Chemical Thermodynamics Question 361
Question: The difference between the reaction enthalpy change $ ({\Delta_{r}}H) $ and reaction internal energy change $ ({\Delta_{r}}U) $ for the reaction: $ 2C_6H_6( l )+15O_2( g )\xrightarrow{{}}12CO_2( g )+6H_2O( l ) $ at $ 300K $ is $ (R=8.314,Jmo{{l}^{-1}}{{K}^{-1}}) $
Options:
A) $ 0J,mo{{l}^{-1}} $
B) $ 2490Jmo{{l}^{-1}} $
C) $ -2490Jmo{{l}^{-1}} $
D) $ -7482Jmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ \Delta H=\Delta U+\Delta n_{g}RT $ For the reaction $ \Delta n_{g}=12-15=-3 $
$ \Delta H-\Delta U=-3\times 8.314\times 300 $
$ =-7482Jmo{{l}^{-1}} $
