constant pressure Endothermic reaction since H = + 9720 cal. n = 1 - 0 =1  H = E + nRT 9720 = E + 1 x 2 x 373  E = 8974 cal.

  39. Example 3: 6.4g of naphthalene C10H8 when burned under constant volume gave 123 KJ at 20°C, calculate E and H. Solution: at constant volume the heat of combustion is equal to E gave means exothermic C10H8(s) + 12 O2(g) → 10CO2(g) + 4H2O(l) 6.4g at constant volume → -123KJ 128g at constant volume →E E = = = -2460KJ = - 2460 x 103 J H = E + nRT = - 2460 x 103 + (- 2 x 8.3 x 293) = - 2460 X10-3 - 4863.8 J = - 2464863 J = - 2464.863 KJ

  40. Example • The heat of combustion of ethylene at 17° C and at constant volume is -332.19 kcal. Calculate the heat of combustion at constant pressure considering water to be in liquid state (R = 2 cal.). The chemical equation for the combustion of ethylene is C2H4 + 3 O2 = 2CO2(g) + 2H2O (1) 1 mole 3 moles 2moles negligible volume No. of moles of the products = 2 No. of moles of the reactants = 4 Δn =(2-4) =-2 ΔH = ΔE + Δn RT ΔH = -332.19 +[ 2 x I0-3x -2 x 290] =-333.3 kcal Given that ΔE=-332. 19 kcal. T= 273+17= 290k R=2cal=2xlO-3kcals.

  41. Example • The heat of combustion of carbon monoxide at constant volume and at 17° C is -283.3 Kj. Calculate its heat of combustion at constant pressure(R= 8.314 J degree-1 mole-1). CO(g) + ½ O2(g) →CO2(g) 1 mole ½ mole 1 mole No. of mles of products = 1 No. of moles of reactants =1.5 n = No. of moles of products - No. of moles of reactants =1-1.5 =-0.5 ΔH = ΔE + Δn x RT ΔH= -283.3 + (-0.5x (8.314x10-3) x 290] = - 283.3-1.20 =-284.5 KJ Heat of combustion of CO at constant pressure is -284.5 kJ. Given that : ΔE =-283.3 kJ T = (273+17) = 290 K. R = 8.314 J or 8.314x10-3 KJ

  42. Heat of combustion • The change in enthalpy (heat evolved) when 1 mole of a substance is completely burnt in presence of excess oxygen. Organic compound + O2(g) → CO2(g) + H2O(l) • The thermochemical equation must be balanced firstly.

  43. Example 4:  H for the combustion of liquid heptane C7H16 into CO2(g) and H2O(l) is - 1151 kcal/mole at 20°C calculate  E. Solution: C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(i) n = 7-11 = -4 From thermochemical equation  H =  E +  nRT E =  H - A nRT = - 1151 x 10-3 - (- 4x2x293) = - 1151 x 103 - (-2344) cal./mol. = -1151 x 103 + 2344 = -1148.656 x103 cal/mo

  44. Example 5: The heat of combustion of benzoic acid C6H5COOH, into CO2(g) and H2O(i) at constant pressure is - 78.2kJ/mole at 27°C calculate heat of combustion at constant volume? Solution: C6H5COOH(S) + 7.5 O2(g) → 7CO2(g) + 3H2O(i)  H = - 78.2 x 103J n = 7-7.5 = -0.5  H =  E + nRT  E =  H - nRT  E = - 73200 - (-0.5 x 8.3 x (27 + 273)) = -81935 J

  45. Try to solve: 3.2 g of naphthalene C10H8 solid when burnt in excess O2 gas into CO2(g) and H2O(l) under constant volume gives 1423 KJ at 20°C calculate  E and  H? (M.Wt = 128). Calculate ∆H of reaction?

  46. Heat of Formation : (Enthalpies of Formation) H°f • is the enthalpy change for the formation of one mole of the substance from its elements, at standard pressure (1 atm) and a specified temperature (25°C) H2(g) + ½ O2(g) → H2O(l) H°f = -285.8 KJ C (graphite) + 2H2(g) → CH4(g) Ho f = -74.9 KJ

  47. Example 6: Calculate the  H° of the reaction Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) given that: H°f CO2(g) = - 393.5 KJ/mol  H°f Fe2O3 = -822.2 KJ/mol H°f CO(g) = - 110.5KJ/mol Solution:  Ho = (products) - (reactants). = [(3X-393.5) + (2x0)] - [(lx-822.2) + (3x-110.5)]J = -1180.5 + 1153.7 = -26.8 KJ.

  48. Example 7: Give the following data: B2H6(g) + 6 H2O(l) → 2H3BO3(s) + 6 H2(g)  H° = - 493.4 KJ • H°f of H3BO3(S) is - 1088.7 KJ/mol, and •  H°f of H2O(l) is - 285.9 KJ/mol Calculate the standard enthalpy of formation of B2H6

  49. Solution: = 2 H°f H3BO3 - [ H°f B2H6 + 6 H°f H2O] - 493.4 KJ = 2x- 1088.7 KJ- [1x H°f B2H6 + 6X -285.9KJ] = - 2177.4KJ - [H°f B2H6 - 1715.4KJ] - 493.4KJ =- 462.0KJ -  HofB2H6 H°fB2H6 = + 31.4KJ/mol

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