Giáo trình Lý thuyết cháy - Bùi Tuyên
Tóm tắt Giáo trình Lý thuyết cháy - Bùi Tuyên: ... y v x u ρρρ τ ρ (2.10) dưới dạng vector Bảo toàn động lượng mvQ = Bảo tồn động năng: E = 1/2 mv2 Phương trình chuyển động: F = m.a Phương trình chuyển động Bernoulli cho dòng chảy đoạn nhiệt chất nước: Bernoulli's Equation is basically a statement of the conservati... có tổn thất nhiệt hoá học (một phần ít nhiên liệu chuyển hoá không hết ra sản phẩm do cân bằng hoá học), tổn thất nhiệt do cách nhiệt không lý tưởng và cuối cùng là tổn thất qua các chất pha loãng (không tham gia phản ứng như N2 chẳng hạn) Ta cĩ thể... thời gian đọc hàng núi sách, anh chợt nhận ra mình tồn loay loay với những cuốn sách hạng ba hạng tư, thời giờ đã mất, mà kiến thức thu được chẳng bao nhiêu, thì người đáng để anh ta buơng lời trách mĩc lại là chính bản thân anh - ối oăm là ở chỗ đĩ! Thường nhìn vào khoa học, người ta dễ ...
ess. Exothermic processes occur when heat energy is transferred out of a system. H2O(g) ---> H2O(l) q = -40.7 kJ/mol Enthalpy (ΔH) is the heat transferred in a physical or chemical process occurring at constant pressure. The quantity of heat associated with a physical change (e.g., heat of vaporization, qvap) can be calculated as the product of the amount (mole) of the substance and the enthalpy of the phase change (e.g., enthalpy of vaporiztion, qvap). qvap = (mole)(ΔHvap) Example 7 How much heat is required to vaporize 25.0 g of carbon disulfide at 25oC? The heat of vaporization (ΔHvap) for carbon disulfide is +27.4 kJ/mol. qvap = (mole)(ΔHvap) = (25.0 g)(1mol/76.15 g)(27.4 kJ/mol) = 9.00 kJ Example 8 Liquid butane, C4H10, is stored in cylinders to be used as a fuel. Suppose 31.4 g of butane gas is removed from a cylinder. How much heat must be provided to vaporize this much gas? The heat of vaporization of butane is 21.3 kJ/mol. qvap = (mole)(ΔHvap) = (31.4 g)(1 mol/ 58.14 g)(21.3 kJ/mol) = 11.5 kJ Heats of Reaction and Enthalpy Change, ΔH A thermochemical equation is a balanced chemical reaction equation (including phase labels) with the enthalpy of reaction value written directly after the equation. 2Na(s) + 2H2O(l) ---> 2NaOH(aq) + H2(g) ΔH = -367.5 kJ A negative value (-) for ΔH indicates an exothermic reaction, that is heat is produced by the reaction system. CH4(g) + 2 O2(g) ---> CO2(g) + 2 H20(g) ΔH = - 890 kJ A positive value (+) for ΔH indicates an endothermic reaction, that is heat is absorbed by the reaction system. CH3OH(l) ---> CO(g) + 2H2(g) ΔH = +90.7 kJ The value of ΔH depends upon the phase of the reactants and the products. 2H2 (g) + O2 (g) ---> 2H2O(g) ΔH = -483.7 kJ 2H2(g) + O2(g) ----> 2H2O(l) ΔH = -571.7 kJ Stoichometric Calculations Involving Thermochemical Equations 1. When a thermochemical equation is multiplied by any factor, the value of ΔH for the new equation is obtained by multiplying the value of ΔH in the original equation by that same factor. 2. When a chemical equation if reversed, the value of ΔH is reversed in sign. EXAMPLE 3 Using the following thermochemical equation, calculate how much heat is associated with the decomposition of 4.00 moles of NH4Cl. NH3(g) + HCl(g) ---> NH4Cl(s) ΔH = - 176 kJ (4.00 mol NH4Cl)(+176 kJ/1 mol NH4Cl) = +704 kJ EXAMPLE 4 How much heat is associated with the synthesis of 45.0 g of NH3 according to the following equation? 4 NO(g) + 6 H2O(l) ---> 4 NH3(g) + 5 O2(g) ΔHrxn = +1170 kJ (45.0 g NH3)(1 mol/17.04 g NH3) = 2.641 mol NH3 (2.641 mol NH3)(+1170kJ/4 mol NH3) = +772 kJ EXAMPLE 5 Calculate the mass of ethane, C2H6, which must be burned to produce 100 kJ of heat. 2 C2H6(g) + 7 O2(g) ---> 4 CO2(g) + 6 H2O(l) ΔH = - 3120 kJ (-100 kJ)(2 mol C2H6 / -3120 kJ) = 0.0641 mol C2H6 (0.0641 mol C2H6)(30.08 g C2H6/1 mol C2H6) = 1.93 g C2H6 Calorimetry Calorimetry is the measurement of heat changes. A calorimeter is an apparatus used to measure the quantity of thermal energy gained or lost during physical or chemical changes. Bomb Calorimetry: Reactions at Constant Volume The amount of heat given off by the combustion of a fuel can be determined very accurately in the so-called bomb calorimeter, which consists of a combustion chamber (the "bomb") set in another chamber filled with water. Heat generated by combustion of the fuel is transmitted to the water, raising its temperature. The calorie content of food is tested this way. Bomb calorimeters are usually composed of two parts, the bomb itself wherein the reaction takes place and the water which surrounds the bomb. The calculations for determining the heat of reaction require that the heat change of the bomb and the water be considered individually. qcalorimeter = heat capacity of the calorimeter x ΔT qsystem = qcalorimeter + qrxn Because no heat enters or leaves the system qsystem = 0 qrxn = -qcalorimeter= - Ccalorimeter(ΔT) or another way of looking at it is that the heat associated with the reaction will be transferred to the calorimeter so that the heat will have the same magnitude but opposite sign. The amount of heat involved in a reaction can be determined from: qrxn = qwater + qbomb qwater = measured heat change of the water. qbomb = measured heat change of bomb. Thus, qwater and qbomb must be formed before qrxn can be determined. qwater = (specific heat of water)(mass of water)(ΔT) qcalorimeter = (heat capacity of the bomb)(ΔT) EXAMPLE 11 Exactly 3.00 g of graphite is burned to CO2 in a copper calorimeter. The initial temp is 25.0oC and the final temp was 36.3oC. The heat capacity of the calorimeter is 8.887 kJ/oC. What is the thermochemical equation for the combustion of graphite? A. Calculate the heat transferred to the calorimeter. qcalorimeter = C(ΔT) = (8.887 kJ/oC)(36.3oC - 25.0oC) = +100.4 kJ The energy change for the reaction is equal in magnitude and opposite in sign to the heat energy required to raise the temperature of the calorimeter. This means that qrxnis -100.4 kJ. B. Derive the thermochemical equation for the combustion of graphite. C(s) + O2(g) ---> CO2(g) ΔH = ? (-100.4 kJ)/[(3.00 g C)(1 mol C/12.01 g C)] = -401.9 kJ/mol C (1 mole C(s))(-401.9 kJ/mol C) = -401.9 kJ C(s) + O2(g) ---> CO2(g) ΔH = -401.9 kJ EXAMPLE 12 A certain calorimeter has a heat capacity of 16.35 kJ/oC. What is the final temperature of the calorimeter if 17.038 g of kerosene (C12H26) is burned in the calorimeter? The initial temperature of the calorimeter is 25.00 oC. The heat of combustion of kerosene is shown in the equation below. C12H26(l) + 37/2 O2(g) ---> 12 CO2(g) + 13 H2O(l) ΔH = -7513 kJ A. (17.038g C12H26)(1 molC12H26/170.38g C12H26)(-7513 kJ/mol C12H26) = -751.3 kJ B. qrxn = -qcal = -C(ΔT) -751.3 kJ = -(16.35 kJ/oC)(Tf - 25.00oC) Tf = 70.95oC Hess’s Law For a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps. In other words, no matter how you go from given reactants to products (whether in one step or several), the enthalpy change for the overall chemical change is the same. Hess's law states that the heat transferred in a given change is the same whether the change takes place in a single step or in several steps. ΔHrxn for a given reaction is the same whether that reaction takes place directly through one step or via several different reactions. S(s) + O2(g) ---> SO2(g) ΔH = -296 kJ SO2(g) + 1/2 O2(g) ---> SO3(g) ΔH = -98.9 kJ ____________________________________________ S(s) + 1 1/2 O2(g) ---> SO3(g) ΔH = -394.9 kJ In order to use Hess's Law: 1) If a rxn is reversed, the sign of ΔH is reversed. S(s) + O2(g) ---> SO2(g) ΔH = -296 kJ SO2(g) ---> S(s) + O2(g) ΔH = +296 kJ 2) If the coefficients in a balanced equation are multiplied by some number, the value of ΔH must be multiplied by that same number. S(s) + O2(g) ---> SO2(g) ΔH = -296 kJ 2 S(s) + 2 O2(g) ---> 2 SO2(g) ΔH = (2)(-296 kJ) = -592 kJ EXAMPLE 13 Calculate ΔH for the vaporization of water from liquid using the thermochemical equations given. H2O(l) ---> H2O(g) ΔH = ? a) H2(g) + 1/2 O2(g) --->H2O(l) ΔH = -286 kJ b) H2(g) + 1/2 O2(g) ---> H2O(g) ΔH = -242 kJ H2O(l) ---> H2(g) + 1/2 O2(g) ΔH = +286 kJ H2(g) + 1/2 O2(g) ---> H2O(g) ΔH = -242 kJ _____________________________________ H2O(l) ----> H2O(g) ΔH = +44 kJ EXAMPLE 14 Calculate ΔH for the following reaction using the thermochemical equations given. 2 C(s) + H2 (g) ---> C2H2 (g) ΔH = ? a) C(s) + O2 (g) ---> CO2 (g) ΔH = -393.5 kJ b) H2 (g) + 1/2 O2 (g) ---> H2O(l) ΔH = -285.8 k J c) 2 C2H2 (g) + 5 O2 (g) ---> 4 CO2 (g) + 2 H2O(l) ΔH = -2598.8 kJ Solution: 2 C(s) + 2 O2 (g) ---> 2 CO2 (g)ΔH = -787.0 kJ H2 (g) + 1/2 O2 (g) ---> H2O(l)ΔH = -285.8 k J 2 CO2 (g) + H2O(l) ---> C2H2 (g) + 2 1/2 O2 (g) ΔH = +1299.4 kJ _____________________________________________________ 2 C(s) + H2 (g) ---> C2H2 (g)ΔH = +226.6 kJ Standard Enthalpy of Formation, ΔHof The change in enthalpy that accompanies the formation of one mole of a substance from its elements with all substances in their standard state at 25oC and 1 atm is referred to as standard enthalpy of formation, ΔHof. Standard state is the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atm pressure and 25oC. Standard conditions are indicated by a superscript degree sign (o). For a gas, the standard state is a pressure of exactly 1 atmosphere. For a substance present in solution, the standard state is a concentration of exactly 1 M. For a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid. For an element, the standard state is the form in which the element exists under conditions of 1 atm and 25oC. The standard enthalpies of formation of the elements in their standard states are zero. The standard enthalpy change, ΔHo, for any chemical reaction is found by subtracting the sum of the heats of formation of the reactants from the sum of the heats of formation of the products. This is shown in the following equation. ΔHo= ΣΔHof (Products) - ΣΔHof (Reactants) [NOTE: "Σ" is the symbol which means "sum".] EXAMPLE 15 Using the standard heats of formation, calculate the standard heat of reaction, ΔHo , for the following reaction. 2 NO2 (g) ---> N2O4 (g) ΔHo = ? ΔHof NO2 (g)= +33.9 kJ/mol; ΔHof N2O4 (g) = +9.7 kJ/mol ΔHo= ΣΔHof (Products) - ΣΔHof (Reactants) ΔHo = [(1 mol N2O4)(9.7 kJ/mol)] - [(2 mol NO2)(33.9 kJ/mol)] = -58.1 kJ EXAMPLE 16 Using the standard heats of formation, calculate the standard heat of reaction, ΔHo , for the following reaction. C7H16 (l)+ 11 O2 (g) ----> 7 CO2 (g) + 8 H2O(l) ΔHo = ? ΔHof C7H16 (l) = -198.8 kJ/mol; ΔHof CO2 (g) = -393.5 kJ/mol; ΔHof H2O(l) = -285.9 kJ/mol ΔHo= ΣΔHof (Products) - ΣΔHof (Reactants) ΔHo = [(7 mol CO2)(-393.5 kJ/mol) + (8 mol H2O(l))(-285.9 kJ/mol)] - [(1 mol C7H16 (l))(-198.8 kJ/mol)] = -5041.7 kJ - (-198.8 kJ) = -4842.9 kJ CHEMISTRY 101 THERMOCHEMISTRY SAMPLE PROBLEMS 1. How much heat is liberated when 6.00 mol of Fe2O3(s) are reacted according to the following reaction? (-21.4 kcal) 3 Fe2O3(s) + CO(g) --> 2 Fe3O4(s) + CO2(g) ΔH = -10.7 kcal 2. What amount of heat is associated with the reaction of 43.77 g of NaCl with sulfuric acid? (+23.81 kJ) H2SO4(l) + 2 NaCl(s) --> Na2SO4(s) + 2 HCl(g) ΔH = +63.60 kJ 3. A 466 g sample of water is heated from 8.50C to 74.60C. Calculate the amount of heat absorbed by the water. The specific heat of water is 4.184 goC. (129 kJ) 4. An electrical heater is used to supply 25.0 J of energy to a 25.0 g sample of Ag originally at 22oC. Calculate the final temperature. The specific heat of Ag is 0.2349 goC. (26oC) 5. Assuming no heat loss to the surrounding or to the container, calculate the final temperature when 100 g of silver at 40.0oC is immersed in 60.0 g of water at 10.0oC. The specific heat of silver is 0.2349 J/goC and for water is 4.184 J/goC. (12.6oC) 6. Determine ΔHofor the following reaction of burning ethyl alcohol in oxygen: C2H5OH(l) + 3 O2(g) ---> 2 CO2(g) + 3 H2O (l) ΔH = ? (-327.2 kcal) ΔHf of C2H5OH(l) = -65.9 kcal/mol; ΔHf of CO2(g) = -94.1 kcal/mol; ΔHf of H2O(l) = -68.3 kcal/mol 7. From the following equations and the enthalpy changes, calculate the enthalpy of reaction of iron(III) oxide, Fe2O3, with carbon monoxide: Fe2O3(s) + 3 CO(g) ---> 2 Fe(s) + 3 CO2(g) ΔH = ? (-26.7 kJ) (a) CO(g) + 1/2 O2(g) ---> CO2(g) ΔH = -283.0 kJ (b) 2 Fe(s) + 3/2 O2(g) ---> Fe2O3(s) ΔH = -822.3 kJ 8. Using the thermochemical equations given, calculate the heat of hydrogenation of acetylene. C2H2(g) + 2 H2(g) --> C2H6(g) ΔH = ? (-312 kJ) (a) 2 C2H2(g) + 5 O2(g) --> 4 CO2 + 2 H2O(g) ΔH = -2602 kJ (b) 2 C2H6(g) + 7 O2(g) --> 4 CO2 + 6 H2O(g) ΔH = -3123 kJ (c) 2 H2(g) + O2(g) --> 2 H2O(g) ΔH = -572 kJ 9. The combustion of 1 mol of benzene, C6H6(l), to produce CO2(g) and H2O(l) liberates 3271 kJ. Given the the ΔHf of CO2(g) and H2O(l) are -394 kJ/mol and -286 kJ/mol respectively, calculate the heat of formation of benzene. (+49 kJ/mol) CALORIMETRY 10. A quantity of 1.435 g of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.17C to 25.84C. If the quantity of water surrounding the calorimeter was exactly 2000 g and the heat capacity of the bomb calorimeter was 1.80 kJ/C, calculate the heat of combustion of naphthalene on a molar basis; that is , find the molar heat of combustion. (-5.15 x 103 kJ/mol) 11. A quantity of 1.00 x 102 mL of 0.500 M HCl is mixed with 1.00 x 102 mL of 0.500 M NaOH in a constant-pressure calorimeter having a heat capacity of 335 J/C. The initial temperature of the HCl and NaOH solutions is the same, 22.50C, and the final temperature of the mixed solution is 24.90C. Calculate the heat change for the neutralization reaction NaOH(aq) + HCl(aq) ---> NaCl(aq) + H2O(l) Assume that the densities and specific heats of the solutions are the same as for water (1.00g/mL and 4.182 J/gC, respectively). (-56.2 kJ/mol) WORK 12. A certain gas initially at room temperature undergoes an expansion in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands (a) against vacuum and (b) against a constant pressure of 1.2 atm. 1 L.atm = 101.325 J ((a) 0 (b) -4.8 Latm = -4.9 x 102 J) 13. The work done when a gas is compressed in a cylinder is 462 J. During this process, there is a heat transfer of 128 J from the gas to the surroundings. Calculate the energy change for this process. (334 J) 14. The oxidation of nitric oxide to nitrogen dioxide is a key step in the formation of smog: 2NO(g) + O2(g) ---> 2NO2(g) ΔH = -113.1 kJ If 6.00 moles of NO react with 3.00 moles of O2 at 1.00 atm and 25C to form NO2, calculate the work done (in kilojoules) against a pressure of 1.00 atm. What is the ΔU for the reaction? Assume the reaction to go to completion. 1 L.atm = 101.325 J (7.39 x 103J = 7.39 kJ, ΔU = -331.9 kJ) Formula Name ΔHf (kcal/mol) H2 Hydrogen 0.0 CH4 Methane -17.9 C2H6 Ethane -20.0 C3H8 n-Propane -25.0 C4H10 n-Butane -30.0 C5H12 n-Pentane -35.1 C6H14 n-Hexane -40.0 C7H16 n-Heptane -44.9 C8H18 n-Octane -49.8 C9H20 n-Nonane -54.8 C10H22 n-Decane -59.6 C4H10 2-Methylpropane (Isobutane) -32.1 C5H12 2,2-Dimethylpropane (Neopentane) -40.1 C5H12 2-Methylbutane (Isopentane) -36.9 C6H14 2,2-Dimethylbutane -44.5 C6H14 2-Methylpentane (Isohexane) -41.8 C6H14 3-Methylpentane -41.1 C6H14 2,3-Dimethylbutane -42.5 C7H16 2,2,3-Trimethylbutane -49.0 C7H16 3-ethylpentane -45.3 C8H18 2,2,3,3-Tetramethylbutane -53.9 C7H16 2-Methylhexane -46.5 C7H16 3-Methylhexane -45.7 C7H16 2,2-Dimethylpentane -49.2 C7H16 3,3-Dimethylpentane -48.1 C7H16 2,3-Dimethylpentane -47.3 C7H16 2,4-Dimethylpentane -48.2 C8H18 2-Methylheptane -51.5 C8H18 2,2-Dimethylhexane -53.7 C8H18 2,3-Dimethylhexane -55.1 C8H18 2,4-Dimethylhexane -52.4 C8H18 2,5-Dimethylhexane -53.2 C8H18 3,3-Dimethylhexane -52.6 C8H18 3,4-Dimethylhexane -50.9 C8H18 3-Ethyl-2-Methylpentane -50.4 C8H18 3-Ethyl-3-Methylpentane -51.4 C8H18 2,2,3-Trimethylpentane -52.6 C8H18 2,2,4-Trimethylpentane (Isooctane) -53.5 C8H18 2,3,3-Trimethylpentane -51.7 C8H18 2,3,4-Trimethylpentane -51.9 C9H20 3,3-Diethylpentane -55.7 C9H20 2,2,3,3-Tetramethylpentane -56.7 C9H20 2,2,3,4-Tetramethylpentane -56.6 C9H20 2,2,4,4-Tetramethylpentane -57.8 C9H20 2,3,3,4-Tetramethylpentane -56.4 Miscellaneous Compounds Species Phase(Matter) Chemical Formula ΔHfo (kJ/mol) Aluminum Solid Al 0 Aluminum Chloride Solid AlCl3 -705.63 Aluminum Oxide Solid Al2O3 -1675.7 Barium Chloride Solid BaCl2 -858.6 Barium Carbonate Solid BaCO3 -1213 Barium Oxide Solid BaO -548.1 Barium Sulfate Solid BaSO4 -1473.2 Beryllium Solid Be 0 Beryllium Hydroxide Solid Be(OH)2 -902.5 Boron Trichloride Solid BCl3 -402.96 Bromine Liquid Br2 0 Bromine Gas Br 111.884 Bromine Gas Br2 30.91 Bromine Trifluoride Gas BrF3 -255.60 Hydrobromic Acid Gas HBr -36.29 Calcium Solid Ca 0 Calcium Gas Ca 178.2 Calcium(II) Ion Gas Ca2+ 1925.90 Calcium carbide Solid CaC2 -59.8 Calcium Carbonate(Calcite) Solid CaCO3 -1207.6 Calcium Chloride Solid CaCl2 -795.8 Calcium Phosphate Solid Ca3(PO4)2 -4132 Calcium Fluoride Solid CaF2 -1219.6 Calcium Hydride Solid CaH2 -186.2 Calcium hydroxide Solid Ca(OH)2 -986.09 Calcium hydroxide Aqueous Ca(OH)2 -1002.82 Calcium Oxide Solid CaO -635.09 Calcium Sulfate Solid CaSO4 -1434.52 Calcium Sulfide Solid CaS -482.4 Benzene Liquid C6H6 48.95 Benzoic acid Solid C7H6O2 -385.2 Carbon(Graphite) Solid C 0 Carbon(Diamond) Solid C 1.8 Carbon Gas C 716.67 Carbon Dioxide Gas CO2 -395.893 Carbon disulfide Liquid CS2 89.41 Carbon disulfide Gas CS2 116.7 Carbon Monoxide Gas CO -110.525 Carbon tetrachloride Liquid CCl4 -139.5 Carbon tetrachloride Gas CCl4 -103.18 Carbonyl Chloride(Phosgene) Gas COCl2 -218.8 Ethane Gas C2H6 -83.85 Ethanol Liquid C2H5OH -277.0 Ethanol Gas C2H5OH -235.3 Ethene Gas C2H4 52.47 Ethyne Gas C2H2 226.73 Methane Gas CH4 -74.87 Methanol(Methyl Alcohol) Liquid CH3OH -238.4 Methanol(Methyl Alcohol) Gas CH3OH -201.0 Methyl Trichloride(Chloroform) Liquid CHCl3 -134.47 Methyl Trichloride(Chloroform) Gas CHCl3 -103.18 Propane Liquid C3H8 -104.7 Caesium Solid Cs 0 Caesium(I) ion Gas Cs+ 457.964 Caesium chloride Solid CsCl -443.04 Chlorine Gas Cl2 0 Chromium Solid Cr 0 Copper Solid Cu 0 Fluorine Fluorine Gas F2 0 Hydrogen Gas H2 0 Water Liquid H2O -285.83 Water Gas H2O -241.83 Hydrogen Peroxide Liquid H2O2 -187.78 Hydrogen Cyanide Gas HCN +130.5 Hydrogen Iodide Gas HI +26.5 Hydrofluoric acid Gas HF -269 Hydrochloric acid Gas HCl -92.30 Iodine Solid I2 0 Iodine Gas I2 62.438 Magnesium Carbonate Solid MgCO3 -1111.69 Magnesium Chloride Solid MgCl2 -641.8 Magnesium hydroxide Solid Mg(OH)2 -924.54 Magnesium Oxide Solid MgO -601.24 Magnesium sulfate Solid MgSO4 -1278.2 Manganese Manganese(II) Oxide Solid MnO -384.9 Manganese(IV) Oxide Solid MnO2 -519.7 Mercury Mercury(II) Oxide (red) Solid HgO -90.83 Mercury Sulfide (red, cinnabar) Solid HgS -58.2 Nickel Nitrogen Ammonia Aqueous NH3 -80.8 Ammonia Gas NH3 -45.90 Nitrogen Dioxide Gas NO2 33.1 Nitrogen Monoxide Gas NO 90.29 Oxygen Monoatomic oxygen Gas O 249 Ozone Gas O3 143 Phosphorus Phosphorus trichloride Liquid PCl3 -320 Phosphorus pentachloride Solid PCl5 -440 Potassium Potassium Bromide Solid KBr -392.2 Potassium Chlorate Solid KClO3 -397.73 Potassium chloride Solid KCl -436.68 Potassium Fluoride Solid KF -562.6 Potassium Perchlorate Solid KClO4 -430.12 Silicon Silica (Quartz) Solid SiO2 -910.86 Silver Silver Bromide Solid AgBr -99.5 Silver Chloride Solid AgCl -127.01 Silver Iodide Solid AgI -62.4 Silver Oxide Solid Ag2O -31.1 Silver Sulfide Solid Ag2S -31.8 Sodium Sodium Carbonate Solid Na2CO3 -1130.77 Sodium chloride Aqueous NaCl -407.27 Sodium chloride Solid NaCl -411.12 Sodium chloride Liquid NaCl -385.92 Sodium chloride Gas NaCl -181.42 Sodium Fluoride Solid NaF -569.0 Sodium Hydroxide Aqueous NaOH -469.15 Sodium Hydroxide Solid NaOH -425.93 Sodium nitrate Aqueous NaNO3 -446.2 Sodium nitrate Solid NaNO3 -424.8 Sulfur Hydrogen Sulfide Gas H2S -20.63 Sulfur Dioxide Gas SO2 -296.84 Sulfuric acid Liquid H2SO4 -814
File đính kèm:
- giao_trinh_ly_thuyet_chay_bui_tuyen.pdf