Chemical Thermodynamics Question 478
Question: For a gaseous reaction:
$ 2A_2(g)+5B_2(g)\to 2A_2B_5(g) $ at $ 27^{o}C $
the heat change at constant pressure is found to be $ -,50160,J $ . Calculate the value of internal energy change $ (\Delta E) $ . Given that R = 8.314 J/K mol.
Options:
A) - 34689 J
B) - 37689 J
C) - 27689 J
D) - 38689 J
Show Answer
Answer:
Correct Answer: B
Solution:
- $ 2A_2( g )+5B_2( g )g\to 2A_2B_5( g );\Delta H=-50160J $
$ \Delta n=2-( 5+2 )=-5,mol $
$ \Delta H=\Delta E+{{( \Delta n )}_{g}}RT $
$ -50160=\Delta E+{{( \Delta n )}_{g}}RT $
$ \Delta E=-50160-( -5 )( 8.314 )( 300 ) $
$ =-,50160+12471=-,37689J $
