Chemical Thermodynamics Question 398
Question: For the reaction $ 2C_6H_5CO_2H( s )+15O_2( g )\xrightarrow{{}} $
$ 14CO_2( g )+6H_2O( g ) $
$ \Delta U{}^\circ =-772.7kJmo{{l}^{-1}} $ at 298 K. Calculate $ \Delta H{}^\circ $
Options:
A) $ +760.3kJmo{{l}^{-1}} $
B) $ -760.3kJmo{{l}^{-1}} $
C) $ +670.3kJmo{{l}^{-1}} $
D) $ -790.3kJmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \Delta {{H}^{{}^\circ }}=\Delta {{U}^{{}^\circ }}+\Delta nRT $
$ \Delta {{H}^{{}^\circ }}=-772.7+\frac{5\times 8.314\times 298}{1000} $
$ =-760.3kJ,mo{{l}^{-1}} $
