Chemical Thermodynamics Question 93
Question: $ H_2(g)+\frac{1}{2}O_2(g)\to H_2O(l) $ ; $ \Delta H $ at 298 K = - 285.8 kJ The molar enthalpy of vaporisation of water at 1 atm and $ 25^{o}C $ is 44 kJ. The standard enthalpy of formation of 1 mole of water vapour at $ 25^{o}C $ is [KCET 1999]
Options:
A) - 241.8 kJ
B) 241.8 kJ
C) 329.8 kJ
D) -329.8 kJ
Show Answer
Answer:
Correct Answer: A
Solution:
$ H_2+\frac{1}{2}O_2\to H_2{O _{(l)}};,\Delta H=-285.8,KJ $
$ H_2{O _{(l)}}\to H_2{O _{(g)}};,\Delta H=44,KJ $
$ \therefore H_2+\frac{1}{2}O_2\to H_2{O {(g)}}; ,\Delta H^{o}=-241.8,KJ $
BETA
