At a pressure of 1ATM, _____ KJ of heat is needed to vaporize a 32.3 g sample of liquid ethanol at its normal boiling point of 78.4 C. Boiling point (78.4 C), specific heat (2.46 J/g C), and Hvap(78.4 C)=38.6 KJ/mol are provided. Please include how to solve step by step so I can understand for similar problems. Thanks
find the number of moles in 32.3 g
... the molar mass is 46.1 g
multiply by the heat of vaporization (Hvap)
To solve this problem, we can break it down into two steps: calculating the heat required to raise the temperature of the liquid ethanol from its boiling point to its boiling point, and calculating the heat required to vaporize the ethanol at its boiling point.
Step 1: Calculating the heat required to raise the temperature
The formula for calculating the heat required to raise the temperature of a substance is:
q = m * C * ΔT
Where:
q = heat (in joules)
m = mass of the substance (in grams)
C = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
In this case, we want to calculate the heat required to raise the temperature from the boiling point (78.4°C) to the boiling point (78.4°C). Since there is no change in temperature, ΔT is 0.
q = m * C * ΔT
q = 32.3 g * 2.46 J/g°C * 0
q = 0 J
Therefore, no heat is required to raise the temperature of the liquid ethanol to its boiling point.
Step 2: Calculating the heat required to vaporize the liquid ethanol
The formula for calculating the heat required to vaporize a substance is:
q = n * ΔHvap
Where:
q = heat (in joules)
n = number of moles of the substance
ΔHvap = enthalpy of vaporization (in J/mol)
First, we need to calculate the number of moles of ethanol. We can use the molar mass of ethanol (C₂H₅OH) to convert the mass (32.3 g) to moles.
molar mass of ethanol (C₂H₅OH) = 12.01 g/mol (C) + 2(1.008 g/mol) (H) + 16.00 g/mol (O) = 46.07 g/mol
moles of ethanol = mass of ethanol / molar mass of ethanol
moles of ethanol = 32.3 g / 46.07 g/mol
moles of ethanol = 0.702 mol
Now, we can calculate the heat required to vaporize the ethanol:
q = n * ΔHvap
q = 0.702 mol * 38.6 kJ/mol
Note: We need to convert kJ to J since the molar enthalpy of vaporization is given in kJ/mol.
q = 0.702 mol * 38.6 kJ/mol * 1000 J/1 kJ
q = 27017 J
Therefore, 27017 J or 27.017 kJ of heat is needed to vaporize the 32.3 g sample of liquid ethanol at 1 ATM of pressure.
To calculate the amount of heat needed to vaporize a sample of liquid ethanol at its boiling point, we can use the formula:
q = m * Hvap
where:
q is the amount of heat (in joules) needed to vaporize the substance
m is the mass of the substance (in grams)
Hvap is the molar enthalpy of vaporization (in joules per mole)
First, let's convert the given mass of ethanol from grams to moles. We can do this by using the molar mass of ethanol.
The molar mass of ethanol (C2H5OH) =
2 * molar mass of carbon + 6 * molar mass of hydrogen + molar mass of oxygen
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + 16.00 g/mol
= 46.07 g/mol
Now, let's convert the mass of ethanol to moles:
moles = mass / molar mass
moles = 32.3 g / 46.07 g/mol
moles ≈ 0.702 moles
Next, we need to calculate the amount of heat needed to vaporize the ethanol. We will use the molar enthalpy of vaporization (Hvap) given.
Since Hvap is given in KJ/mol, we need to convert it to J/mol by multiplying by 1000:
Hvap = 38.6 KJ/mol * 1000 J/KJ
Hvap = 38,600 J/mol
Now, let's calculate the amount of heat needed:
q = m * Hvap
q = 0.702 moles * 38,600 J/mol
q ≈ 27,061.20 J
Finally, let's convert the amount of heat from joules to kilojoules:
1 KJ = 1000 J
27,061.20 J / 1000 J/KJ ≈ 27.06 KJ
Therefore, at a pressure of 1 ATM, approximately 27.06 KJ of heat is needed to vaporize a 32.3 g sample of liquid ethanol at its normal boiling point of 78.4°C.