Exercise 17.40 with Feedback Consider the evaporation of methanol at 25.0°C: CH3

Question

Exercise 17.40 with Feedback Consider the evaporation of methanol at 25.0°C: CH3OH(l) CH3OH(g) You may want to reference (LA p ages 714-719) Section 17.7 while completing this problem. Part A Find AG at 25.0°C. Express the free energy change in kilojoules to one decimal place. kJ AG Submit My Answers Give U The free energy change in the previous part is calculated based on the assumption of standard conditions For Parts B through D, find AG at 25.0 C under the given nonstandard conditions. Part B PCHsoH 159.0 mmHg Express the free energy change in kilojoules to one decimal place. AG kJ

Explanation / Answer

PART A

Find G° at 25 °C. The superscript "°" means that you are considering the conditions T = 298 K, P = 1 atm and you must use the "" or write "deltaG".

CHOH() CHOH(g)

You didn’t give any information, and its not possible to answer your question without some information. In this case, I looked up the G's of formation, from literature value.

G°f(CHOH()) = –166.36 kJ/mol,

G°f(CHOH(g)) = –162.00 kJ/mol

Therefore, G°(vap) = –162.00 – (–166.36) kJ/mol

                                  = +4.4 kJ/mol

Since G° is positive, that means that if the vapor pressure of methanol at 298 K were 1.0 atm, the equilibrium will run the other way, that is, it will condense to form more liquid until it reaches the equilibrium vapor pressure.

PART B:

G = G° + RTlnQ

In this case Q = P/P, P is the vapor pressure and P = 1 atm = 760 mm

G = G° + RTlnQ

      = +4.4 kJ/mol + (8.314 J/mol K)(298 K) ln(159 mm/760 mm)

      = +4400 J/mol + (2477.57 J/mol) ln(0.21)

      = +4400 J/mol + (2477.57 J/mol) (- 1.61)

      = +4400 J/mol - 3989 J/mol

      = +411 J/mol

      = 0.411 kJ/mol

PART C:

G = G° + RTlnQ

In this case Q = P/P, P is the vapor pressure and P = 1 atm = 760 mm

G = G° + RTlnQ

      = +4.4 kJ/mol + (8.314 J/mol K)(298 K) ln(104 mm/760 mm)

      = +4400 J/mol + (2477.57 J/mol) ln(0.14)

      = +4400 J/mol + (2477.57 J/mol) (- 1.97)

      = +4400 J/mol - 4881 J/mol

      = +481 J/mol

      = 0.481 kJ/mol

PART B:

G = G° + RTlnQ

In this case Q = P/P, P is the vapor pressure and P = 1 atm = 760 mm

G = G° + RTlnQ

      = +4.4 kJ/mol + (8.314 J/mol K)(298 K) ln(11 mm/760 mm)

      = +4400 J/mol + (2477.57 J/mol) ln(0.0145)

      = +4400 J/mol + (2477.57 J/mol) (- 4.23)

      = +4400 J/mol - 10480 J/mol

      = 6080 J/mol

      = 6.08 kJ/mol

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