1 / 69

General Chemistry

General Chemistry. M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology. فصل چهارم:. استوکیومتری II و معادلات شیمیائی. Contents. 4-1 Chemical Reactions and Chemical Equations 4-2 Chemical Equations and Stoichiometry 4-3 Chemical Reactions in Solution

eaglem
Télécharger la présentation

General Chemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. General Chemistry M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology

  2. فصلچهارم: استوکیومتریII و معادلات شیمیائی

  3. Contents 4-1 Chemical Reactions and Chemical Equations 4-2 Chemical Equations and Stoichiometry 4-3 Chemical Reactions in Solution 4-4 Determining the Limiting reagent

  4. 4-1 Chemical Reactions and Chemical Equations As reactants are converted to products, we observe: • Color change • Precipitate formation • Gas evolution • Heat absorption or evolution Chemical evidence may be necessary.

  5. Formation of AlBr3

  6. Chemical Reaction Nitrogen monoxide + oxygen → nitrogen dioxide Step 1: Write the reaction using chemical symbols. Step 2: Balance the chemical equation. 2 1 2 NO + O2 → NO2

  7. Molecular Representation

  8. Balancing Equations Example: • Never introduce extraneous atoms to balance. Nitrogen monoxide + oxygen → nitrogen dioxide NO + O2 → NO2 + O • Never change a formula for the purpose of balancing an equation. NO + O2 → NO3

  9. Balancing Equation Strategy • Balance elements that occur in only one compound on each side first. • Balance free elements last. • Balance unchanged polyatomics as groups. • Fractional coefficients are acceptable and can be cleared at the end by multiplication.

  10. Chemical Equations • The physical states: • Solid (s) • liquid (l) • gas (g) • and aqueous (aq) P4 (s) + 6Cl2 (g) 4PCl3 (l)

  11. 4-2 Chemical Equations and Stoichiometry • Stoichiometry includes all the quantitative relationships involving: • atomic and formula masses • chemical formulas. • The coefficients in front of the compounds in a balanced equation are called stoichiometric coefficients • Mole ratio is a central conversion factor.

  12. Mass Relationships From this we can calculate mass of one compound required to complete the reaction if the mass of the other compound is given

  13. Write the Chemical Equation: Balance the Chemical Equation: H2 + O2→ H2O 2 2 Use the stoichiometric factor or mole ratio in an equation: 2 mol H2O nH2O = 2.72 mol H2× = 2.72 mol H2O 2 mol H2 Example 4-3 Relating the Numbers of Moles of Reactant and Product. How many moles of H2O are produced by burning 2.72 mol H2 in an excess of O2?

  14. Example 4-6 Additional Conversion Factors in a Stoichiometric Calculation: Volume, Density, and Percent Composition. An alloy used in aircraft structures consists of 93.7% Al and 6.3% Cu by mass. The alloy has a density of 2.85 g/cm3. A 0.691 cm3 piece of the alloy reacts with an excess of HCl(aq). If we assume that all the Al but none of the Cu reacts with HCl(aq), what is the mass of H2 obtained?

  15. Write the Chemical Equation: Balance the Chemical Equation: Al + HCl → AlCl3 + H2 2 6 2 3 Example 4-6

  16. Plan the strategy: cm3 alloy → g alloy → g Al → mole Al → mol H2 → g H2 Write the Equation and Calculate: 97.3 g Al 2.85 g alloy mH2 = 0.691 cm3 alloy × × × 100 g alloy 1 cm3 3 mol H2 1 mol Al 2.016 g H2 × × = 0.207 g H2 2 mol Al 26.98 g Al 1 mol H2 Example 4-6 2 Al + 6 HCl → 2 AlCl3 + 3 H2 We need 5 conversion factors!

  17. Combustion Reactions • Combustion reaction – burning of a substance. Substance combines with oxygen to form carbon dioxide and water • C8H18 (l) + O2(g) CO2(g) + H2O (l)

  18. Balancing Combustion Reactions • Write correct formulas for the reactants and products • Balance the carbon atoms • Balance the hydrogen atoms • Balance the oxygen atoms • Verify that the number of atoms of each element is balanced

  19. Example 4-2 Writing and Balancing an Equation: The Combustion of a Carbon-Hydrogen-Oxygen Compound. Liquid triethylene glycol, C6H14O4, is used a a solvent and plasticizer for vinyl and polyurethane plastics. Write a balanced chemical equation for its complete combustion.

  20. 15 6 7 2 Example 4-2 Chemical Equation: C6H14O4 + O2→ CO2 + H2O 6 7 6 1. Balance C. 2. Balance H. 3. Balance O. 4. Multiply by two 2 C6H14O4 + 15 O2→ 12 CO2 + 14 H2O and check all elements.

  21. Yields • Theoretical yield – the maximum amount of product that can be formed from a chemical reaction • Actual yield – the amount of product that is formed in the laboratory

  22. بازده واکنش میزان مطلوبیت یک واکنش معمولا با بازده آن گزارش می شود. مقدار عملی 100 x = درصد بازده واکنش مقدار نظری

  23. بازده واکنش مثال: تیتانیم فلزی قوی، سبک وزن و مقاوم در برابر خوردگی است که از آن برای صنایع هواپیما و دوچرخه سازی استفاده می شود. تیتانیم در واکنش تیتانیم کلرید با منیزیم مذاب بین C 950 تا C 1150 به دست می آید. TiCl4(g) + 2 Mg(l)  Ti(s) + 2 MgCl2(l) در یک فرآیند صنعتی 3.54x107 g از نمک تیتانیم کلرید با مقدار اضافی منیزیم وارد واکنش شده است. الف) تولید تیتانیم چند گرم است؟ ب) اگر در عمل 7.91x106 g تیتانیم تولید شده باشد، درصد بازده را حساب کنید.

  24. الف: ب:

  25. Theoretical, Actual and Percent Yield • When actual yield = % 100, then the reaction is said to be quantitative. • Side reactions reduce the percent yield. • By-products are formed by side reactions.

  26. Limiting Reagents • The goal of chemical reactions is to produce the highest amount of product possible • So, one reagent will probably be in excess • This allows for the complete reaction of one reagent, even though some others remain unreacted (NH4)2PtCl4(s) + 2 NH3(aq) 2 NH4Cl(aq) + Pt(NH3)2Cl2(s) $100/g $0.01/g All of the expensive reagent is used up, leaving the cheap unreacted ammonia

  27. عامل محدود کننده • عامل محدود کننده آن ماده ای است که به تمامی مصرف شده و درپایان واکنش چیزی از آن باقی نماند. و آن محدود کننده تولید یا مصرف بقیه مواد واکنش است. • در حل مسئله ابتدا باید عامل محدود کننده مشخص شود تا بقیه مواد نسبت به آن سنجیده شوند.

  28. عامل محدود کننده کدام است؟ 4NH3 4N2 6H2 2N2

  29. عامل محدود کننده • در واکنش بین 73 گرم کلریدریک اسید و 80 گرم سود چه مقدار نمک ایجاد می شود؟ • در واکنش بین 98 گرم سولفوریک اسید و 160 گرم سود چقدر نمک ایجاد می شود؟

  30. مثال محلول سولفوریک اسید و سرب استات واکنش داده و سرب سولفات جامد و استیک اسید محلول می دهند. اگر 15.0 گرم از هریک از مواد اولیه باهم مخلوط شوند، حساب کنید چند گرم سرب سولفات تولید می شود. همچنین حساب کنید پس از انجام واکنش چند گرم ماده اضافی باقی می ماند؟ H2SO4 + Pb(CH3COO)2 PbSO4 + 2 CH3COOH

  31. حل:

  32. Limiting Reagent

  33. Chemical Equations & Chemical Analysis • Analytical chemists try to identify substances in a mixture, and try to measure the quantities of the components. • Mostly it is done with instrumental methods • It is essential to use chemical reactions and stoichiometry

  34. Quantitative Analysis of a Mixture • Usually depends on one of the two following ideas. Idea 1: • A substance, present in unknown amount, can be allowed to react with a known quantity of another substance. If the stoichiometric ratio for their reaction is known, the unknown can be determined. CH3CO2H(aq) + NaOH(aq) NaCH3CO2(aq) + H2O(l) Know the amount of NaOH so we can determine the amount of acetic acid

  35. Quantitative Analysis of a Mixture Idea 2 • A material of unknown composition can be converted to one or more substances of known composition. Those substances can be identified, their amounts determined, and these amounts related to the amount of the original, unknown substance. C7H5NO3S + X convert to SO42- Na2SO4+ other Na2SO4(aq) + BaCl2(aq) BaSO4(s) + 2 NaCl(aq) 1 mol Na2SO41 mol S 1 mol SO42-1 mol BaSO4

  36. Chemical Reactions in Solution • Close contact between atoms, ions and molecules necessary for a reaction to occur. • Solvent • We will usually use aqueous (aq) solution. • Solute • A material dissolved by the solvent.

  37. Chemical Reactions in Solution • بسیاری از واکنشها در محلول انجام می شود. • محاسبات استوکیومتری برای این گونه واکنشها بر مبنای حجمهای محلول های به کار رفته و غلظت این محلول ها است. • غلظت یک محلول ، مقدار ماده حل شده در مقدار معینی حلال ، یا مقدار ماده حل شده موجود در مقدار معینی از محلول است. • چند روش برای بیان غلظت وجود دارد.

  38. If 0.444 mol of urea is dissolved in enough water to make 1.000 L of solution the concentration is: 0.444 mol urea curea = = 0.444 M CO(NH2)2 1.000 L Molarity # Mole of solute        Molarity (M) = Volume of solution (L)

  39. مولاریته -M • توجه داشته باشید که تعریف مولاریته بر مبنای یک لیتر محلول است ، و نه بر مبنای یک لیتر حلال . • یک محلول 1.0 M شامل 1.0 مول ماده حل شده در 1 L محلول است . • یک محلول 1.5 M شامل 1.5 مول ماده حل شده در1 L محلول است. • یک محلول 3.0 M شامل3.0 مول ماده حل شده در1 L محلول است.

  40. مولاریته –M(ادامه) مولاریته -M • برای یک محلول 3.0 M : • 1000 mL، شامل 3.0 مول ماده حل شده است. • 500 mL ، شامل 1.5 مول ماده حل شده است. • 2000 mL، شامل 6.0 مول ماده حل شده است. • درهرسه نمونه غلظت 3.0 M است.

  41. Preparation of a Solution Weigh the solid sample. Dissolve it in a volumetric flask partially filled with solvent. Carefully fill to the mark.

  42. 194.02 g 0.250 mol mK2CrO4 = 0.2500 L × × = 12.1 g 1.00 mol 1.00 L Example 4-6 Calculating the mass of solute in a solution of known molarity. We want to prepare exactly 0.2500 L (250 mL) of an 0.250 M K2CrO4 (MW=194.02) solution in water. What mass of K2CrO4 should we use? Plan strategy: Volume → moles → mass We need 2 conversion factors! Write equation and calculate:

  43. Mi × Vi Mf × Vf n M = V Mi× Vi = ni = nf= Mf × Vf Mi× Vi Vi = Mi Mf = Vf Vf Solution Dilution

  44. Vi Mf Plan strategy: Mi Vf Mf = Vi = Vf Mi Calculate: 1.000 L 0.0100 mol VK2CrO4 = 0.2500 L × × = 0.0100 L 0.250 mol 1.00 L Example 4-10 Preparing a solution by dilution. A particular analytical chemistry procedure requires 0.0100 M K2CrO4. What volume of 0.250 M K2CrO4 should we use to prepare 0.250 L of 0.0100 M K2CrO4?

  45. Solution formation by Dilution

  46. Oxidation States Metals tend to lose electrons. Na Na+ + e- Non-metals tend to gain electrons. Cl + e-Cl- We use the Oxidation State to keep track of the number of electrons that have been gained or lost by an element.

  47. Rules for Oxidation States • The oxidation state (OS) of an individual atom in a free element is 0. • The total of the OS in all atoms in: • Neutral species is 0. • Ionic species is equal to the charge on the ion. • In their compounds, the alkali metals and the alkaline earths have OS of +1 and +2 respectively. • In compounds the OS of fluorine (F) is always –1

  48. Rules for Oxidation States • In compounds, the OS of hydrogen (H) is usually +1 • In compounds, the OS of oxygen (O) is usually –2. • In binary (two-element) compounds with metals: • Halogens have OS of –1, • Group 6A have OS of –2 and • Group 5A have OS of –3.

  49. Example: Assigning Oxidation States. What is the oxidation state of the underlined element in each of the following? a) P4; b) Al2O3; c) MnO4-; d) NaH • P4 is an element. P OS = 0 • Al2O3:O is –2. O3 is –6. Since (+6)/2=(+3), Al OS = +3. • MnO4-: net OS = -1, O4 is –8. Mn OS = +7. • NaH: net OS = 0, rule 3 beats rule 5, Na OS = +1 and H OS = -1.

More Related