Question

please answer me in 6 Marks
please answer both parts (a) and (b)
Both parts are so important​

Answer 1

Answer:

hloo mate...

here's ur answer..

a) the higher the enthalpy of formation of a substance, the higher the energy level of the substance. The higher the energy level of the substance, the less stable it becomes, as it can lose more energy by reacting to from lower energy products.

[n-butane(∆fH°=-120 KJ mol-1) and iso-butane (∆fH°=-130 KJ mol-1)]

b) The bond enthalpy of N-H in NH3 is, 557.6 KJ/mole

Explanation :

The balanced chemical reaction are,

(1)

(2)

(3)

as given in the question..

The bond dissociation reaction of ammonia will be,

NH3(g) -------> 3H + N

Now adding reverse reaction of reaction 1, thrice of reaction 2 and reaction 3, we get the bond enthalpy of N-H.

The expression of bond dissociation enthalpy is,

del H = del H1 + 3× del H2 + del H3/3

del H = 46 + 3× 218 + 973 / 3

= 557.6 KJ/mol

Therefore, the bond enthalpy of N-H in NH3 is, 557.6 KJ/mole.

HOPE IT HELPS U....

IT TOOK A REALLY HARD WORK TO DO..