Ch. 8: Chemical Reactions and Chemical Quantities Dr. Namphol Sinkaset Chem 200: General Chemistry I.

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Presentation transcript:

Ch. 8: Chemical Reactions and Chemical Quantities Dr. Namphol Sinkaset Chem 200: General Chemistry I

I. Chapter Outline I.Introduction II.Chemical Change III.Writing and Balancing Equations IV.Reaction Stoichiometry V.Limiting Reactants and % Yield VI.Three Types of Reactions

I. Reaction Chemistry The stereotypical chemist sits in a lab and runs reactions w/ glassware and chemicals. In reaction chemistry, we are concerned with two things. 1)What will happen if we mix this and that? 2)How much reactant do we need or how much product will form?

II. Chemical Change Chemical change occurs when atoms rearrange, such that the original compound becomes a different one. Contrast with physical change, in which the composition remains the same. These are closely related to chemical and physical properties.

II. Chemical/Physical Properties A chemical property is only displayed when a substance changes its composition.  e.g. flammability – when a substance exhibits flammability, it becomes one or more different substances A physical property is displayed without changing composition.  e.g. smell or color

II. Chemical/Physical Change

III. Chemical Reactions Chemical changes occur via chemical reactions, of which there are several types. We represent chemical reactions with chemical equations. Chemical equations must be balanced. Why?

III. Balancing Equations Steps for balancing: 1)Translate into formulas, if necessary. Reactants on left, products on right. 2)Balance atoms one at a time. Start w/ most complicated compound or element that’s NOT O or H. Single elements are easy to balance last. 3)Can ONLY change coefficients, not subscripts!! Adjust to smallest whole numbers, if necessary. 4)Check work! 5)Specify states.

III. Balancing Practice e.g. Balance the following. a)Solid calcium carbonate reacts with nitric acid to form carbon dioxide gas, liquid water, and aqueous calcium nitrate. b)PCl 3 + HF  PF 3 + HCl c)C 4 H 10(g) + O 2(g)  CO 2(g) + H 2 O (g)

III. Last Thoughts on Balancing Coefficients distribute over all atoms in a formula. Never change a formula when trying to balance. e.g. MgO cannot be changed to MgO 2 to balance O atoms. Never add other reactants/products to try to achieve balance. A balanced equation stays balanced when multiplied by a constant.

IV. Calculations w/ Equations A balanced equation allows calculations of amounts of reactants or products. If you know the # of moles of one substance in a balanced equation, you know the # of moles of any of the other substances. Numerical relationships between chemical amounts in a balanced chemical equation are called reaction stoichiometry.

IV. Mole Ratios Many stoichiometric relationships exist in any one balanced equation. e.g. C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (l)  1 mole C 3 H 8 reacts w/ 5 moles O 2  1 mole C 3 H 8 produces 3 moles CO 2  5 moles O 2 leads to 4 moles H 2 O These can be converted to mole ratios. 1 mole C 3 H 8 5 moles O 2 1 mole C 3 H 8 3 moles CO 2 5 moles O 2 4 moles H 2 O

IV. Mass-to-Mass Conversions How much CO 2 produced from burning gasoline? 2 C 8 H O 2  16 CO H 2 O

IV. Sample Problem How many grams of iron form when 135 g of aluminum reacts according to the reaction below? Fe 2 O 3(s) + 2Al (s)  Al 2 O 3(s) + 2Fe (l)

IV. Sample Problem How many kg of HNO 3 forms, assuming adequate O 2 and H 2 O, if 16 kg of NO 2 reacts according to the equation below? NO 2(s) + O 2(s) + H 2 O (l)  HNO 3(aq)

V. Limiting Reactants Consider the following reaction.  2CO (g) + O 2(g)  2CO 2(g) Submicroscopically, 2 molecules CO react with 1 molecule O 2 to give 2 molecules CO 2. How much CO 2 forms if we have 2 molecules CO and 10 molecules of O 2 ? We call CO the limiting reactant.

V. Sample Problem How many grams of Al 2 S 3 can be formed according to the reaction below if 10.0 g Al is reacted with 15.0 g S 8 ? 16Al (s) + 3S 8(s)  8Al 2 S 3(s)

V. Sample Problem What is the maximum amount of silicon dioxide that can be produced when 168 g of CaSiO3 reacts with 125 g of CO2 with adequate water according to the reaction below? 2CO 2(g) + CaSiO 3(s) + H 2 O (l)  SiO 2(s) + Ca(HCO 3 ) 2(aq)

V. Real-life Reactions In reality, we cannot have complete conversion to products. Even if there was complete conversion, difficult to actually collect all of the product.

V. Reaction Yields Stoichiometry gives us theoretical yield. What is collected in lab is the actual yield. The efficiency of a reaction is commonly expressed as percent yield.

V. Sample Problem Elemental iron can be obtained using the reaction below. If 167 g of Fe2O3 reacts with 85.8 g of CO to produce 72.3 g of Fe, what is the percent yield? Fe 2 O 3(s) + 3CO (g)  2Fe (s) + 3CO 2(g)

VI. Three Chemical Reactions We will now look briefly at 3 examples of chemical reactions:  Combustion reactions  Alkali metal reactions  Halogen reactions

VI. Combustion Reactions A combustion reaction involves the reaction of a substance with O 2 to form one or more oxygen-containing compounds.  One of the products is often H 2 O.  Combustion reactions also emit heat.

VI. Typical Combustion Reactions In the combustion reactions you will commonly see, the reactant will only have C, H, and O.  All of the C becomes CO 2 and all of the H becomes H 2 O.  CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g)  If there are other elements in the reactant, they would form an oxide.

VI. Sample Problem Write a balanced reaction for the complete combustion of liquid C 2 H 5 SH.

VI. Alkali Metal Reactions Alkali metals have ns 1 valence e- configs, so they are quite reactive. They react with nonmetals.  2M + X2  2MX They react with water.  2M (s) + 2H 2 O (l)  2M + (aq) + 2OH - (aq) + H 2(g)

VI. Na/Cl 2 and Alkali Metals/H 2 O

VI. Halogen Reactions Halogens have ns 2 np 5 outer e- configs, so they are quite reactive. React w/ metals to form metal halides.  2Fe (s) + 3Cl 2(g)  2FeCl 3(s) React w/ H 2 to form hydrogen halides.  H 2(g) + I 2(g)  2HI (g) React w/ each other to form interhalogens.  Br 2(l) + F 2(g)  2BrF (g)