Calculate the total amount of heat released in units of kilojoules when 25.8 g of ethanol vapor (C2H5OH) at 92.0◦C is cooled to -10.0◦ C. The molar heat capacities of ethanol vapor and liquid ethanol are 65.6 J/◦C.mol and 112.3 J/◦C.mol. The boiling point of ethanol is 78.0 C, the freezing point of ethanol is -112.6 C, the enthalpy of vaporization (∆Hvap) of ethanol is 37.4 kJ /mol and the enthalpy of fusion (∆Hfus) of ethanol is 5.02 kJ/mol.
And where are you stuck on this?
To calculate the total amount of heat released, we need to consider two parts: the heat released during cooling from the initial temperature of 92.0°C to the boiling point of ethanol and the heat released during the condensation of ethanol vapor at the boiling point to -10.0°C.
Let's break down the steps and calculations:
1. Calculate the heat released during cooling from 92.0°C to the boiling point:
First, calculate the heat released to cool the ethanol vapor from 92.0°C to the boiling point of 78.0°C, using the molar heat capacity of ethanol vapor (65.6 J/°C.mol):
q1 = m * C * ΔT1
where:
q1 = heat released during cooling
m = mass of ethanol vapor (25.8 g)
C = molar heat capacity of ethanol vapor (65.6 J/°C.mol)
ΔT1 = change in temperature (92.0 - 78.0) = 14.0°C
Substituting the given values:
q1 = (25.8 g / (C2H5OH molar mass)) * (65.6 J/°C.mol) * (14.0°C)
To find the C2H5OH molar mass, we need to consider the atomic masses of carbon (12.01 g/mol), hydrogen (1.01 g/mol), and oxygen (16.00 g/mol):
C2H5OH molar mass = (2 * carbon atomic mass) + (6 * hydrogen atomic mass) + oxygen atomic mass
Calculate the molar mass and substitute it into the equation. Let's assume it's approximately 46.07 g/mol:
q1 = (25.8 g / 46.07 g/mol) * (65.6 J/°C.mol) * (14.0°C)
2. Calculate the heat released during condensation:
Next, calculate the heat released during the condensation of ethanol vapor at its boiling point (78.0°C) to -10.0°C (assuming it becomes a liquid). This involves two steps: cooling the vapor to its boiling point and then condensing it.
a. Heat released during cooling from 78.0°C to the boiling point (assuming ethanol is still a vapor):
q2a = m * C * ΔT2a
where:
q2a = heat released during cooling
m = mass of ethanol vapor (25.8 g)
C = molar heat capacity of ethanol vapor (65.6 J/°C.mol)
ΔT2a = change in temperature (78.0°C - boiling point)
Substitute and calculate the value of q2a.
b. Heat released during condensation at the boiling point:
q2b = m * ΔHvap
where:
q2b = heat released during condensation
m = moles of ethanol vapor (moles = mass / molar mass)
ΔHvap = enthalpy of vaporization of ethanol (37.4 kJ/mol)
Calculate the moles of ethanol vapor and substitute it into the equation.
3. Calculate the heat lost during the final cooling:
Finally, calculate the heat released during cooling from the boiling point (-10.0°C) to the final temperature (-10.0°C), where ethanol is assumed to be a liquid. Use the molar heat capacity of liquid ethanol (112.3 J/°C.mol) in the following equation:
q3 = m * C * ΔT3
where:
q3 = heat released during cooling
m = mass of ethanol liquid (use the mass of ethanol vapor since it is assumed to condense completely)
C = molar heat capacity of liquid ethanol (112.3 J/°C.mol)
ΔT3 = change in temperature (final temperature - boiling point)
Substitute the values and calculate q3.
4. Calculate the total heat released:
The total heat released is the sum of q1, q2a, q2b, and q3:
Total heat released = q1 + q2a + q2b + q3
Substitute the calculated values and sum them up to find the total heat released.
Note: It's important to double-check the accuracy of the assumed molar mass and ensure that the units are consistent throughout the calculations.