Chemical Thermodynamics Question 424
Question: The bond dissociation energies of gaseous $ H_2,,Cl_2 $ and $ HCl $ are 104, 58 and 103 kcal respectively. The enthalpy of formation of $ HCl $ gas would be [MP PET 1997; MP PMT 1999, 2001]
Options:
A) - 44 kcal
B) 44 kcal
C) - 22 kcal
D) 22 kcal
Show Answer
Answer:
Correct Answer: C
Solution:
Aim: $ \frac{1}{2}H_2+\frac{1}{2}Cl_2\to HCl $
$ \Delta H=\sum B.E{. _{\text{(Products)}}}-\sum B.E{. _{\text{(Reactants)}}} $
$ =B.E.(HCl,)-[ \frac{1}{2}B.E.(H_2)+\frac{1}{2}B.E.(Cl_2) ] $
$ =-103-[ \frac{1}{2}(,-104,)+\frac{1}{2}(,-58) ] $
$ =-103-(,-52-29,)=-22kcal $ .
BETA
