Question

1.
The bond dissociation energies of X, Y, and XY are
in the ratio of 1 : 0.5 : 1. AH for the formation of
XY is -200 kJ mol-1. The bond dissociation energy
of X, will be
[NEET-2018]
(1) 200 kJ mol-1 (2) 100 kJ mol-1
(3) 400 kJ mol-1 (4) 800 kJ mot!​

Answer 1

Answer:

The correct answer is option (4).

Explanation:

Ratio of bond dissociation of $X_2,Y_2 \& YX$

$1:0.5:1$

$X_2+Y_2\rightarrow 2XY$

Let the bond dissociation of $X_2,Y_2 \& YX$ be:

$\Delta H_{X-X}=q$

$\Delta H_{Y-Y}=0.5q$

$\Delta H_{X-Y}=q$

Enthalpy of formation of XY = $\Delta H_{f,XY}=-200 kJ./mol$

$\Delta H_{f,XY}=[1mol\times \Delta H_{Y-Y}+1 mol\times \Delta H_{X-X}]-[2\times \Delta H_{X-Y}]$

$-200 kJ/mol\times 2mol=q+0.5q-2q$

$-400=0.5q$

q = 800 kJ

In reaction 1 mole of $X_2$ dissociating, so the bond dissociation enthalpy of X, will be:

$\Delta H_{X-X}=q=800 kJ/mol$