Chemical Thermodynamics Question 386
Question: The molar enthalpies of combustion of isobutene and n-butane are $ -2870kJmo{{l}^{-1}} $ and $ -2875kJmo{{l}^{-1}} $ respectively at 298 K and 1 atm. Calculate $ \Delta H{}^\circ $ for the conversion of 1 mole of n-butane to 1 mole of isobutane
Options:
A) $ -8kJmo{{l}^{-1}} $
B) $ +8kJmo{{l}^{-1}} $
C) $ -5748kJmo{{l}^{-1}} $
D) $ +5748kJmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ Isobutane+oxygen\to CO_2+H_2O $
$ \Delta H=-2870,kJmo{{l}^{-1}} $ ……(i) $ n-butane+oxygen\to CO_2+H_2O $
$ \Delta H=-2878,kJmo{{l}^{-1}} $ ….(ii) (iii) (i); n-butane - Isobutane, $ \Delta H=( -2878+2870 ) $
$ =-8,kJmo{{l}^{-1}} $ .
