Solveeit Logo
Login

Question

Question: The heat of atomization of \(P{H_3}_{\left( g \right)}\) is \(228kcalmo{l^{ - 1}}\) and that of \({P...

The heat of atomization of PH3(g)P{H_3}_{\left( g \right)} is 228kcalmol−1228kcalmo{l^{ - 1}} and that of P2H4(g){P_2}{H_4}_{\left( g \right)} is 355kcalmol−1355kcalmo{l^{ - 1}} . The energy of P−PP - P bond (kcal) will be:
A.102102
B.5151
C.2626
D.204204

Explanation

Solution

The heat of atomization is defined as the energy required to dissociate the one mole of substance into atoms. Heat of atomization is also called enthalpy of atomization. It is represented by the symbol ΔHa\Delta {H_a} .

Complete step by step answer:
Heat of atomization or enthalpy of atomization is defined as the energy that is required to dissociate substances into atoms or in other words we can say the energy that is required to break a bond.
The formula for enthalpy of atomization is given as follows:
Enthalpy of atomization=sum of bond enthalpy of each bond……(1)
Given data:
a)Enthalpy of atomization of PH3=228P{H_3} = 228 kcalmol−1kcal mol^{-1}.
PH3P{H_3} consists of three P−HP - H bonds.
So from equation (1), we will calculate enthalpy of atomization for PH3P{H_3} .
Enthalpy of atomization=sum of bond enthalpy of each bond
Substituting the values we get,
228=3×228 = 3 \times bond enthalpy
Bond enthalpy =2283 = \dfrac{{228}}{3}
Bond enthalpy =76kJ/mol = 76kJ/mol …..(2)
b) Enthalpy of atomization of P2H4=355{P_2}{H_4} = 355 kcalmol−1kcal mol^{-1}.
P2H4{P_2}{H_4} consists of four P−HP - H bonds and one bond.
So, Enthalpy of atomization of P2H4={P_2}{H_4} = 4×4 \times bond enthalpy of P−HP - H +1×+ 1 \times bond enthalpy of P−PP - P
We know the bond enthalpy of P−HP - H bond from equation (2).
Substituting the value we get,
Enthalpy of atomization of P2H4={P_2}{H_4} = 4×4 \times bond enthalpy of P−HP - H +1×+ 1 \times bond enthalpy of P−PP - P
355=(4×76)355 = (4 \times 76) +1×+ 1 \times Bond enthalpy of P−PP - P
355=304355 = 304 +1×+ 1 \times Bond enthalpy of P−PP - P
Bond enthalpy of P−P=355−304P - P = 355 - 304
Bond enthalpy of P−P=51kJ/molP - P = 51kJ/mol

So the correct answer is option b) bond enthalpy of P−P=51kJ/molP - P = 51kJ/mol.

Note:
The standard heat of atomization can be endothermic or exothermic. Enthalpy change does not depend on how the reaction has taken place. Enthalpy of atomization for diatomic molecules is the same as that of enthalpy of bond dissociation.