Question
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!
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!
Answers
DrBob222
All of those look ok to me. Thanks for showing your work.