Chemical Thermodynamics Question 487
Question: Calculate the heat change in the reaction $ 4,NH_3(g)+3O_2(g)\to 2N_2(g)+6H_2O(\ell ) $ at 298 K, given that the heats of formation at 298 K for $ NH_3(g) $ and $ H_2O(\ell ) $ are - 46.0 and - 28.60 kJ $ mo{{l}^{-1}} $ respectively.
Options:
A) - 1861 kJ
B) - 1361 kJ
C) - 1261 kJ
D) - 1532 kJ
Show Answer
Answer:
Correct Answer: D
Solution:
- $ \Delta H^{o} $ for the reaction $ 4NH_3( g )+3O_2( g )\to 6H_2O( l )+2N_2( g ) $
$ \Delta H^{o}=\Delta H_{f}^{o}( products )-\Delta H_{f}^{o}( reactants ) $
$ ={ 6H_{f}^{o}[ H_2O( l ) ]+\Delta {{H}^{\circ }}_{f}[ N_2( g ) ] } $
$ -{ 4\Delta H[ NH_3( g ) ]+3\Delta H_{f}[ O_2( g ) ] } $
$ \Delta H_{f}^{o}[ H_2O( l ) ]=-286.0kJmo{{l}^{-1}} $
$ \Delta H_{f}^{o}[ NH_2( g ) ]=0 $ and $ \Delta H_{f}^{o}[ N_2( g ) ] $
= 0 (by convention) $ \Delta H^{o}={ 6( -286 )+2( 0 ) }-{ 4( -46.0 )+3( 0 ) } $
$ =-1716+184=-1532,kJ $
