Organic Chemistry Assignment Report

Chemistry 1
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Chemistry 2
Q2
Part a
i). Ethane
Ethane + Oxygen → Carbon (IV) +water
2C2H6 (g)+7O2 (g) → 4O2(g)+6H2O(l)
ii. Ethene
Ethene + Oxygen → Carbon (IV) +water
C2H4 (g)+3O2 (g) → 2O2 (g)+2H2O (l)
iii. Hydrogen
Hydrogen + Oxygen → Water
2H2 (g) + O2(g)→ 2H2O (l)
Part b
2C2H6 (g) + 7O2 (g) → 4O2(g)+6H2O(l)
C+ H2+ O2
∆ Hc= ( 6 ×285.5 ) + ( 4 ×−393 ) + ( 2 ×−83.6 ) −1713−1572+167.2
∆ Hc=3117.8

Chemistry 3
C2H4 (g) + 3O2 (g) → 2O2 (g)+2H2O (l)
C+H2+O2
∆ Hc= ( 2×−393 ) + ( 2× 285.5 ) −52
∆ Hc=−1304
Hydrogen
2H2 (g) + O2(g) → 2H2O (l)
H2+O2
∆ Hc=2×−285.5
∆ Hc=¿-571
Q3
CH4 (g)+Cl2(g) CH3Cl + HCl
C+H2+Cl2
∆ Hc=∆ H products−∆ H reactants
∆ Hc=−13.5+ ( −92.3 ) −(−74.8)

Chemistry 4
∆ Hc=−31
Q4
4NH3+ 3O2 2N2 + 6H2O
H2 + ½ O2
∆ Hc=−1530−288
∆ Hc=1818
Q five
B2H6(g) + 3O2(g) -----------> B2O3(s) + 3H2O(g)
H0rxn = [ ∆H0f(B2O3) + 3
∆H0f(H2O)] - [
∆H0f(B2H6) + 3 ∆H0f(O2)]
∆H0rxn = [-1270 + 3(-242)] - [31.4 + 3 (0)]
∆H0rxn = - 1996 - 31.4
∆H0rxn = - 2027 KJ /mol
now second part
C6H6(g) + 7.5O2(g) ----------> 6CO2(g) + 3H2O(g)
∆H0rxn = [6 ∆H0f(CO2) + 3 ∆H0f(H2O)] - [∆H0f(C6H6) + 7.5∆H0f(O2)]
∆H0rxn = [6 (-393) + 3(-242)] - [83.9 + 7.5(0)]

Chemistry 5
∆H0rxn = [-2358 - 726] - [83.9]
∆H0rxn = - 3167.9 KJ/mol
Q6
C2H5OH + 3O2 2CO2 +3H2O ∆ H 1=1368
Reaction for formation of CO2
C+O2 CO2 ∆ H 1=−393.7
Reaction for formation of H2O is given below
H2+½O2 H2O ∆ H 1=−285.9
Reaction for formation of ethanol
2C+3H2+½O C2H5OH
2C+3H2+3/2 O2- C2H5OH-3O2 2CO2+3H2O-2CO2-3H2O
= (-393.2 ×2 ¿+ ( 3 × 285.9 ) −(−1368)
= -277.1kJ mol-1
The enthalpy of formation of ethanol is -277.1kJ mol-1
Q 7
Part a
CH4+Br CH3Br+HBr
∆ H f =∆ Hproduct−∆ Hreactant

Chemistry 6
∆ H = (3× 413 ¿+ ( 280 ) +366−193+(4 × 413)
∆ H =1239+280+366−193+1652
∆ H = 40kJmol-1
Part b
CH4 +F2 CH3F+HF
∆ H = ( 3 ×413 )+ 425+565− ( 4 × 413 ) +152
∆ H =1239+425+565−1652+152
∆ H =2229−1804
∆ H =425
Part c
∆ H =∆ Hproduct−∆ Hreactant
∆ H =366−1845
∆ H =−1479
Part d
C3H8+H2 C3H6
∆ H =611+347+ ( 6 × 413 )=3436
∆ H = ( 2× 347 ) + ( 8 × 413 ) +435=4433
∆ H =3436−4433=−997

Chemistry 7
∆ H =−997
Part e
C3H8+Br C3H6Br2 +H2
∆ H0 f = ( 2 ×347 )+ (6 × 413 )+ ( 2 ×280 )+870=4602
∆ H0 f = ( 2 ×347 )+ ( 8× 413 ) +(2 ×193)=4384
∆ H0 f =4602−4384
Part f
CH4+2O2 CO2+2H2O
∆ H0 f = ( 2 ×805 )+ (2 × 464 ×2 )− ( 4 × 413 )+(498 × 2)
∆ H0 f =1610+1856−1652+ 996
∆ H0 f =3466−2648
∆ H0 f =818
Part g
C2H4+H2 CH6
∆ H0 f =347+ ( 6 × 413 )−611+(4 × 413)
∆ H0 f =347+2478−611+1652
∆ H0 f =2825−2263
∆ H0 f =562

Chemistry 8
Part h
C2H6+3O2 2O2 + 2H2O
∆ H0 f = ( 2 ×805 )+ (2 × 464 ×2 )− ( 2× 347 ) + ( 6 × 413 ) +(3× 498)
∆ H0 f =3220+1850−694+2478+1494
∆ H0 f =5076−4666
∆ H0 f =410
Q 8
N2+3H2 2NH3
2x-945+(436) = 46
2 X
2 = 2299
2
x=1149.5
But there are 3 bonds of N-H
3x =1149.5
3 x
3 = 1149.5
3
X=383
Q9
-135.5+715+242

Chemistry 9
-135.5+957
= 821.5
Q 10
C4H10+I2 C4H9I+HI
X+ ( ( 3 ×347 )+ ( 9 ×413 )−4130+ ( 3× 347 ) +214=−52
X+1041+3717-4130+1041+214=−52
X+3717-4130+214 = −52
X-413+214= −52
X= −52+199=147
The bond between C-I is therefore 147
Q11
C4+O2 C2H5OH
347+ ( 5 × 435 ) +464+ X −611+ ( 4 × 413 ) =−277
2986-2263+X = −277
X+723= −277
X=−277−723
X= −1000
The enthalpy of the C-O bond is −1000

Chemistry 10
Q12
Q13

Chemistry 11
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