Question

Chapter 8 problems plus review Two important industrial chemicals, ethene (CaHs) and propene (C)He) are produced by a steam cracking process 1. 2C He (g)CHe (g)+ CaHs (g) + CHe (g) H 24KS)1Cs 4) ) 4)+1 (2) a) Draw Lewis Structures for each of the molecules b) Predict the geometry and bond angles about each carbon atom. c) Determine the hybridization about each of the carbon atoms d) Predict the AH based on the bond energies e) Is the reaction exothermic or endothermic? Draw all possible resonance structures for SO. Determine the formal charge on each atom for each drawing. Circle the Lewis structure you believe to be the most likely. 2.

Answer 1

4 1 it i 1) 2 C3H8 H+ + C3 Hbg + Chung) + H2 (9) a) C348 (propane) CH4 (Ethere) €346 (prospera) Chin (meltone) 4 4 1 1 H-G- - -H H-_=_-H H-S-C=4-- H- -1+ и н и H₂ Chydrogen) H-H b) propane, geometry = Tebrahedral Eltone, geometry & Trigonal planar Propene, 1 Carbon = Tetrahedral (sps), 8Carbon atorms Trigonal planar (sp) methane geometry = Tetrahedral c) propane, all carbons are = sp² hybridized Ethene, all carbon are = sp² hybridized propene, I carbon = sp3, 2, 3 comborsatoms = sp2 hybridized methane, carbon atom = sp hybridized d)stran = Bond energy of reactants -Bond energy of producto =f(16*413)+(4x342) 34[C14~*'3 )+ (346)+(2464) **36)] +(80001(1794) = +206 kJ a e) Reaction is endothermic, on it requires energy to takes place. total valenca electrons = 6+ (2x6) =1828 =s=o orsza m rsa: (I) (0) (CT) Formal charge = valency of atom _{lone pair est le bondpair et] I) 0 = 6-(4+2)=0 0 -6-(6+1] =- 1 0 046-4+2) = 0 SE 6-(3+4)=0 SE 6-(2+3)= + = 6-(2+3)=+1 026-(4+3] 0 0 = 6-56+1] =-1 Structurel I) is more stable, as it contains Zero formal change on allatom.