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}} $

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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}} $ .



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