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!
All of those look ok to me. Thanks for showing your work.
Your calculations are almost correct, but there are a couple of errors in your answers. Let's go through each part one by one:
a) Calculating the PV work:
The formula for work is indeed w = -PΔV. However, you made a sign error in your calculation. The gas volume decreases, so the work done by the gas should be negative. The correct calculation is:
w = (-1 atm) * (-22.4 L) * (101.3 J/1 L*atm) = 2270 J or 2.27 kJ (correct)
b) Calculating the heat evolved:
The equation in this case is already balanced, so we can directly use the stoichiometry. The correct calculation is:
44.8 L H2O * (1 mol H2O / 22.4 L H2O) * (-241.8 kJ / 1 mol H2O) = -1083.6 kJ (corrected)
c) The change in internal energy of the system:
The change in internal energy (ΔE) of the system is the sum of heat (q) and work (w). So, the correct calculation is:
ΔE system = q + w = (-1083.6 kJ) + (2.27 kJ) = -1081.33 kJ (corrected)
d) The change in internal energy of the surroundings:
Since the system releases heat to the surroundings, the change in internal energy for the surroundings will have the opposite sign to the system. The correct calculation is:
ΔE surroundings = -ΔE system = -(-1081.33 kJ) = 1081.33 kJ (corrected)
So, the corrected answers are:
a) The PV work = 2.27 kJ
b) The heat evolved = -1083.6 kJ
c) The change in internal energy of the system = -1081.33 kJ
d) The change in internal energy of the surroundings = 1081.33 kJ
Please note that the answers provided here are based on the corrections made to your original calculations.