Question

2. Heat of formation of benzene, assuming no
resonance. Given that
BE (C-C) = 83 kcal
BE (C=C) = 140 kcal
BE (C-H) = 99 kcal
Heat of atomisation of C = 170.9 kcal
Heat of atomisation of H = 104.2 kcal
will be
(1) 39 kcal
(2) 75 kcal
(3) 1263 kcal
(4) 421 kcal​

Answer 1

Answer:

first find total heat of formation of product

then total heat of formation of reactant

heat of formation = heat of reactant -heat of product

Answer 2

Answer:

75 kcal.

Explanation:

Since, the heat of formation is the heat of the reactant - the heat of product. Here the benzene is formed by the addition of the carbon and hydrogen. If the benzene has no resonance, then there will be carbon and hydrogen for the reactant and the benzene alone for the product.

The reaction is 6C+3H2 -> C6H6.

So, the heat of the reactant will be 6*170.9 for carbon and for hydrogen will be 104.2*3 which on solving we will get 1338 kcal.

The heat of the product will be 3(C=C) + 3(C-C) + 6 (C-H) which will be 3*140 + 3*83 + 6*99 which on solving we will get 1263 kcal.

So, the heat of formation will be 1338-1263=75 kcal