Combusting 44.8L of H2 with 22.4 L O2 (both at STP) results in the formation of 44.8 L of H2O if the reaction is maintained at the same conditions (STP) and goes to completion. Given the heat of formation of water (= -241.8 kK/mol) calculate
a) the PV work
b) the heat evolved
c) the change in internal energy of the system
d) the change in internal energy of the surroundings
Please check:
a)
w= -p delta V
(-1atm)(-22.4 L)x(101.3J/1Lxatm)=
Final Answer: 2.27 x 10^3 J or 2.27 kJ
b)
44.8LxH2Ox(1molH2O/22.4L)x(-241.8kJ/1molH2O)=
Final Answer: -483.6 kJ
c) delta E of system= q+ w= -481.3 kJ + 2.27 kJ
= -481.3 kJ
d) delta E of surroundings= +481.3 kJ
Please check and thank you!
1 answer
All of those look ok to me. Thanks for showing your work.
1 answer