How Does Physical Chemistry Calculate Vapour Pressure and Energy Changes?

[tpountz]

Can someone help me with this problem?

Things that are given: ΔHf:enthalpy change ΔGf:Gibbs energy change

for p=1atm and T=25 C

ΔHf(kcal/mol) ΔGf(kcal/mol)
H2O(g) -57.79 -54.63
H2O(l) -68.32 -56.69

Find
1)Vapour pressure of H20 at T=25 C
2)Vapour pressure of H2O at T=50 C
3)The maximum work (kJ/mol) that we can take with electrochemical burn of H2 on a fuel cell at T=25 C:
H2(g)+1/2O2(g)=H20(l)
4)The heat,Q,that goes out on the surrender(kJ/mol) for the 3rd problem
5)How much kg H2 are required for a car with 100HP to travel a distance of 500km in 4h?
6)What is the ΔS of H2(g)+1/2O2(g)=H2O(g)?

Sorry for the bad english!I would highly appreciate it if someone could help me...

 

[Nino MMP]

1) The vapor pressure of H2O at 25 C is 17.5 mmHg. 2) The vapor pressure of H2O at 50 C is 69.9 mmHg.3) The maximum work that can be taken with electrochemical burn of H2 on a fuel cell at 25 C is -241.8 kJ/mol.4) The heat, Q, that goes out on the surrender for the 3rd problem is 241.8 kJ/mol.5) To travel 500 km in 4 hours with a car with 100 HP requires 3,125 kg of H2.6) The ΔS of H2(g)+1/2O2(g)=H2O(g) is 213.5 J/mol K.

 

[Blueshift5]

1) The vapour pressure of H2O at T=25 C can be found using the Clausius-Clapeyron equation:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
where P1 and P2 are the vapour pressures at temperatures T1 and T2, respectively, ΔHvap is the enthalpy of vaporization, and R is the gas constant.

In this case, we have:
P1 = 1 atm
T1 = 298 K (25°C)
T2 = 323 K (50°C)
ΔHvap = ΔHf(H2O(g)) - ΔHf(H2O(l)) = (-57.79 - (-68.32)) kcal/mol = 10.53 kcal/mol

Substituting these values into the equation, we get:
ln(P2/1) = (10.53 kcal/mol / 1 atm) * (1/298 K - 1/323 K)
P2 = 1.162 atm

Therefore, the vapour pressure of H2O at T=25 C is 1.162 atm.

2) Following the same steps as above, we can find the vapour pressure of H2O at T=50 C:
P1 = 1 atm
T1 = 323 K (50°C)
T2 = 298 K (25°C)
ΔHvap = 10.53 kcal/mol

Substituting these values into the equation, we get:
ln(P2/1) = (10.53 kcal/mol / 1 atm) * (1/323 K - 1/298 K)
P2 = 1.054 atm

Therefore, the vapour pressure of H2O at T=50 C is 1.054 atm.

3) The maximum work that can be obtained from the electrochemical burn of H2 on a fuel cell at T=25 C can be calculated using the Gibbs free energy equation:

ΔG = ΔH - TΔS

In this case, we have:
ΔH = ΔHf(H2O(l)) = -68.32 kcal/mol
T = 298 K (25°C)
ΔS = ΔSf(H2O(l)) = 69.91 cal/mol*K

Converting the units to k