Chemistry chapter 11
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Chemistry Chapter 11. Writing Chemical Equations. In a chemical reaction, one or more substances (the reactants) change into one or more new substances (the products). Word Equations

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Chemistry Chapter 11

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Chemistry chapter 11

Chemistry Chapter 11


Chemistry chapter 11

  • Writing Chemical Equations


Chemistry chapter 11

  • In a chemical reaction, one or more substances (the reactants) change into one or more new substances (the products).


Chemistry chapter 11

  • Word Equations

  • How do you describe what happens in a chemical reaction? Recall from Chapter 2 the shorthand method for writing a description of a chemical reaction. In this method, the reactants were written on the left and the products on the right. An arrow separated them. You read the arrow as yields, gives, or reacts to produce.

  • Reactants → products


Chemistry chapter 11

  • To write a word equation, write the names of the reactants to the left of the arrow separated by plus signs; write the names of the products to the right of the arrow, also separated by plus signs.


Chemistry chapter 11

  • Three common chemical reactions are. a When methane gas burns, it combines with oxygen to form carbon dioxide and water. b Iron turns to red-brown rust (iron(III) oxide) in the presence of oxygen in the air. c Hydrogen peroxide decomposes to water and oxygen when used as an antiseptic.


Chemistry chapter 11

  • Methane + oxygen → carbon dioxide + water


Chemistry chapter 11

  • Methane + oxygen → carbon dioxide + water


Chemistry chapter 11

  • Hydrogen peroxide → water + oxygen


Chemistry chapter 11

  • chemical equation

  • an expression representing a chemical reaction; the formulas of the reactants (on the left) are connected by an arrow with the formulas for the products (on the right)


Chemistry chapter 11

  • skeleton equation

  • a chemical equation that does not indicate the relative amounts of reactants and products


Chemistry chapter 11

  • The first step in writing a complete chemical equation is to write the skeleton equation. Write the formulas of the reactants to the left of the yields sign (arrow) and the formulas of the products to the right.


Chemistry chapter 11

  • To add more information to the equation, you can indicate the physical states of substances by putting a symbol after each formula. Use (s) for a solid, (l ) for a liquid, (g) for a gas, and (aq) for a substance in aqueous solution (a substance dissolved in water).


Chemistry chapter 11

  • Iron plus oxygen gas yields Iron III oxide

  • Fe(s) + O2(g) → Fe2O3(s)


Chemistry chapter 11

  • catalyst

  • a substance that increases the rate of reaction by lowering the activation-energy barrier; the catalyst is not used up in the reaction


Chemistry chapter 11

  • A catalyst is neither a reactant nor a product, so its formula is written above the arrow in a chemical equation


Chemistry chapter 11

  • Hydrogen peroxide decomposes to form water and oxygen gas. a Bubbles of oxygen appear slowly as decomposition proceeds. b With the addition of the catalyst manganese(IV) oxide (MnO2), decomposition speeds up. The white “smoke” is condensed water vapor. (show on board)


Chemistry chapter 11

  • Balancing Chemical Equations


Chemistry chapter 11

  • coefficients

  • a small whole number that appears in front of a formula in a balanced chemical equation


Chemistry chapter 11

  • balanced equation

  • a chemical equation in which mass is conserved; each side of the equation has the same number of atoms of each element


Chemistry chapter 11

  • To write a balanced chemical equation, first write the skeleton equation. Then use coefficients to balance the equation so that it obeys the law of conservation of mass.


Chemistry chapter 11

  • In every balanced equation, each side of the equation has the same number of atoms of each element.

  • Sometimes a skeleton equation may already be balanced.


Chemistry chapter 11

  • carbon burns in the presence of oxygen to produce carbon dioxide. This equation is balanced. One carbon atom and two oxygen atoms are on each side of the equation.


Chemistry chapter 11

  • Rules for writing and balancing chemical equations


Chemistry chapter 11

  • The reaction of copper metal with an aqueous solution of silver nitrate is described by this skeleton equation. How can it be balanced?

  • Cu(s) + AgNO3(aq) → Cu(NO3)2(aq) + Ag(s)


Chemistry chapter 11

  • Key Concept How do you write a word equation?Hint

  • (8) Key Concept How do you write a skeleton equation?Hint

  • (9) Key Concept Describe the steps in writing a balanced chemical equation.Hint


Chemistry chapter 11

  • 10)Write skeleton equations for these reactions.

  • Heating copper(II) sulfide in the presence of diatomic oxygen produces pure copper and sulfur dioxide gas.

  • When heated, baking soda (sodium hydrogen carbonate) decomposes to form the products sodium carbonate, carbon dioxide, and water.


Chemistry chapter 11

  • (11)Write and balance equations for the following reactions.

  • Iron metal and chlorine gas react to form solid iron(III) chloride.

  • Solid aluminum carbonate decomposes to form solid aluminum oxide and carbon dioxide gas.

  • Solid magnesium reacts with aqueous silver nitrate to form solid silver and aqueous magnesium nitrate.


Chemistry chapter 11

  • 12)Balance the following equations.

  • SO2 + O2 → SO3

  • Fe2O3 + H2 → Fe + H2O

  • P + O2 → P4O10

  • Al + N2 → AlN


Chemistry chapter 11

  • When hydrogen and oxygen are mixed, a spark will initiate a rapid reaction. The product of the reaction is water. This is the equation for the burning of hydrogen:


Chemistry chapter 11

  • Classifying reactions

  • The five general types of reaction are combination, decomposition, single-replacement, double-replacement, and combustion


Chemistry chapter 11

  • combination reaction

  • a chemical change in which two or more substances react to form a single new substance; also called a synthesis reaction


Chemistry chapter 11

  • 2Mg(s) + O2(g) → 2MgO(s)

  • Notice that in this reaction, as in all combination reactions, the product is a single substance (MgO), which is a compound. The reactants in this combination reaction (Mg and O2) are two elements


Chemistry chapter 11

  • When a Group A metal and a nonmetal react, the product is a compound consisting of the metal cation and the nonmetal anion.

  • 2K(s) + Cl2(g) → 2KCl(s)


Chemistry chapter 11

  • When two nonmetals react in a combination reaction, more than one product is often possible.

  • S(s) + O2(g) → SO2(g) sulfur dioxide

  • 2S(s) + 3O2(g) → 2SO3(g) sulfur trioxide


Chemistry chapter 11

  • More than one product may also result from the combination reaction of a transition metal and a nonmetal.

  • Fe(s) + S(s) → FeS(s) iron(II) sulfide

  • 2Fe(s) + 3S(s) → Fe2S3(s) iron(III) sulfide


Chemistry chapter 11

  • decomposition reaction

  • a chemical change in which a single compound is broken down into two or more simpler products


Chemistry chapter 11

  • single-replacement reaction

  • a chemical change in which one element replaces a second element in a compound; also called a displacement reaction


Chemistry chapter 11

  • Zn(s) + Cu(NO3)2(aq) → Cu(s) + Zn(NO3)2(aq)


Chemistry chapter 11

  • You can identify a single-replacement reaction by noting that both the reactants and the products consist of an element and a compound. In the equation above, zinc and copper change places. The reacting element Zn replaces copper in the reactant compound Cu(NO3)2. The products are the element Cu and the compound Zn(NO3)2.


Chemistry chapter 11

  • Whether one metal will displace another metal from a compound depends upon the relative reactivities of the two metals.


Chemistry chapter 11

  • activity series

  • a list of elements in order of decreasing activity; the activity series of halogens is Fl, Cl, Br, I


Chemistry chapter 11

  • A halogen can also replace another halogen from a compound. The activity of the halogens decreases as you go down Group 7A of the periodic table—fluorine, chlorine, bromine, and iodine.


Chemistry chapter 11

  • Bromine is more active than iodine, so this reaction occurs:

  • Br2(aq) + NaI(aq) → NaBr(aq) + I2(aq)

  • But bromine is less active than chlorine, so this reaction does not occur:

  • Br2(aq) + NaCl(aq) → No reaction


Chemistry chapter 11

  • double-replacement reaction

  • a chemical change that involves an exchange of positive ions between two compounds


Chemistry chapter 11

  • Double-replacement reactions are also referred to as double-displacement reactions. They generally take place in aqueous solution and often produce a precipitate, a gas, or a molecular compound such as water.


Chemistry chapter 11

  • For a double-replacement reaction to occur, one of the following is usually true.


Chemistry chapter 11

  • (1) One of the products is only slightly soluble and precipitates from solution. For example, the reaction of aqueous solutions of sodium sulfide and cadmium nitrate produces a yellow precipitate of cadmium sulfide.

  • Na2S(aq) + Cd(NO3)2(aq) → CdS(s) + 2NaNO3(aq)


Chemistry chapter 11

  • (2) One of the products is a gas. Poisonous hydrogen cyanide gas is produced when aqueous sodium cyanide is mixed with sulfuric acid.

  • 2NaCN(aq) + H2SO4(aq) → 2HCN(g) + Na2SO4(aq)


Chemistry chapter 11

  • (3) One product is a molecular compound such as water. Combining solutions of calcium hydroxide and hydrochloric acid produces water.

  • Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l )


Chemistry chapter 11

  • combustion reaction

  • a chemical change in which an element or a compound reacts with oxygen, often producing energy in the form of heat and light


Chemistry chapter 11

  • The complete combustion of a hydrocarbon produces carbon dioxide and water.


Chemistry chapter 11

  • 2C8H18(l ) + 25O2(g) → 16CO2(g) + 18H2O(l )


Chemistry chapter 11

  • 2Mg(s) + O2(g) → 2MgO(s)

  • S(s) + O2(g) → SO2(g)


Chemistry chapter 11

  • The number of elements and/or compounds reacting is a good indicator of possible reaction type and thus possible products


Chemistry chapter 11

  • What are the five types of chemical reactions?


Chemistry chapter 11

  • What are the keys to predicting the products of the five general types of reactions?Hint


Chemistry chapter 11

  • Classify each reaction and balance the equations.

  • C3H6 + O2 → CO2 + H2OHint

  • Al(OH)3 → Al2O3 + H2OHint

  • Li + O2 → Li2OHint

  • Zn + AgNO3 → Ag + Zn(NO3)2Hint


Chemistry chapter 11

  • (25)Which of the five general types of reaction would most likely occur, given each set of reactants? What are the probable products?

  • an aqueous solution of two ionic compoundsHint

  • a single compoundHint

  • two elementsHint

  • oxygen and a compound of carbon and hydrogenHint


Chemistry chapter 11

  • 26)Complete and balance an equation for each reaction.

  • CaI2 + Hg(NO3)2 → (HgI2 precipitates.)

  • Al + Cl2 →

  • Ag + HCl →

  • C2H2 + O2 →

  • MgCl2 →


Chemistry chapter 11

  • (27)What are the three types of products that result from double-replacement reactions?


Chemistry chapter 11

m

  • Reactions in Aqueous Solution


Chemistry chapter 11

  • Net Ionic Equations


Chemistry chapter 11

  • when sodium chloride dissolves in water, it separates into sodium ions (Na+(aq)) and chloride ions (Cl−(aq)). Similarly, silver nitrate dissociates into silver ions (Ag+(aq)) and nitrate ions (NO3−(aq)).


Chemistry chapter 11

  • AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)


Chemistry chapter 11

  • complete ionic equation

  • an equation that shows dissolved ionic compounds as dissociated free ions


Chemistry chapter 11

  • Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + NO3−(aq)

  • Notice that the nitrate ion and the sodium ion appear unchanged on both sides of the equation. The equation can be simplified by eliminating these ions because they don’t participate in the reaction.


Chemistry chapter 11

  • spectator ion

  • an ion that is not directly involved in a chemical reaction; an ion that does not change oxidation number or composition during a reaction


Chemistry chapter 11

  • net ionic equation

  • an equation for a reaction in solution showing only those particles that are directly involved in the chemical change


Chemistry chapter 11

  • Ag +(aq) + Cl−(aq) → AgCl(s)

  • In writing balanced net ionic equations, you must make sure that the ionic charge is balanced. For the previous reaction, the net ionic charge on each side of the equation is zero and is therefore balanced.


Chemistry chapter 11

  • But consider the skeleton equation for the reaction of lead with silver nitrate.

  • Pb(s) + AgNO3(aq) → Ag(s) + Pb(NO3)2(aq)

  • The nitrate ion is the spectator ion in this reaction. The net ionic equation is this.


Chemistry chapter 11

  • Pb(s) + Ag+(aq) → Ag(s) + Pb2+(aq) (unbalanced)


Chemistry chapter 11

  • Why is this equation unbalanced? Notice that a single unit of positive charge is on the reactant side of the equation. Two units of positive charge are on the product side. Placing the coefficient 2 in front of Ag+(aq) balances the charge. A coefficient of 2 in front of Ag(s) rebalances the atoms.

  • Pb(s) + 2Ag+(aq) → 2Ag(s) + Pb2+(aq) (balanced)


Chemistry chapter 11

  • A net ionic equation shows only those particles involved in the reaction and is balanced with respect to both mass and charge.


Chemistry chapter 11

  • Predicting the Formation of a Precipitate


Chemistry chapter 11

  • You can predict the formation of a precipitate by using the general rules for solubility of ionic compounds.


Chemistry chapter 11

  • 2Na+(aq) + CO32−(aq) + Ba2+(aq) + 2NO3−(aq) → ?


Chemistry chapter 11

  • When these four ions are mixed, the cations could change partners. If they did, the two new compounds that would form are NaNO3 and BaCO3. These are the only new combinations of cation and anion possible.


Chemistry chapter 11

  • To find out if an exchange will occur, refer to Table 11.3, which gives guidelines for determining whether ion combinations are soluble. Recall that sodium is an alkali metal. Rows 1 and 2 tell you that sodium nitrate will not form a precipitate because alkali metal salts and nitrate salts are soluble.


Chemistry chapter 11

  • Row 5 indicates that carbonates in general are insoluble. Barium carbonate will precipitate. In this reaction Na+ and NO3− are spectator ions. The net ionic equation for this reaction is as follows.

  • Ba2+(aq) + CO32−(aq) → BaCO3(s)


Chemistry chapter 11

  • (30) Key Concept What is a net ionic equation?Hint

  • (31) Key Concept How can you predict the formation of a precipitate in a double-replacement reaction?Hint


Chemistry chapter 11

  • 32)Write a balanced net ionic equation for each reaction.

  • Pb(NO3)2(aq) + H2SO4(aq) → PbSO4(s) + HNO3(aq)

  • Pb(C2H3O2)2(aq) + HCl(aq) → PbCl2(s) + HC2H3O2(aq)

  • Na3PO4(aq) + FeCl3(aq) → NaCl(aq) + FePO4(s)

  • (NH4)2S(aq) + Co(NO3)2(aq) → CoS(s) + NH4NO3(aq)


Chemistry chapter 11

  • (33)Write a balanced net ionic equation for each reaction. Identify the spectator ions in each reaction.

  • HCl(aq) + AgNO3(aq) →

  • Pb(C2H3O2)2(aq) + LiCl(aq) →

  • Na3PO4(aq) + CrCl3(aq) →


Chemistry chapter 11

  • (33)Write a balanced net ionic equation for each reaction. Identify the spectator ions in each reaction.

  • HCl(aq) + AgNO3(aq) →

  • Pb(C2H3O2)2(aq) + LiCl(aq) →

  • Na3PO4(aq) + CrCl3(aq) →


Chemistry chapter 11

  • (35)Will a precipitate form when the following aqueous solutions of ionic compounds are mixed?

  • AgNO3 and Na2SO4

  • NH4Cl and Ba(NO3)2

  • CaCl2 and K2SO4

  • Pb(NO3)2 and HCl


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