From data below, calculate the total heat (J) needed to convert 0.757 mol of gaseous ethanol (C2H6O) at 151°C and 1 atm to liquid ethanol at 25.0°C and 1 atm.

Boiling point at 1 atm 78.5°C
cgas 1.43 J/g·°C
cliquid 2.45 J/g·°C
ΔH°vap 40.5 kJ/mol

It's easier to work these problems in steps. There are two types of steps you use.

For phase changes (solid to liquid or liquid to gas OR the reverse) you use
q = mass x delta H.
delta H is heat fusion for solid to liquid or reverse.
delta H is heat vaporization for liquid to gas or reverse.

The second type is for heating/cooling within a phase; i.e., heating a liquid from 20 C to 100 C or cooling a liquid from 100 c to 20 C.

To calculate the total heat needed to convert gaseous ethanol at 151°C to liquid ethanol at 25.0°C, we need to consider the heat required to cool the ethanol from 151°C to its boiling point (78.5°C), the heat required to vaporize it at its boiling point, and the heat required to cool the vapor to 25.0°C.

Step 1: Calculate the heat required to cool the ethanol from 151°C to its boiling point:
The specific heat capacity of gaseous ethanol (cgas) is given as 1.43 J/g·°C. Since we have 0.757 mol of ethanol, we need to convert it to grams.
Molar mass of ethanol (C2H6O) = 2*(12.01 g/mol) + 6*(1.01 g/mol) + 16.00 g/mol = 46.07 g/mol
Mass of 0.757 mol of ethanol = 0.757 mol * 46.07 g/mol = 34.85 g
Temperature change = 151°C - 78.5°C = 72.5°C
Heat required to cool the ethanol = mass * specific heat capacity * temperature change
= 34.85 g * 1.43 J/g·°C * 72.5°C = 3591.2545 J

Step 2: Calculate the heat required to vaporize the ethanol at its boiling point:
The molar enthalpy of vaporization (ΔH°vap) is given as 40.5 kJ/mol. Since we have 0.757 mol of ethanol, we can directly multiply it by the molar enthalpy of vaporization to find the heat required.
Heat required to vaporize ethanol = 0.757 mol * 40.5 kJ/mol * 1000 J/1 kJ = 30668.5 J

Step 3: Calculate the heat required to cool the vapor to 25.0°C:
The specific heat capacity of liquid ethanol (cliquid) is given as 2.45 J/g·°C. We need to convert the mass of the vapor to grams.
Molar mass of ethanol (C2H6O) = 46.07 g/mol (as calculated before)
The mass of the vapor can be determined using the ideal gas equation: PV = nRT
where:
P = 1 atm
V = unknown
n = 0.757 mol
R = 0.0821 L·atm/mol·K (ideal gas constant)
T = 78.5°C + 273.15 K = 351.65 K
Rearranging the equation to solve for V:
V = nRT / P = (0.757 mol * 0.0821 L·atm/mol·K * 351.65 K) / (1 atm) ≈ 23.35 L

Since the volume is given in liters, we can assume the density of ethanol vapor is approximately 1 g/mL. Therefore, the mass of the vapor is the same as its volume, which is 23.35 g.
Temperature change = 78.5°C - 25.0°C = 53.5°C
Heat required to cool the vapor = mass * specific heat capacity * temperature change
= 23.35 g * 2.45 J/g·°C * 53.5°C = 2972.5075 J

Step 4: Sum up the heats from the above steps to find the total heat:
Total heat = heat required to cool the ethanol + heat required to vaporize + heat required to cool the vapor
= 3591.2545 J + 30668.5 J + 2972.5075 J = 37331 J (rounded to four significant figures)

Therefore, the total heat needed to convert 0.757 mol of gaseous ethanol at 151°C and 1 atm to liquid ethanol at 25.0°C and 1 atm is approximately 37331 J.