Chapter 3. Equations for Chemical Reactions

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1 hapter 3 Equations for hemical Reactions

2 hemical Reactions Reactions involve rearrangement and exchange of atoms to produce new pure substances. Reactants Products hemical Equations Shorthand way of describing a reaction Provides information about the reaction 1. formulas of reactants and products 2. states of reactants and products 3. relative numbers of reactant and product molecules

3 ombustion of Methane Methane gas reacts with oxygen gas to produce carbon dioxide gas and gaseous water. H4(g) + O2(g) O2(g) + H2O(g) This equation reads 1 molecule of H4 gas combines with 1 molecule of O2 gas to make 1 molecule of O2 gas and 1 molecule of H2O gas. H H H H + O O O + O H O H

4 What about conservation of mass?? H H H H + O O O + O H O H H + 2 O O + 2 H + O H + 3 O

5 ombustion of Methane, Balanced To show the reaction obeys the Law of onservation of Mass, the equation must be balanced. H4(g) + O2(g) O2(g) + H2O(g) H4(g) + O2(g) O2(g) + 2 H2O(g) H4(g) + 2 O2(g) O2(g) + 2 H2O(g) 1 molecule of H4 gas combines with 2 molecules of O2 gas to make 1 molecule of O2 gas and 2 molecules of H2O gas.

6 Symbols Used in Equations Symbols used to indicate state after chemical: (g) = gas; (l) = liquid; (s) = solid (aq) = aqueous = dissolved in water Energy symbols used above the arrow for conditions for reactions: Δ = heat hν = light shock = mechanical elec = electrical

7 Steps in Balancing Equations 1. In compounds balance elements other than H and O. a. Balance elements which occur only once on each side of the equation. b. Start with the elements which occur the most. c. Balance polyatomic ions which do not change in the reaction. 2. Be prepared to rebalance if something changes!!! 3. Balance H. 4. Balance O. 5. Balance elements which appear in their elemental forms.

8 When aluminum metal reacts with air, it produces a white, powdery compound, aluminum oxide. aluminum(s) + oxygen(g) aluminum oxide(s) Al(s) + O2(g) Al2O3(s) 2 Al(s) + O2(g) Al2O3(s) 2 Al(s) + 3 O2(g) 2 Al2O3(s) 4 Al(s) + 3 O2(g) 2 Al2O3(s)

9 Solid phosphorous (P4) reacts with hydrogen gas to produce phosphorous trihydride. P4 (s) + H2 (g) > PH3 (g) P4 (s) + H2 (g) > 4 PH3 (g) P4 (s) + 6 H2 (g) > 4 PH3 (g)

10 Solid potassium chlorate decomposes to produce oxygen gas and potassium chloride. KlO3 (s) > O2 (g) + Kl (s) 2 KlO3 (s) > 3 O2 (g) + Kl (s) 2 KlO3 (s) > 3 O2 (g) + 2 Kl (s)

11 Aqueous sulfuric acid reacts with solid sodium cyanide to produce aqueous sodium sulfate and hydrogen cyanide gas. H2SO4 (aq) + NaN (s) > Na2SO4 (aq) + HN(g) H2SO4 (aq) + 2 NaN (s) > Na2SO4 (aq) + HN(g) H2SO4 (aq) + 2 NaN (s) > Na2SO4 (aq) + 2 HN(g)

12 Aqueous potassium phosphate reacts with aqueous calcium nitrate to produce solid calcium phosphate and aqueous potassium nitrate. K3PO4 (aq) + a(no3)2 (aq) > a3(po4)2 (s) + KNO3 (aq) K3PO4 (aq) + 3 a(no3)2 (aq) > a3(po4)2 (s) + KNO3 (aq) K3PO4 (aq) + 3 a(no3)2 (aq) > a3(po4)2 (s) + 6 KNO3 (aq) 2 K3PO4 (aq) + 3 a(no3)2 (aq) > a3(po4)2 (s) + 6 KNO3 (aq)

13 Write a balanced equation for the combustion of butane, 4H10. 4H10 (g) + O2 (g) O2 (g) + H2O (g) 4H10 (g) + O2 (g) 4 O2 (g) + H2O (g) 4H10 (g) + O2 (g) 4 O2 (g) + 5 H2O (g) 4H10 (g) + 13/2 O2 (g) 4 O2 (g) + 5 H2O (g) 2 4H10 (g) + 13 O2 (g) 8 O2 (g) + 10 H2O (g)

14 Organic hemistry

15 lassifying ompounds Organic vs. Inorganic In the18 th century, compounds from living things were called organic; compounds from the nonliving environment were called inorganic. Organic compounds easily decomposed and could not be made in the 18 th -century lab. Inorganic compounds were very difficult to decompose, but could be synthesized.

16 Modern lassification of ompounds Organic vs. Inorganic Today we commonly make organic compounds in the lab and find them all around us. Organic compounds are mainly made of and H, sometimes with O, N, P, S, halogens, and trace amounts of other elements. The main element that is the focus of organic chemistry is carbon.

17 arbon Bonding in Organic ompounds arbon atoms bond almost exclusively covalently in organic compounds. When bonds, it forms four covalent bonds. arbon is unique in that it can form limitless chains of atoms, both straight and branched, and rings of atoms.

18 arbon Bonding in Organic ompounds

19 lassifying Organic ompounds There are two main categories of organic compounds, hydrocarbons and functionalized hydrocarbons.

20 Hydrocarbons Most fuels are mixtures of hydrocarbons

21 Families of Organic ompounds Alkanes O H Aldehydes O Alkenes Ketones Alkynes O O H arboxylic Acids Aromatics O O Esters O H Alcohols N Amines O O Ethers N Amides

22 hapter 4 hemical Quantities and Aqueous Reactions

23 Stoichiometry The study of the numerical relationship between chemical quantities in a chemical reaction

24 Making Pizza The number of pizzas you can make depends on the amount of the ingredients you use. 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza This relationship can be expressed mathematically 1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza If you want to make more than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make.

25 Predicting Amounts from Stoichiometry According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose? 6H12O6 + 6 O2 6 O2 + 6 H2O mol 6 H 12 O 6 mol H 2 O 1 mol 6H12O6 : 6 mol H2O

26 The amounts of any other substance produced or consumed in a chemical reaction can be determined from the amount of just one substance. Moles of A oefficients Moles of B

27 Estimate the mass of O2 produced in 2007 by the combustion of 3.5 x g of octane (8H18). 8H18(l) + O2(g) O2(g) + H2O(g) 2 8H18(l) + 25 O2(g) 16 O2(g) + 18 H2O(g) g 8 H 18 mol 8 H 18 mol O 2 g O 2

28 How many grams of glucose can be synthesized from 37.8 g of O2 in photosynthesis? 6 O2 + 6 H2O 6H12O6+ 6 O2 g O 2 mol O 2 mol 6 H 12 O 6 g 6 H 12 O 6 1 mol g 1 mol H O mol O g 1 mol 37.8 g = 25.8 g O 2 H 6 1 mol O g O O mol 6 H 12 O 6 mol O g 6 H 12 O 1 mol H O

29 How many grams of O2 can be made from the decomposition of g of PbO2? 2 PbO2(s) 2 PbO(s) + O2(g) (PbO2 = 239.2, O2 = 32.00) g PbO 2 mol PbO 2 mol O 2 g O 2

30 Stoichiometry Road Map Grams of A Grams of B Molar Mass Molar Mass Moles of A oefficients Moles of B Avogadro s Number Avogadro s Number Particles of A Particles of B

31 More Making Pizzas 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza What would happen if we only had 4 crusts, 15 oz. tomato sauce, and 10 cu cheese? Limiting reagent Theoretical yield

32 The Limiting Reactant For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others. When this reactant is used up, the reaction stops and no more product is made.

33 Limiting and Excess Reactants in the ombustion of Methane H4(g) + O2(g) O2(g) + H2O(g) H4(g) + 2 O2(g) O2(g) + 2 H2O(g)

34 If we have five molecules of H4 and eight molecules of O2, which is the limiting reactant?

35 If we have five molecules of H4 and eight molecules of O2, which is the limiting reactant?

36 How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 moles of N2 in the reaction 3 Si + 2 N2 Si3N4? Limiting reactant Theoretical yield

37 More Making Pizzas Let s now assume that as we are making pizzas, we burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical reaction is called the actual yield. We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of pizzas we actually make. In chemical reactions, we call this the percent yield.

38 When 28.6 kg of reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO2(s) + 2 (s) Ti(s) + 2 O(g) kg kg TiO 2 } smallest amount is from limiting reactant smallest mol Ti

39 When 28.6 kg of reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO2(s) + 2 (s) Ti(s) + 2 O(g) ollect needed relationships: 1000 g = 1 kg Molar Mass Ti = g/mol Molar Mass = g/mol Molar Mass TiO2 = g/mol 1 mole TiO2 : 1 mol Ti 2 mole : 1 mol Ti

40 When 28.6 kg of reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO2(s) + 2 (s) Ti(s) + 2 O(g) 1.19 x 10 3 mol Ti 1.10 x 10 3 mol Ti limiting reactant Theoretical yield

41 When 28.6 kg of reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO2(s) + 2 (s) Ti(s) + 2 O(g) theoretical yield percent yield

42 How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(ii) oxide? 2 NH3(g) + 3 uo(s) N2(g) + 3 u(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield? g NH 3 mol NH 3 mol N 2 g uo mol uo mol N 2? g N 2

43 How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(ii) oxide? 2 NH3(g) + 3 uo(s) N2(g) + 3 u(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield? Theoretical yield