∆G° for the process benzene (l) benzene (g) is 3.7 Kj/mol at 60 °C , calculate the vapor pressure of benzene at 60 °C [R=0.0821 atm. L/mol. K = 8.314 J/mol. K] 1) 0.38 atm
2) 5.1 atm
3) 11.1 atm
4) 0.26 atm
5) 0.54 atm

4) 0.26 atm

Explanation:

In the process:

Benzene(l) → Benzene(g)

ΔG° for this process is:

ΔG° = -RT ln Q

Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm

ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm

1.336 = ln P(benzene) / 1atm

0.26atm = P(benzene)

Right answer is:

they are all solids.

12 is it do you think so