Question

Heat of atomisation of NH3 and N2H4 are x kcal-1 mol-1 and y kcal mol-1 respectively. Average bond angle of N-N bond is:
A.$\dfrac{{3y - 4x}}{3}kcalmo{l^{ - 1}}$
B.$\dfrac{{3y - x}}{5}kcalmo{l^{ - 1}}$
C.$\dfrac{{2y - 4x}}{4}kcalmo{l^{ - 1}}$
D.None of these

Answer 1

Hint: The change in the enthalpy of the reaction when one mole of the bonds in an atom are broken to obtain the atoms in the gaseous phase is known as Enthalpy of atomisation. For example: the energy required to break the methane molecule so as to get the C and H atoms in the gas phase.New products are always formed in a spontaneous reaction. Some processes involve the evolution of energy while some involve the absorption of energy, energy can be given in the form of heat, light and radiations in order to carry out the reaction. These changes in the energy is largely due to the enthalpy of atomisation.
Enthalpy of atomization is denoted by the symbol $\Delta_{at}H$The enthalpy of sublimation of a solid in its elemental state is similar to the enthalpy of atomization of the element in the solid state and thus the atom becomes monoatomic in such phase.

Complete answer:
Heat of atomisation of NH3 and N2H4 are x kcal-1 mol-1 and y kcal mol-1 respectively. Thus the average bond angle can be calculated as follows:
$ N{H_3} \to N(g) + 2H(g);\Delta {H_1} = xkcalmo{l^{ - 1}} \\
  {N_2}{H_4} \to 2N(g) + 4H(g);\Delta {H_2} = ykcalmo{l^{ - 1}} \\
  \Delta {H_1} = 3 \times {e_{N - H}} = x..................equation(i) \\
  \Delta {H_2} = 4 \times {e_{N - H}} + {e_{N - N}} = y.........equation(ii) \\
    \\
$
Since it is clear from the question that bond length between the N-H bond will be 2x H1, since the three atoms are involved in the balanced chemical equation and thus we arrived at equation 1, While the 4 atoms are involved in the reaction 2 and the bond length for both the Bonds N-H and N-N are needed, thus the equation 2 is derived in such a way.
From equation (i) and (ii):
$
    \\
    \\
  y = 4.\dfrac{x}{3} + {e_{N - N}} \\
  {e_{N - N}} = y - \dfrac{{4x}}{3} \\
   = \dfrac{{3y - 4x}}{3}kcalmo{l^{ - 1}} \\
$.

So, the correct answer is Option A.

Note:
Such enthalpy changes are always taken against a standard and 1 mole number of atoms is considered as a standard to contrast and compare the appropriate reactions in the field of physical chemistry. These energy changes are easy to quantify if calculated for the 1 mole number of atoms.