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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 •Surrounding: the part of the universe which surround the system. System: part of universe selected for the thermodynamic study. HEAT HEAT HEAT HEAT 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). 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 the thermodynamic system (or simply the system.) • The surroundings are everything outside of the thermodynamic system. Exothermic If E < 0, Efinal < Einitial cellular respiration of glucose Burning fossil fuels is an exothermic reaction Endothermic If E > 0, Efinal > Einitial Photosynthesis is an endothermic reaction (requires energy input from sun) What is Energy? Internal Energy E Kinetic energy (EK) Energy due to motion Potential energy (EP) Energy due to position (stored energy) 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. • 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). HEAT CAPACITY • J°C-1 Specific Heat x mass • The amount of heat which required to raise the temperature of the substance by 1°C. Which has the larger heat capacity? 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 Specific heat capacity is the quantity of energy required to change the temperature of a 1g sample of something by 1oC Specific Heat Capacity (Cs) J / oC / g = Heat capacity Mass = J / oC g Specific Heat Capacity How much energy is transferred due to T difference? The heat (q) “lost” or “gained” is related to a) b) sample mass change in T and c) specific heat capacity Specific heat capacity = heat lost or gained by substance (J) (mass, g)(T change, K) Specific Heat Capacity Substance Spec. Heat (J/g•K) H2O 4.184 Ethylene glycol 2.39 Al 0.897 glass 0.84 Aluminum 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? Specific heat capacity = heat lost or gained by substance (J) (mass, g)(T change, K) 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? heat gain/lose = q = (sp. ht.)(mass)(∆T) 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. 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 James Joule 1818-1889 FIRST LAW OF THERMODYNAMICS heat energy transferred ∆E = q + w work done by the system energy change Energy is conserved! َمحْ فُوظة heat transfer in (endothermic), +q heat transfer out (exothermic), -q SYSTEM ∆E = q + w w transfer in (+w) w transfer out (-w) • 1st Law of Thermodynamics: • Energy can neither created not destroyed, only transformed from one form to another Exothermic Endothermic 6.2 Enthalpy Diagrams • Values of H are measured experimentally. • Negative values indicate exothermic reactions. • Positive values indicate endothermic reactions. A decrease in enthalpy during the reaction; H is negative. An increase in enthalpy during the reaction; H is positive. Exothermic Examples • 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 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 Changes in Internal Energy • If E > 0, Efinal > Einitial –Therefore, the system absorbed energy from the surroundings. –This energy change is called endergonic. Changes in Internal Energy • If E < 0, Efinal < Einitial –Therefore, the system released energy to the surroundings. –This energy change is called exergonic. Thermochemical equations • factors which affects the quantity of heat evolved or absorbed during a physical or a chemical transformation. 1. Amount of the reactants and products 2. Physical state of the reactants and products 3. Temperature 4. Pressure Thermochemical equations It must essentially: 1.be balanced; 2.give the value of ΔE or ΔH corresponding to the quantities of substances given by the equation; 3.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. 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 ????? 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°. 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 Calculation of ΔH from ΔE • When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) H = E + PV At constant pressure and temperature The enthalpy of a chemical is measured in kilojoules per mole (kJmol-1). H = E + PV H = E For solid and liquid • At constant volume Liquids and solids are neglected in calculation of n n = moles of gaseous products - moles of gaseous reactants. • In case of gases H = E + PV (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 R = 1.987 cal. (=2 cal.) or = 8.314 joules 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. 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. 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 = = 128 x - 123 = -2460KJ = - 2460 x 103 J 6.4 H = E + nRT = - 2460 x 103 + (- 2 x 8.3 x 293) = - 2460 X10-3 - 4863.8 J = - 2464863 J = - 2464.863 KJ 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. 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 Given that : ΔE =-283.3 kJ T = (273+17) = 290 K. R = 8.314 J or 8.314x10-3 KJ 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. 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. 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 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 103 J n = 7-7.5 = -0.5 H = E + nRT E = H - nRT E = - 73200 - (-0.5 x 8.3 x (27 + 273)) = -81935 J • 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? 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) C (graphite) + 2H2(g) → CH4(g) H°f = -285.8 KJ Ho f = -74.9 KJ Hf o superscript o means standard state = 25oC and 1 atm pressure subscript f means formation from most stable elements 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. 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 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 The equations representing the variation of heat changes of reaction with temperature are known as Kirchoff's equations. VARIATION OF HEAT (OR ENTHALPY) OF REACTION WITH TEMPERATURE KIRCHOFF'S EQUATIONS. 1. At constant volume, internal energies of the reactants and products. ΔE=E2-E1 Differentiating this equation with respect to temperature at constant volume, we get KIRCHOFF'S EQUATIONS. heat capacities But we have already seen that heat capacities of the products and reactants Integrating between temperature T l and T2 , we have ΔE2- ΔE1 = Δ E2 -ΔE1 = Δ Cv (T2-T1) (3) KIRCHOFF'S EQUATIONS. 2. At constant pressure ΔH=H2-H1 And finely …. ΔH2- ΔH1= ΔCp (T2-T1) (6) Example 3 – 3 The heat of reaction ½ H2 + ½ Cl2 = HCl at 27 C° is -22.1 Kcal. Calculate the heat of reaction at 77°C. The molar heat capacities at constant pressure at 27° C for hydrogen, chlorine and HCl are 6.82, 7.70 and 6.80 cal mol-1, respectively. Here ½ H2 + ½Cl- → HCl Δ H = -22.1 Kcal Δ Cp = Heat capacities of products - Heat capacities of reactants = 6.80-[½(6.82) +½ (7.70)] T2 = 273 + 77 =350 K = 6.80 - 7.26 = -0.46 x 10-3 Kcal T1 = 273 + 27 = 300 K T2-T1= (350-300)K=50K ΔH2- ΔH1= ΔCp (T2-T1) ΔH2 - (-22.1) = (-0.46 x 10 -3) x 50= -21.123 Example 3 – 4 • Calculate the standard heat of formation of propane (C3H8) if its heat of combustion is 2220.2 KJ mol-1. The heats of formation of CO2 (g) and H2O (L) are - 393.5 and -285.8 KJ mol1 respectively. C3H8 (g) + 5 O2 (g) → 3CO2 (g) + 4 H2O (L) Δ Hc = 2220.2 KJ C (s) + O2 (g) → CO2(g) ΔHf=- 393.5 KJ H2(g)+½O2(g) →H2O (L) ΔHf=-285.8KJ We should manipulate these equations in a way so as to get the required equation 3 C (s) + 4 H2 (g) → C3H8 (g) ΔH=? Multiplying equation (ii) by 3 and equation (iii) by 4 and adding up we get 3C(s) + 3O2 (g) → 3 CO2 (g) ΔH = -1180.8 KJ 4H2 (g) +2O2(g) → 4H20(L) ΔH = -1143.2 KJ 3C(s) + 4H2 (g) + 502(g) → 3CO2(g) +4H2O(L) ΔH= - 2323.7 KJ subtracting equation (i) from equation (iv), we have 3C(s)+4H2(g) → C3H8(g) Δ H= -103.5 KJ Heat of transition • the change in enthalpy which occurs when one mole of an element changes from one allotropic form to another. • P white → P red • S monoclipic → S rhombic ΔH= -1.028 kcal ΔH=-0016 Kcal Hess's law of constant heat summation If a reaction is carried out in a series of steps, ΔH for the reaction is the sum of ΔH’s for each of the steps. A →B + q1 ΔH1= q1 B → C + q2 ΔH2 = q2 C → Z + q3 ΔH3 = q3 A →z + Q1 ΔH1 = -Q1 The total evolution of heat = q1 +q2 +q3 =Q2 According to Hess's law Q1= Q2 We can find H(a) by subtracting H(b) from H(c) Hess’s Law & Energy Level Diagrams Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed. Hess’s Law & Energy Level Diagrams Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed. Illustrations of Hess's law • (1) Burning of carbon to CO2 1st way: C(S) + 02(g) → CO2 (g) ΔH = -94.05kcal (-393.50 KJ) 2nd way: C(s)+½O2(g) →CO(g) ΔH =-26.42 kcal and CO2 (g) + ½ O2 (g) → CO2 (g) ΔH = -67.71 kcal Overall change C (s) + O2 (g) →CO2 (g) ΔH = -94.13 kcal Illustrations of Hess's law (2) Formation of sodium by Hydroxide from Na 1st way: 2Na (s) + ½ O2 (g) → Na2O (s) Na2O + H2O (L) →2 NaOH (aq) 2 Na (s) + H2O (L) + ½ O2 (g) → 2 NaOH (aq) 2nd way 2 Na (s) + 2 H2O (L) → 2 NaOH (aq) + H2 (g) H2 (g) + ½ O2 (g) → H2O (g) 2 Na (s) + H2O (L) + ½ O2 (g) → 2 NaOH (aq) ΔH = -100 kcal Δ H = -56 kcal ΔH = - 156 Kcal ΔH = - 88 Kcal ΔH = - 68.5 Kcal ΔH = - 156 Kcal Example 8: Calculate the heat of hydrogenation of acetylene 1) C2H2(g) + 2 H2(g) → C2H6(g) H1 = ? At 25°C and 1 atm., given that; 2) 2C2H2(g) + 5 O2(g) → 4CO2 + 2H2O(l) H2 = - 2602 KJ 3) 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(I) H3 =-3123KJ 4) H2(g)+ ½O2(g) →H2O(l) H4= -286KJ Solution: 5) C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l) 2 ) علي2( بقسمة H5 = -2602/2 = -1301KJ 6) 2H2(g) + O2(g) → 2H2O(I) 2 في4 بضرب H6 = 2 (-286KJ) = - 572 KJ 7) 2CO2(g) + 3H2O (l) → 2C2H6(g) + 7/2 O2(g)2 ) علي3( عكس و قسمة H7 = + 3123/2 = +1561KJ Adding equation 5,6 and 7 gives C2H2(g) + 2H2(g) + 7/2O2(g) + 2CO2 + 3H2O(I) → 2CO2(g) + 3H2O(|) + C2H6(g) + 7/2 O2(g) Canceling elements that are the same on both sides, we get C2H2(g) + 2H2(g) → C2H6(g) H1 = H5 +H6 + H7 = (-1301 KJ) + (-572 KJ) + (1561 KJ) = -312 KJ Calorimetry • We measure heat flow using calorimetry. • A calorimeter is a device used to make this measurement. • A “coffee cup” calorimeter may be used for measuring heat involving solutions. A “bomb” calorimeter is used to find heat of combustion; the “bomb” contains oxygen and a sample of the material to be burned. Bomb Calorimeter Coffee-cup calorimeter.