a Ca s Cl 2 g CaCl 2 s b Na s 12 H 2 g Cgraphite 32 O 2 g NaHCO 3 s c Cgraphite

# A ca s cl 2 g cacl 2 s b na s 12 h 2 g cgraphite 32 o

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a) Ca( s ) + Cl 2 ( g ) CaCl 2 ( s ) b) Na( s ) + 1/2 H 2 ( g ) + C(graphite) + 3/2 O 2 ( g ) NaHCO 3 ( s ) c) C(graphite) + 2 Cl 2 ( g ) CCl 4 ( l ) d) 1/2 H 2 ( g ) + 1/2 N 2 ( g ) + 3/2 O 2 ( g ) HNO 3 ( l ) 6.52 a) 1/2 H 2 ( g ) + 1/2 I 2 ( s ) HI( g ) b) Si( s ) + 2 F 2 ( g ) SiF 4 ( g ) c) 3/2 O 2 ( g ) O 3 ( g ) d) 3 Ca( s ) + 1/2 P 4 ( s ) + 4 O 2 ( g ) Ca 3 (PO 4 ) 2 ( s ) 6-8
6.53 The enthalpy change of a reaction is the sum of the f H Δ D of the products minus the sum of the f H Δ D of the reactants. Since the f H Δ D values (Appendix B) are reported as energy per one mole, use the appropriate coefficient to reflect the higher number of moles. = m[ rxn H Δ D f H Δ D (products)] – n[ f H Δ D (reactants)] a) = [2 rxn H Δ D f H Δ D (SO 2 ( g )) + 2 f H Δ D (H 2 O( g ))] – [2 f H Δ D (H 2 S( g )) + 3 f H Δ D (O 2 ( g ))] = [2 mol(–296.8 kJ/mol) + 2 mol(–241.826 kJ/mol)] – [2 mol(–20.2 kJ/mol) + 3(0.0)] = –1036.8 kJ b) The balanced equation is CH 4 ( g ) + 4 Cl 2 ( g ) CCl 4 ( l ) + 4 HCl( g ) rxn H Δ D = [1 f H Δ D (CCl 4 ( l )) + 4 f H Δ D (HCl( g ))] – [1 f H Δ D (CH 4 ( g )) + 4 f H Δ D (Cl 2 ( g ))] = [1 mol(–139 kJ/mol) + 4 mol(–92.31 kJ/mol)] – [1 mol(–74.87 kJ/mol) + 4 mol(0)] = –433 kJ rxn H Δ D 6.54 a) = [ 1 mol ( rxn H Δ D f H Δ D SiF 4 ( g )) + 2 mol ( f H Δ D H 2 O( l ))] – [1 mol ( f H Δ D SiO 2 ( s )) + 4 mol ( f H Δ D HF( g ))] = [1 mol (–1614.9 mol kJ ) + 2 mol (–285.840 mol kJ )] – [1 mol (–910.9 mol kJ ) + 4 mol (–273 mol kJ )] = –184 kJ b) 2 C 2 H 6 ( g ) + 7 O 2 ( g ) 4 CO 2 ( g ) + 6 H 2 O( g ) = [4 mol ( rxn H Δ D f H Δ D CO 2 ( g )) + 6 mol ( f H Δ D H 2 O( g ))] – [2 mol ( f H Δ D C 2 H 6 ( g )) + 7 mol ( f H Δ D O 2 ( g ))] = [4 mol (–393.5 mol kJ ) + 6 mol (–241.826 mol kJ )] – [2 mol (–84.667 mol kJ ) + 7 mol (0)] = –2855.6 kJ (or –1427.8 kJ for reaction of 1 mol of C 2 H 6 ) 6.55 = m[ rxn H Δ D f H Δ D (products)] – n[ f H Δ D (reactants)] Cu 2 O( s ) + 1/2 O 2 ( g ) 2 CuO( s ) = –146.0 kJ rxn H Δ D = [2 mol ( rxn H Δ D f H Δ D CuO( s ))] – [1 mol ( f H Δ D Cu 2 O( s )) + 1/2 mol ( f H Δ D O 2 ( g ))] –146.0 kJ = [2 mol ( f H Δ D CuO( s ))] – [1 mol (–168.6 mol kJ ) + 1/2 mol (0)] –146.0 kJ = 2 mol ( f H Δ D CuO( s )) + 168.6 kJ f H Δ D CuO( s ) = mol 2 kJ 314.6 = –157.3 kJ/mol 6.56 C 2 H 2 ( g ) + 5/2 O 2 ( g ) 2 CO 2 ( g ) + H 2 O( g ) = –1255.8 kJ rxn H Δ D = [2 mol ( rxn H Δ D f H Δ D CO 2 ( g )) + 1 mol ( f H Δ D H 2 O( g ))] – [1 mol ( f H Δ D C 2 H 2 ( g )) + 5/2 mol ( f H Δ D O 2 ( g ))] –1255.8 kJ = [2 mol (–393.5 kJ/mol) + 1 mol (–241.826 kJ/mol)] – [1 mol ( f H Δ D C 2 H 2 ( g )) + 5/2 mol(0.0)] –1255.8 kJ = –787.0 kJ – 241.8 kJ – 1 mol ( f H Δ D C 2 H 2 ( g )) f H Δ D C 2 H 2 ( g ) = mol 1 kJ 227.0 = 227.0 kJ/mol 6.57 a) 4 C 3 H 5 (NO 3 ) 3 ( l ) 6 N 2 ( g ) + 10 H 2 O( g ) + 12 CO 2 ( g ) + O 2 ( g ) b) = [6 mol ( rxn H Δ D f H Δ D N 2 ( g )) + 10 mol ( f H Δ D H 2 O( g )) + 12 mol ( f H Δ D CO 2 ( g )) + 1 mol ( f H Δ D O 2 ( g ))] – [4 mol ( f H Δ D C 3 H 5 (NO 3 ) 3 ( l ))] –2.29 x 10 4 kJ= [6 mol (0) +10 mol (–241.826 mol kJ ) +12 mol (–393.5 mol kJ ) +1 mol (0)] – [4 mol ( f H Δ D C 3 H 5 (NO 3 ) 3 ( l ))] 6-9