How much energy (in kilojoules) is released when 29.0 g of ethanol vapor at 88.0 ∘C is cooled to -10.0 ∘C?

Ethanol has mp = -114.5 ∘C, bp = 78.4 ∘C, ΔHvap = 38.56 kJ/mol and ΔHfusion = 4.60 kJ/mol. The molar heat capacity is 113 J/(K⋅mol) for the liquid and 65.7 J/(K⋅mol) for the vapor.
Please explain/show work.

To determine the amount of energy released when the ethanol vapor is cooled, we need to consider the following steps:

Step 1: Calculate the energy released due to the cooling of the vapor from 88.0 °C to its boiling point.
Step 2: Calculate the energy released due to the condensation of ethanol vapor to liquid at its boiling point.
Step 3: Calculate the energy released due to the cooling of the liquid from its boiling point to -10.0 °C.

Step 1: Calculate the energy released due to cooling of ethanol vapor from 88.0 °C to boiling point (78.4 °C):
First, calculate the change in temperature: ΔT = T_final - T_initial
ΔT = 78.4 °C - 88.0 °C = -9.6 °C

Next, calculate the energy released using q = n * C * ΔT, where:
q = energy released in Joules (J)
n = number of moles of ethanol
C = molar heat capacity of the vapor (65.7 J/(K⋅mol))
ΔT = change in temperature in Kelvin (K) (since the units of heat capacity are in J/(K⋅mol))

To convert the given mass of ethanol vapor (29.0 g) to moles, we need to use the molar mass of ethanol:
Molar mass of ethanol (C2H5OH) = 46.07 g/mol
Number of moles (n) = mass / molar mass = 29.0 g / 46.07 g/mol

Now, convert the change in temperature to Kelvin:
ΔT_K = ΔT + 273.15
ΔT_K = -9.6 °C + 273.15 = 263.55 K

Substitute the values into the equation: q = n * C * ΔT_K
q = (29.0 g / 46.07 g/mol) * (65.7 J/(K⋅mol)) * 263.55 K

Step 2: Calculate the energy released due to the condensation of ethanol vapor to liquid at its boiling point:
The heat of vaporization (ΔHvap) given is 38.56 kJ/mol. We need to convert it to J/mol for our calculation.
ΔHvap = 38.56 kJ/mol * 1000 J/kJ = 38,560 J/mol

The number of moles of ethanol that condense will be the same as in Step 1 since there is no change in the substance:
Number of moles (n) = 29.0 g / 46.07 g/mol

Now, calculate the energy released due to condensation: q = n * ΔHvap
q = (29.0 g / 46.07 g/mol) * 38,560 J/mol

Step 3: Calculate the energy released due to cooling of the liquid from its boiling point (-10.0 °C) to the final temperature.
In this step, we will make use of the molar heat capacity of the liquid (113 J/(K⋅mol)) instead of the vapor.

First, calculate the change in temperature: ΔT = T_final - T_initial
ΔT = -10.0 °C - 78.4 °C = -88.4 °C

Convert the change in temperature to Kelvin: ΔT_K = ΔT + 273.15
ΔT_K = -88.4 °C + 273.15 = 184.75 K

Substitute the values into the equation: q = n * C * ΔT_K
q = (29.0 g / 46.07 g/mol) * (113 J/(K⋅mol)) * 184.75 K

Finally, add up the results from all three steps to obtain the total energy released:
Total energy released = q1 + q2 + q3

Solve each equation and add the results together to find the total energy released in joules (J). Divide by 1000 to convert it to kilojoules (kJ).