NEET Solved Paper 2018 Question 40
Question: The bond dissociation energies of $ {X_2}\text{,}{Y_2} $ and $ XY $ are in the ratio of 1 : 0.5 : 1. $ \Delta H $ for the formation of XY is $ \text{–200 kJ mo}{l^{\text{–1}}} $ . The bond dissociation energy of $ {X_2} $ will be [NEET - 2018]
Options:
A) $ \text{800 kJ mo}{l^{\text{–1}}} $
B) $ \text{100 kJ mo}{l^{\text{–1}}} $
C) $ \text{200 kJ mo}{l^{\text{–1}}} $
D) $ \text{400 kJ mo}{l^{\text{–1}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
The reaction for $ {\Delta_f}\text{H }{}^\circ\text{ (XY)} $ $ \frac{1}{2}X _2(g)+\frac{1}{2}Y _2(g)\xrightarrow{{}}XY(g) $
Bond energies of $ X _2,Y _2 $ and $ XY $ are $ X,\frac{X}{2},X $ respectively
$ \therefore $ $ \Delta H=( \frac{X}{2}+\frac{X}{4} )-x=-200 $ On solving, we get
$ \Rightarrow -\frac{X}{2}+\frac{X}{4}=-200 $
$ \Rightarrow \text{X=800}\text{kJ/mole} $
BETA
