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}} $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक