Question

1. Use the average bond enthalpies from the table to calculate the overall ΔH for this reaction. Hint: Don't forget the C-C bond in CH₃CH₃.

2 CH₃CH₃ + 7 O₂ → 4 CO₂ + 6 H₂O

Selected Average Bond Energies
Bond	Avg. Bond Energy (kJ/mol)	Bond	Avg. Bond Energy (kJ/mol)
C-H	415	H-H	436
C-C	350	H-O	464
C=C	611	H-N	389
C-O	360	N-O	222
C=O	736	N2	946
C=O in CO2	804	Cl2	243
C-N	305	Br2	193
H-Cl	431	O2	498
H-Br	366	C-Br	276
N-O	201	NO triple	631

2. Given the following known reactions:

R + N → 2 P = - 38.6 kJ/mol reaction

R → 1/2 P + M = - 20.2 kJ/mol reaction

Give the value of ΔH for the unknown reaction:

2M + 2 N → 3 P   = ?

Answer 1

Solution:

1) For the given reaction,

2CH3CH3 + 7O2 = 4CO2 + 6H2O

ΔH = Bond enthalpies of reactants - Bond enthalpies of products

= (2 x C-C + 6 x C-H + 7 O=O) - ( 8 x C=O + 12 x O-H)

= 2 mol x 350 kJ /mol + 6 mol x 415 kJ/mol + 7 mol x 498 kJ/mol ) - ( 8 mol x 736 kJ/mol + 12 mol x 464 kJ/mol)

= 700 kJ + 2490 kJ + 3486 kJ - 5888 kJ - 5568 kJ

= -4780 kJ

2) The given reactions are:

R + N = 2P , ΔH = -38.6 kJ/mol ----(1)

R = 1/2 P + M , ΔH = -20.2 kJ/mol ----(2)

Multiply equation 1 and 2 by 2 and inverting equation 2 we get,

2R + 2N = 4P , ΔH = - 77.2 kJ/mol ---(3)

P + 2M = 2R , ΔH = + 40.4 kJ/mol ----(4)

On adding equation 3 and 4, we get desired equation as,

2N  + 2M = 3P ,

ΔH = - 77.2 + 40.4 = - 36.8 kJ/mol