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Chemistry Unit 7: Chemical Equations Evidence of a chemical reaction: odor heat sound light gas emitted color change A reaction has occurred if the chemical and physical properties of the reactants and products differ. For a reaction to occur, particles of reactants must collide, and with sufficient energy collision theory activation energy: energy needed to start a reaction Chemical reactions release or absorb energy. exothermic reactions The reaction in an oxyacetylene torch is exothermic. endothermic reactions Photosynthesis is an endothermic reaction. catalyst: speeds up reaction wo/being consumed … it lowers the activation energy (AE) Energy without catalyst AE with catalyst AE time time Examples: enzymes catalyze biochemical reactions catalytic converters convert CO into CO2 Reaction Conditions and Terminology Certain symbols give more info about a reaction. (s) = solid (l) = liquid (g) = gas (aq) = aqueous (dissolved in H2O) NaCl(s) NaCl(aq) More on aqueous… -- “soluble” or “in solution” also indicate that a substance is dissolved in water (usually) -- acids are aqueous solutions Other symbols… (i.e., clues about the reaction) means...“yields” or “produces” heat is added to the reaction means ______ MgCO3(s) MgO(s) + CO2(g) Temp. at which we perform rxn. might be given. C6H5Cl + NaOH 400oC C6H5OH + NaCl The catalyst used might be given. C2H4(g) + H2(g) Pt C2H6(g) precipitate: a solid product that forms in an aqueous solution reaction Na2CO3 (aq) + Ca(NO3)2 (aq) Na1+ Na1+ CO32– clear Na2CO3 solution Ca2+ CaCO3(s) + 2 NaNO3(aq) NO31– NO31– clear Ca(NO3)2 solution ppt “chunks” “sinkies” “floaties” cloudy solution containing CaCO3(s) and NaNO3(aq) Factors that influence the rate of a reaction To make reaction rate increase… concentration of reactants particle size temperature mechanical mixing pressure catalyst nature of reactants use one N/A In a reaction: atoms are rearranged mass AND are conserved energy charge Balancing Chemical Equations law of conservation of mass = same # of atoms of each type on each side of equation EX. solid iron reacts with oxygen gas to yield solid iron (III) oxide Fe3+ O2– ___Fe(s) + ___O2(g) ___Fe2O3(s) If all coefficients are 1… + ___O 1 2(g) 1 ___Fe(s) 1 ___Fe 2O3(s) + If we change subscripts… 1 3(g) 1 ___Fe 2(s) + ___O + 1 ___Fe 2O3(s) subscript changes the substance. Changing a ___________ coefficients To balance, modify only _____________. superscripts Right now, _______________ don’t enter into our “balancing” picture. 4 3 2(g) + ___O ___Fe(s) 2 ___Fe 2O3(s) + Hint: Start with most complicated substances first and leave simplest substances for last. solid sodium reacts w/oxygen to form solid sodium oxide 4 1 2(g) ___Na(s) + ___O Na1+ O2– 2 ___Na 2O(s) + Aqueous aluminum sulfate reacts w/aqueous calcium chloride to form a white precipitate of calcium sulfate. The other compound remains in solution. Al3+ SO42– Ca2+ Cl1– 1_ Al2(SO4)3 (aq) + 3_ CaCl2 (aq) 3 _ AlCl3 (aq) _ CaSO4(s) + 2 Methane gas (CH4) reacts with oxygen to form carbon dioxide gas and water vapor. Furnaces burn primarily methane. 1 2 O2(g) _ CH4(g) + _ 1 2 H2O(g) _ CO2(g) + _ 1 CaC2(s) + _ 1 H2O(l) _ 1 _ C2H2(g) + 1 _ CaO(s) _ 3 CaSi2 + _ 2 SbI3 _ Si + 2 _ Sb + 3 _ CaI2 6 _ 2 Al + _3 6 CH3OH _1 2 Al(CH3O)3 + _3 H2 1 C2H2(g) + _ 2 5 O2(g) ** _ 2 CO2(g) + 1 4 2 _ _ H2O(l) 1 C3H8 + _ 5 O2 ** _ 3 CO2 + _ 4 H2O _ 1 C5H12 + _ 8 O2 ** _ 5 CO2 + _ 6 H2O _ ** = complete combustion of a hydrocarbon yields CO2 and H2O Write equations for the combustion of C7H16 and C8H18. 1 C7H16 + 11 _ _ O2 7 CO2 + _ 8 H2O _ 1 2 C8H18 + 25 _ _ O2 8 CO2 +18 9 H2O 16 _ _ Classifying Reactions four types synthesis: simpler substances combine to form more complex substances A+B AB AB + C ABC oxygen + rhombic sulfur 8 O2 + __ 1 S8 __ sodium + chlorine gas 2 Na + __ 1 Cl2 __ A+B+C ABC sulfur dioxide 8 SO2 __ sodium chloride __ 2 NaCl decomposition: complex substances are broken down into simpler ones AB A+B ABC lithium chlorate Li1+ ClO31– 2 1 _ LiClO3 water 2 H 2O _ AB + C ABC A+B+C lithium chloride + oxygen Li1+ Cl1– 2 1_ LiCl + _ 3 O2 hydrogen gas + oxygen gas 2 _ H2 1 O2 + _ single-replacement: one element replaces another AB + C A + CB sodium chlorine + bromide Na1+ Br1– 1 _ Cl2 + 2 _ NaBr copper (II) aluminum + sulfate Cu2+ SO42– 2 _ Al + 3 _ CuSO4 AB + C B + AC sodium + bromine chloride Na1+ Cl1– 2 _ Br2 _ NaCl + 1 aluminum ? + copper sulfate Al3+ SO42– 1_ Al2(SO4)3 + 3 _ Cu double-replacement: AB + CD iron (III) + chloride Fe3+ Cl1– 1 _ FeCl3 + AD + CB potassium hydroxide K1+ OH1– 3_ KOH iron (III) potassium ? + hydroxide chloride Fe3+ OH1– K1+ Cl1– 1 _ Fe(OH)3 + 3 _ KCl lead (IV) calcium + nitrate oxide lead (IV) calcium + ? oxide nitrate Pb4+ NO31– Ca2+ O2– 1 _ Pb(NO3)4 + 2 _ CaO Pb4+ O2– Ca2+ NO31– 1_ PbO2 + 2_ Ca(NO3)2 How do we know if a reaction will occur? For single-replacement reactions, use Activity Series. In general, elements above replace elements below. 1 1 FeSO4 _ Ba + _ 3 _ Mg + 2 _ Cr(ClO3)3 1 _ Fe + 1 _ BaSO4 2_ Cr + _ 3 Mg(ClO3)2 _ Pb + _ Al2O3 NR _ NaBr + 1 _ Cl2 2 _ _ Br2 2 NaCl + 1 _ FeCl3 + _ I2 1_ CoBr2 + 1 _ F2 NR 1 CoF2 + 1 _ _ Br2 For double-replacement reactions, reaction will occur water a gas if any product is: a precipitate driving forces Check new combinations to decide. 1 Pb(NO3)2(aq) + _ 2 KI(aq) 1 PbI2(s) + 2 _ _ _ KNO3(aq) Pb2+ NO31– K1+ I1– _ KOH(aq) + _ 2 1 H2SO4(aq) K1+ OH1– H1+ SO42– _ FeCl3(aq) + _ Cu(NO3)2(aq) Fe3+ Cl1– Cu2+ NO31– Pb2+ I1– (?) (ppt) K1+ NO31– (?) (aq) _ K2SO4(aq) + 2 _ H2O(l) 1 K1+ SO42– (?) H1+ OH1– (?) (aq) (water) NR (aq) (aq) Ions in Aqueous Solution Pb(NO3)2(aq) Pb(NO3)2(s) Pb2+(aq) + 2 NO31–(aq) Pb2+ NO3 1– NO31– add water NO31– Pb2+ NO31– dissociation: “splitting into ions” NaI(aq) NaI(s) Na1+(aq) + I1–(aq) Na1+ I1– add water Na1+ I1– Mix them and get the overall ionic equation… Pb2+ NO31– reactants I1– Na1+ NO31– Na1+ I1– 1– 1 2+(aq) + __NO 2 2 1+(aq) + __I 2 1–(aq) __Pb 3 (aq) + __Na yields 1–(aq) + __Na 1+(aq) 1 2 2 + __NO __PbI (s) 3 2 Na1+ Pb2+ I1– NO31– I1– Na1+ products NO31– Cancel spectator ions to get net ionic equation… 2+(aq) + __I 1 2 1–(aq) __Pb 1 __PbI 2(s) I1– Pb2+ I1– Pb2+ I1– I1– Mix together Zn(NO3)2(aq) and Ba(OH)2(aq): clear Zn(NO3)2 solution Zn(NO3)2(aq) Zn2+(aq) + 2 NO31–(aq) clear Ba(OH)2 solution Ba(OH)2(aq) Ba2+(aq) + 2 OH1–(aq) OH1– NO31– Ba2+ Zn2+ NO31– OH1– Mix them and get the overall ionic equation… Zn2+ OH1– NO31– Ba2+ NO31– reactants OH1– 2+(aq) + __NO 1– 2+(aq) + __OH 1–(aq) 1 2 1 2 __Zn 3 (aq) + __Ba yields 1 1–(aq) + __Ba 2+(aq) 2 1 __Zn(OH)2(s) + __NO 3 Ba2+ Zn2+ OH1– products NO31– OH1– NO31– Cancel spectator ions to get net ionic equation… 1–(aq) 1 2+(aq) + __OH 2 __Zn 1 __Zn(OH) 2(s) OH1– Zn2+ Zn2+ OH1– OH1– OH1– Polymers and Monomers polymer: a large molecule (often a chain) made of many smaller molecules called monomers Polymers can be made more rigid if the chains are linked together by way of a cross-linking agent. Monomer Polymer H H O amino acids………… proteins H–N–C–C–O–H nucleotides (w/NR bases A,G,C,T/U)….. nucleic acids styrene……………… polystyrene PVA…………………. “slime” polyvinyl alcohol Quantitative Relationships in Chemical Equations 4 Na(s) + O2(g) 2 Na2O(s) Particles 4 atoms 1 m’cule 2 m’cules Moles 4 mol 1 mol 2 mol Grams 4g 1g 2g ** Coefficients of a balanced equation represent # of particles OR # of moles, but NOT # of grams. When going from moles of one substance to moles of another, use coefficients from balanced equation. mass vol. Use coeff. from balanced equation in crossing this bridge MOL mass vol. MOL part. part. SUBSTANCE “B” (unknown) SUBSTANCE “A” (known) 4 Na(s) + 1 O (g) 2 Na O(s) 2 2 4 Na(s) + O2(g) 2 Na2O(s) Na Na 2O Na Na O22O How many moles oxygen will react with 16.8 moles sodium? 1 mol O2 16.8 mol Na = 4.2 mol O2 4 mol Na How many moles sodium oxide are produced from 87.2 moles sodium? 2 mol Na2O 87.2 mol Na = 43.6 mol Na2O 4 mol Na ( ( ) ) How many moles sodium are required to produce 0.736 moles sodium oxide? 0.736 mol Na2O ( ) 4 mol Na 2 mol Na2O = 1.472 mol Na 0 0