KEAM 2012 Physics and Chemistry Paper

Given,
(i) C6H6+‌

15

2

O2⟶6CO2+3H2O
∆H=−3270kJmol−1
(ii) C(graphite) +O2⟶CO2
∆H=−394kJmol−1
(iii) H2+‌

1

2

O2⟶H2O
∆H=−286kJmol−1
Multiplication of Eq. (ii) by 6 and Eq. (iii) by 3 , gives
(iv) 6C(s)+6O2⟶6CO2
∆H=−394×6kJmol−1
=−2364kJmol−1
(v) 3H2+‌

3

2

O2⟶3H2O
∆H=−286×3kJmol−1
=−858kJmol−1
On inverting Eq, (i), we get
(vi) 6CO2+3H2O⟶C6H6+‌

15

2

O2
∆H=+3270kJmol−1
Addition of Eq. (iv), (v) and (vi) gives
6C(s)+3H2⟶C6H6
∆H=+3270+(−2364−858)
=+48kJmol−1
Thus, the standard enthalpy of formation of C6H6,(∆fHC6H6)is+48kJmol−1