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 needed to convert gaseous ethanol to liquid ethanol, you can follow these steps:
1. Calculate the heat required to cool the gaseous ethanol from 300°C to its boiling point (78.5°C):
Mass of ethanol gas = 0.539 mol x molar mass of ethanol = 0.539 x 46 g/mol = 24.794 g
Heat = mass x specific heat capacity x change in temperature
Heat = 24.794 g x 1.43 J/g·°C x (78.5°C - 300°C) = -7851.89 J (Note: the negative sign indicates heat loss)
2. Calculate the heat required to condense the gaseous ethanol at its boiling point (78.5°C):
Heat of vaporization = 40.5 kJ/mol x 1000 J/1 kJ = 40,500 J/mol
Heat = moles of ethanol gas x heat of vaporization
Heat = 0.539 mol x -40,500 J/mol = -21829.5 J (Note: the negative sign indicates heat released during condensation)
3. Calculate the heat required to cool the liquid ethanol from its boiling point (78.5°C) to 25.0°C:
Mass of liquid ethanol = 0.539 mol x molar mass of ethanol = 0.539 x 46 g/mol = 24.794 g
Heat = mass x specific heat capacity x change in temperature
Heat = 24.794 g x 2.45 J/g·°C x (25.0°C - 78.5°C) = -32929.765 J (Note: the negative sign indicates heat loss)
4. Calculate the total heat needed:
Total heat = heat for cooling + heat of vaporization + heat for further cooling
Total heat = -7851.89 J + -21829.5 J + -32929.765 J = -62610.155 J
So, 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 -62610 J.
To calculate the total heat needed to convert gaseous ethanol to liquid ethanol, we need to consider the following steps:
Step 1: Calculate the heat required to cool the gaseous ethanol from 300°C to its boiling point at 78.5°C.
To do this, we can use the specific heat capacity formula:
q1 = n * c * ∆T
Where:
q1 is the heat required
n is the number of moles
c is the specific heat capacity of ethanol gas (in J/g·°C)
∆T is the change in temperature (in °C)
In this case, n = 0.539 mol, c = 1.43 J/g·°C, and ∆T = 78.5 - 300 = -221.5°C (since we're going from a higher temperature to a lower temperature).
Plugging in the values, we have:
q1 = 0.539 mol * 1.43 J/g·°C * -221.5°C = -171.91 J
Step 2: Calculate the heat required for the phase change from gaseous ethanol to liquid ethanol.
To do this, we can use the enthalpy of vaporization formula:
q2 = n * ∆Hvap
Where:
q2 is the heat required
n is the number of moles
∆Hvap is the enthalpy of vaporization of ethanol (in J/mol)
In this case, n = 0.539 mol and ∆Hvap = 40.5 kJ/mol = 40.5 * 1000 J/mol.
Plugging in the values, we have:
q2 = 0.539 mol * 40.5 * 1000 J/mol = 21829.5 J
Step 3: Calculate the heat required to cool the liquid ethanol from its boiling point (78.5°C) to the final temperature of 25.0°C.
Using the same formula as in Step 1, but now with the specific heat capacity of ethanol liquid (c = 2.45 J/g·°C), we can calculate the heat required (q3):
q3 = n * c * ∆T
In this case, n = 0.539 mol, c = 2.45 J/g·°C, and ∆T = 78.5 - 25 = 53.5°C (since we're going from a higher temperature to a lower temperature).
Plugging in the values, we have:
q3 = 0.539 mol * 2.45 J/g·°C * -53.5°C = -3292.05 J
Finally, to find the total heat required, we add up the three calculated heats:
Total heat = q1 + q2 + q3
Total heat = -171.91 J + 21829.5 J - 3292.05 J
Total heat = 17865.54 J
Therefore, the total heat needed to convert 0.539 mol gaseous ethanol at 300°C and 1 atm to liquid ethanol at 25.0°C and 1 atm is approximately 17865.54 J.