From data below, calculate the total heat (J) needed to convert 0.539 mol gaseous ethanol at 300°C and 1 atm to liquid ethanol at 25.0°C and 1 atm.

Boiling point at 1 atm 78.5°C
c ethanol gas 1.43 J/g·°C
c ethanol liquid 2.45 J/g·°C
ÄH°vap 40.5 kJ/mol

help please Answer in J

my work:
.539 x 46= 24.79
24.79 x 1.43 x -221.50= -7852 J
.539 x -40.5= -21,829 J
24.79 x 2.45 x -53.5 = 32930 J

then -7852 + -21, 829 + -3249= -32930 J

To calculate the total heat (J) needed to convert 0.539 mol of gaseous ethanol at 300°C and 1 atm to liquid ethanol at 25.0°C and 1 atm, you can follow these steps:

1. Calculate the heat required to cool the gaseous ethanol from 300°C to its boiling point at 78.5°C:
Heat = mass_gas × specific heat_gas × ΔT_gas
Here, the mass of gaseous ethanol (C2H5OH) is given as 0.539 mol, and the specific heat of ethanol gas is given as 1.43 J/g·°C. The change in temperature (ΔT_gas) is the difference between the initial temperature (300°C) and the boiling point of ethanol at 1 atm (78.5°C).
Heat_gas = 0.539 mol × 46 g/mol × 1.43 J/g·°C × (78.5°C - 300°C)

2. Calculate the heat required to vaporize the liquid ethanol at its boiling point:
Heat_vap = number of moles × enthalpy of vaporization
The number of moles is given as 0.539 mol, and the enthalpy of vaporization (ÄH°vap) of ethanol is given as 40.5 kJ/mol. Convert it to J/mol by multiplying by 1000.
Heat_vap = 0.539 mol × 40.5 kJ/mol × 1000 J/kJ

3. Calculate the heat released when the vapor condenses and cools to 25.0°C:
Heat = mass_liquid × specific heat_liquid × ΔT_liquid
Here, the mass of liquid ethanol is the same as the mass of gaseous ethanol (0.539 mol × 46 g/mol). The specific heat of ethanol liquid is given as 2.45 J/g·°C. The change in temperature (ΔT_liquid) is the difference between the boiling point of ethanol (78.5°C) and the final temperature (25.0°C).
Heat_liquid = 0.539 mol × 46 g/mol × 2.45 J/g·°C × (25.0°C - 78.5°C)

4. Add up the heats from steps 1, 2, and 3 to get the total heat:
Total heat = Heat_gas + Heat_vap + Heat_liquid

Plugging in the given values:

Heat_gas = 0.539 mol × 46 g/mol × 1.43 J/g·°C × (78.5°C - 300°C)
Heat_vap = 0.539 mol × 40.5 kJ/mol × 1000 J/kJ
Heat_liquid = 0.539 mol × 46 g/mol × 2.45 J/g·°C × (25.0°C - 78.5°C)

Calculate each of these values and then add them together to obtain the total heat in Joules (J).

To calculate the total heat needed to convert gaseous ethanol to liquid ethanol, we need to consider the following steps:

1. Heating the gaseous ethanol from 300°C to its boiling point at 78.5°C.
2. Vaporizing the gaseous ethanol at its boiling point.
3. Cooling the vaporized ethanol to its final temperature of 25.0°C.

Let's calculate the heat required for each step:

Step 1: Heating the gaseous ethanol from 300°C to its boiling point at 78.5°C.

Using the specific heat capacity for ethanol gas (c_ethanol_gas = 1.43 J/g·°C) and the mass of ethanol (0.539 mol x 46 g/mol = 24.79 g), the heat required to heat the gaseous ethanol is:

Heat1 = mass_ethanol x c_ethanol_gas x ΔT1

Heat1 = 24.79 g x 1.43 J/g·°C x (-221.5°C) = -7829.0145 J (note that the change in temperature is negative because we're going from higher to lower temperature)

Step 2: Vaporizing the gaseous ethanol at its boiling point.

The enthalpy of vaporization (ΔH_vap = 40.5 kJ/mol) is given. To calculate the heat required to vaporize the ethanol, we need to multiply the enthalpy of vaporization by the number of moles of ethanol:

Heat2 = moles_ethanol x ΔH_vap

Heat2 = 0.539 mol x (-40.5 kJ/mol) x (1000 J/1 kJ) = -21,829 J (note that the change in enthalpy is negative)

Step 3: Cooling the vaporized ethanol to its final temperature of 25.0°C.

Using the specific heat capacity for ethanol liquid (c_ethanol_liquid = 2.45 J/g·°C) and the mass of ethanol (24.79 g), the heat required to cool the vaporized ethanol is:

Heat3 = mass_ethanol x c_ethanol_liquid x ΔT2

Heat3 = 24.79 g x 2.45 J/g·°C x (-53.5°C) = -32794.6275 J (note that the change in temperature is negative)

Now, to find the total heat required, we sum up the heats obtained in each step:

Total heat = Heat1 + Heat2 + Heat3

Total heat = -7829.0145 J + (-21,829 J) + (-32794.6275 J) = -39,452.651 J

Therefore, the total heat needed to convert 0.539 mol of gaseous ethanol at 300°C and 1 atm to liquid ethanol at 25.0°C and 1 atm is approximately -39,452.651 J.