∆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
1 answer:
Answer:
4) 0.26 atm
Explanation:
In the process:
Benzene(l) → Benzene(g)
ΔG° for this process is:
ΔG° = -RT ln Q
<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>
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:
<h3>4) 0.26 atm
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