chemical reaction

Chapter 4 by Dr. Nor Nadiah Mohamad Yusof Chemical Equation A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction 3 ways of representing the reaction of H2 with O2 to form H2O reactants products How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O2 makes 2 g MgO Balancing Chemical Equations 1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C2H6 NOT C4H12 Balancing Chemical Equations 3. Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 2 carbon on left C2H6 + O2 6 hydrogen on left C2H6 + O2 CO2 + H2O start with C or H but not O 1 carbon on right multiply CO2 by 2 2CO2 + H2O 2 hydrogen on right 2CO2 + 3H2O multiply H2O by 3 Balancing Chemical Equations 4. Balance those elements that appear in two or more reactants or products. C2H6 + O2 2 oxygen on left 2CO2 + 3H2O 7 multiply O2 by 2 4 oxygen + 3 oxygen = 7 oxygen (3x1) on right (2x2) C2H6 + 7 O2 2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O remove fraction multiply both sides by 2 Balancing Chemical Equations 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O 12 4 CO H(2 (2 6) 2) 14 (7xx2) 4(6 C 2)6) 1412 OH (4 x 2x + Reactants 4C 12 H 14 O Products 4C 12 H 14 O Example: Lets balance this equation: 1) AgNO3 + Na2CrO4  AgCrO4 + NaNO3 Answer: 2AgNO3 + Na2CrO4  AgCrO4 + 2NaNO3 2) Al2(SO4)3 + KOH  Al(OH)3 + NaNO3 Answer: Al2(SO4)3 + 6KOH  2Al(OH)3 + 3K2SO4 Questions: 1) C2H5OH (l) + O2(g)  CO2(g) + H2O(g) Answer: C2H5OH + 3O2  2CO2 + 3H2O 2) NH3 (g) + O2(g)  NO(g) + H2O(g) Answer: 4NH3 (g) + 5O2  4NO(g) + 6H2O(g) Mass Changes in Chemical Reactions 1. Write balanced chemical equation 2. Convert quantities of known substances into moles 3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4. Convert moles of sought quantity into desired units Methanol burns in air according to the equation 2CH3OH + 3O2 2CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH3OH moles CH3OH molar mass CH3OH 209 g CH3OH x moles H2O grams H2O molar mass coefficients H2O chemical equation 18.0 g H2O 4 mol H2O 1 mol CH3OH = x x 32.0 g CH3OH 1 mol H2O 2 mol CH3OH 235 g H2O Limiting Reagents 6 red green leftused overup Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe Calculate the mass of Al2O3 formed. g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 124 g Al x mol Fe2O3 1 mol Al 27.0 g Al x mol Al needed 1 mol Fe2O3 2 mol Al Start with 124 g Al 160. g Fe2O3 = x 1 mol Fe2O3 g Al needed 367 g Fe2O3 need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al2O3 2Al + Fe2O3 124 g Al x 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al g Al2O3 Al2O3 + 2Fe 102. g Al2O3 = x 1 mol Al2O3 234 g Al2O3 Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100 Example: Carbon tetrachloride was prepared by reacting 100 g of carbon disulfide with 100 g of chlorine according to the reaction. CS2 + 3CI2 CCI4 + S2CI2 (Molecular weight : CS2 =76.2 , CI2=71.0, CCI4=154 ) Calculate the percentage yield if 65.0 g of CCI4 was obtained from the experiment g CS2 mol CI2 mol CI2 needed g CI2 needed OR g CI2 mol CI2 100 g CS2 x 1 mol CS2 x 76.2 g CS2 Start with 100 g Al mol CS2 needed 3 mol CI2 2 mol CS2 71.0 g CI2 x 1 mol CI2 need 140 g CI2 However we have only 100 g CI2 so CI2 is limiting reagent g CS2 needed =139.76 g CI2 Use limiting reagent (CI2) to calculate amount of product that can be formed. g CI2 mol CI2 mol CCI4 CS2 + 3CI2 100 g CI2 x 1 mol CI2 x 71.0 g CI2 1 mol CCI4 3 mol CI2 g CCI4 CCI4 + S2CI2 154 g CCI4 x 1 mol CCI4 = 72.30 g CCI4 Calculate the percentage yield if 65.0 g of CCI4 was obtained From the experiment Percentage yield = Actual yield X 100 % Theoretical yield 65.0 g Percentage yield = 72.3 g = 90.0 % X 100 % Question: Methanol (CH3OH), also called methyl alcohol, is the simplest alcohol. It is used as a fuel in race cars and is a potential replacement for gasoline. Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5 kg CO (g) is reacted with 8.60 kg H2(g).Calculate the theoretical yield of methanol. If 3.57 x 104 g CH3OH is actually produced, What is the percent yield of methanol?. Answer: 2H2(g) + CO(g)  CH3OH(l) H2 is limiting reagent. Using the molar of CH3OH(32.04g/mol) , we can calculate the theoretical yield in gram = 6.86 x 104 CH3OH Theoretical yield : 52.0 % Chemical Bonding General Features of the Atom • An atom can lose or gain electrons to become an ion, or it can share electrons with another atom to become molecule. Cont’ • When atoms interact to form chemical bond, only their outer regions are in contact. • For this reason ,when we study chemical bonding, we are concerned primarily with valence electrons. • To keep track of valence electrons in a chemical reaction, and to make sure that the total number of electrons does not change, Chemist use a system of dots devised by Lewis and called Lewis dot symbols (Lewis symbols). What is a Valence Electron? • Valence electrons are the outer shell electrons of an atom. • The valence electrons are the electrons that participate in chemical bonding. • The number of valence electrons of an atom is deduced from its electronic configuration. What is a Lewis Symbols? • Lewis dot symbol consist of the symbol of an element and one dot for each valence electron in an atom of the element    Cl  F Li H The Octet Rule Atoms of other element attained stability either by gaining, losing, or sharing electrons until they have duplet (two) or eight(octet) valence electrons. The order in which atomic sub shells are filled in a atom start with the 1s orbital and move downward, following the arrows. Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s Example: The electronic configuration of an oxygen atom is: (a) O (Z=8) Answer: (a) 8 O: 1s2 2s2 2p4 (The number of valence electron=6) 1s 2s   2p     The Lewis symbol is  O   What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s22s22p63s23p5 1s < 2s < 2p < 3s < 3p < 4s 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n=3 l=1 ml = -1, 0, or +1 ms = ½ or -½ Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons that particpate in chemical bonding. Group e- configuration # of valence e- 1A ns1 1 2A ns2 2 3A ns2np1 3 4A ns2np2 4 5A ns2np3 5 6A ns2np4 6 7A ns2np5 7 ns1 5f 4f d1 d5 d10 ns2np1 ns2np2 ns2np3 ns2np4 Ground State Electron Configurations of the Elements ns2 ns2np5 ns2np6 Example: State the number of valence electrons and write the Lewis symbol for: (a) Al (Z=13) (b) N (Z=7) Answer: (a) 13 Al:1s2 2s2 2p6 3s2 3p1 (The number of valence electron=3)  The Lewis symbol is  Al 3s 3p   __ __ (b) 7 N: 1s2 2s2 2p3 1s 2s    2p  (The number of valence electron) =5  The Lewis Symbol is   N  Classification of the Elements Electron Configurations of Cations and Anions Of Representative Elements Na [Ne]3s1 Na+ [Ne] Ca [Ar]4s2 Ca2+ [Ar] Al [Ne]3s23p1 Al3+ [Ne] Atoms gain electrons so that anion has a noble-gas outer electron configuration. Atoms lose electrons so that cation has a noble-gas outer electron configuration. H 1s1 H- 1s2 or [He] F 1s22s22p5 F- 1s22s22p6 or [Ne] O 1s22s22p4 O2- 1s22s22p6 or [Ne] N 1s22s22p3 N3- 1s22s22p6 or [Ne] Question: State the electronic configuration for the cations /anions following element: (a) Al : 1s22s22p63s23p1 (b) F : 1s22s22p5 Answer Answer Al3+ F- : 1s22s22p6 : 1s22s22p6 of the -1 -2 -3 +3 +2 +1 Cations and Anions Of Representative Elements Na+: [Ne] Al3+: [Ne] O2-: 1s22s22p6 or [Ne] F-: 1s22s22p6 or [Ne] N3-: 1s22s22p6 or [Ne] Na+, Al3+, F-, O2-, and N3- are all isoelectronic with Ne What neutral atom is isoelectronic with H- ? H-: 1s2 same electron configuration as He Electron Configurations of Cations of Transition Metals When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals. Fe: [Ar]4s23d6 Fe2+: [Ar]4s03d6 or [Ar]3d6 Fe3+: [Ar]4s03d5 or [Ar]3d5 Mn: [Ar]4s23d5 Mn2+: [Ar]4s03d5 or [Ar]3d5 Ionic bond is the electrostatic force of attraction between a positive ion (Cation) and a negative ion (anion) Li + F 1 22s22p5 1s22s1s e- + Li+ + Li+ F 22s22p6 [He] 1s 1s2[Ne] Li Li+ + e- F F - F - Li+ F - Electrostatic (Lattice) Energy Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. Q+QE=k r Q+ is the charge on the cation Q- is the charge on the anion r is the distance between the ions cmpd MgF2 Lattice energy (E) increases as Q increases and/or as r decreases. lattice energy 2957 Q= +2,-1 MgO 3938 LiF 1036 LiCl 853 Q= +2,-2 r F < r Cl Born-Haber Cycle for Determining Lattice Energy o DHoverall = DHo1 + DHo2 + DHo3 + DHo4 + DHo5 Covalent bonds are formed when atoms share their valence electrons to complete their octet A covalent bond is a chemical bond in which two or more electrons are shared by two atoms. Why should two atoms share electrons? F + 7e- F F F 7e- 8e- 8e- Lewis structure of F2 single covalent bond lone pairs F F single covalent bond lone pairs F F lone pairs lone pairs Lewis structure of water H + O + H single covalent bonds H O H or H O H 2e-8e-2eDouble bond – two atoms share two pairs of electrons O C O or O O C double bonds - 8e8e- 8ebonds double Triple bond – two atoms share three pairs of electrons N N triple bond 8e-8e or N N triple bond Lengths of Covalent Bonds Bond Type Bond Length (pm) C-C 154 CC 133 CC 120 C-N 143 CN 138 CN 116 Bond Lengths Triple bond < Double Bond < Single Bond Comparison of Ionic and Covalent Compounds Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms electron poor region H electron rich region F e- poor H d+ e- rich F d- Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond. Electron Affinity - measurable, Cl is highest X (g) + e- X-(g) Electronegativity - relative, F is highest Classification of bonds by difference in electronegativity Difference Bond Type 0 2 Covalent 0 < and <2 Polar Covalent Ionic Increasing difference in electronegativity Covalent Polar Covalent share e- partial transfer of e- Ionic transfer e- Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H2S; and the NN bond in H2NNH2. Cs – 0.7 Cl – 3.0 3.0 – 0.7 = 2.3 Ionic H – 2.1 S – 2.5 2.5 – 2.1 = 0.4 Polar Covalent N – 3.0 N – 3.0 3.0 – 3.0 = 0 Covalent Classify the following bonds as ionic, polar covalent, or covalent: (a) MgO (b) Cl2O Answer Answer Mg =1.2 O =3.5 3.5 – 1.2 = 2.3 Cl =3.0 Ionic Polar Covalent O =3.5 3.5 – 3.0 = 0.5 Writing Lewis Structures 1. Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. 2. Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge. 3. Complete an octet for all atoms except hydrogen 4. If structure contains too many electrons, form double and triple bonds on central atom as needed. Write the Lewis structure of nitrogen trifluoride (NF3). Step 1 – N is less electronegative than F, put N in center Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5) 5 + (3 x 7) = 26 valence electrons Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms. Step 4 - Check, are # of e- in structure equal to number of valence e- ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons F N F F Write the Lewis structure of the carbonate ion (CO32-). Step 1 – C is less electronegative than O, put C in center Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4) -2 charge – 2e4 + (3 x 6) + 2 = 24 valence electrons Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms. Step 4 - Check, are # of e- in structure equal to number of valence e- ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons Step 5 - Too many electrons, form double bond and re-check # of e- O C O O 2 single bonds (2x2) = 4 1 double bond = 4 8 lone pairs (8x2) = 16 Total = 24 14 ve’s      HOCl 24 ve’s O   Cl C Cl    COCl2  Drawing Lewis Structures H O Cl   CH3OH 14 ve’s O     26 ve’s   ClO3   O Cl O    H  H C O H  H Two possible skeletal structures of formaldehyde (CH2O) H C O H H C H O An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. formal charge on an atom in a Lewis structure = total number total number of valence of nonbonding electrons in electrons the free atom - 1 2 ( total number of bonding electrons The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion. ) H -1 +1 C O formal charge on an atom in a Lewis structure H = C – 4 eO – 6 e2H – 2x1 e12 e- 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 total number total number of valence of nonbonding electrons in electrons the free atom formal charge = 4 - 2 - ½ x 6 = -1 on C formal charge = 6 - 2 - ½ x 6 = +1 on O - 1 2 ( total number of bonding electrons ) H H 0 C formal charge on an atom in a Lewis structure 0 O = C – 4 eO – 6 e2H – 2x1 e12 e- 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 total number total number of valence of nonbonding electrons in electrons the free atom formal charge = 4 - 0 -½ x 8 = 0 on C formal charge = 6 - 4 -½ x 4 = 0 on O - 1 2 ( total number of bonding electrons ) Formal Charge and Lewis Structures 1. For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present. 2. Lewis structures with large formal charges are less plausible than those with small formal charges. 3. Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms. Which is the most likely Lewis structure for CH2O? H -1 +1 C O H H H 0 C 0 O Resonance Structure A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. + + O O O O O O What are the resonance structures of the carbonate (CO32-) ion? - O C O O - O C O O - - - O C O O - Exceptions to the Octet Rule The Incomplete Octet BeH2 BF3 B – 3e3F – 3x7e24e- Be – 2e2H – 2x1e4e- F B H F Be H 3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total = 24 F 9.9 Exceptions to the Octet Rule Odd-Electron Molecules NO N – 5eO – 6e11e- N O The Expanded Octet (central atom with principal quantum number n > 2) SF6 S – 6e6F – 42e48e- F F F S F F F 6 single bonds (6x2) = 12 18 lone pairs (18x2) = 36 Total = 48