How much energy in kilojoules is released when 25.0 g of ethanol (C2H5OH) vapor at 93.0 C is cooled to -10 C?
Ethanol has mp = -114.1 oC, bp = 78.3 oC, ΔHvap = 38.56 kJ/mol, and ΔHfus = 4.93 kJ/mol. The molar heat capacity is 112.3 J/(K.mol) for the liquid and 65.6 J/(K.mol) for the vapor
I answered the questions:
25.0g[(1.7)(93-78.3)+841+2.44(78-(-10))] and got 27.030 kJ but that isn't correct!!
q1 = energy released on vapor cooling from 93 C to 78.3 C
q1 = mass ethanol x specific heat ethanol vapor x (Tfinal - Tinitial) note: Tf is 78.3 and Ti is 93.0.
q2 = energy released on changing from vapor at 78.3 to liquid at 78.3
q2 = mass ethanol x heat vaporization
q3 = energy released on lidquid cooling from 78.3 to -10. Use same formula as in q1 but with correct specific heat.
Total energy released = q1 + q2 + q3
17.83kJ
To determine the energy released when 25.0 g of ethanol vapor at 93.0°C is cooled to -10°C, we need to consider the energy changes during cooling in the liquid and vapor phases.
First, we need to calculate the energy change when the ethanol vapor condenses to a liquid at its boiling point of 78.3°C. This can be calculated using the equation:
ΔHvap = ΔHfus + ΔHl
where ΔHvap is the enthalpy change of vaporization, ΔHfus is the enthalpy change of fusion (melting), and ΔHl is the enthalpy change during cooling from the vapor to the liquid state.
Using the given values, we can calculate:
ΔHl = ΔHvap - ΔHfus
= 38.56 kJ/mol - 4.93 kJ/mol
= 33.63 kJ/mol
Next, we need to calculate the moles of ethanol in 25.0 g. The molar mass of ethanol (C2H5OH) is 46.07 g/mol, so we have:
n = (mass of ethanol) / (molar mass)
= 25.0 g / 46.07 g/mol
≈ 0.543 mol
Now, we can calculate the energy change when the ethanol vapor condenses to a liquid:
Energy change = n × ΔHl
= 0.543 mol × 33.63 kJ/mol
≈ 18.26 kJ
Next, we need to calculate the energy change during cooling of the liquid ethanol. We can use the equation:
Energy change = (mass of liquid ethanol) × (specific heat capacity of liquid) × (change in temperature)
The mass of liquid ethanol is the same as the initial mass of 25.0 g.
The specific heat capacity of liquid ethanol is given as 112.3 J/(K·mol).
Converting the mass to moles:
n = (mass of ethanol) / (molar mass)
= 25.0 g / 46.07 g/mol
≈ 0.543 mol
Converting the given specific heat capacity to J/(g·K):
specific heat capacity = (112.3 J/(K·mol))/(46.07 g/mol)
≈ 2.438 J/(g·K)
Now, we can calculate the energy change during cooling of the liquid ethanol:
Energy change = (mass of liquid ethanol) × (specific heat capacity of liquid) × (change in temperature)
= 25.0 g × 2.438 J/(g·K) × (78.3°C - (-10°C))
≈ 43.04 kJ
Finally, we can sum up the energy changes during vapor condensation and cooling:
Total Energy change = Energy change during vapor condensation + Energy change during cooling
= 18.26 kJ + 43.04 kJ
≈ 61.3 kJ
Therefore, the amount of energy released when 25.0 g of ethanol vapor at 93.0°C is cooled to -10°C is approximately 61.3 kJ.
To find the energy released when 25.0 g of ethanol vapor is cooled from 93.0 °C to -10 °C, we need to account for the different phase changes and temperature changes.
First, let's determine the energy required to cool the vapor from 93.0 °C to its boiling point of 78.3 °C. We can use the molar heat capacity for the vapor phase (65.6 J/(K·mol)), which is given as 65.6 J/(K·mol).
The molar mass of ethanol (C2H5OH) can be calculated as follows:
C: 12.01 g/mol
H: 1.008 g/mol
O: 16.00 g/mol
-------------------
Total: 46.08 g/mol
Next, calculate the moles of ethanol vapor in 25.0 g:
moles = mass/molar mass
moles = 25.0 g / 46.08 g/mol
moles ≈ 0.542 mol
Now, let's determine the energy required to cool the vapor from 93.0 °C to its boiling point of 78.3 °C:
energy = moles × molar heat capacity × temperature change
energy = 0.542 mol × 65.6 J/(K·mol) × (78.3 - 93.0) °C
energy = 0.542 mol × 65.6 J/(K·mol) × (-14.7) °C
energy ≈ -548 J
Next, we need to account for the energy released during the phase change from a vapor to a liquid at its boiling point. The molar enthalpy of vaporization (ΔHvap) is stated to be 38.56 kJ/mol.
energy = moles × ΔHvap
energy = 0.542 mol × 38.56 kJ/mol
energy ≈ 20.93 kJ
Now, we need to determine the energy required to cool the liquid ethanol from its boiling point to the final temperature of -10 °C. For this, we'll use the molar heat capacity for the liquid phase (112.3 J/(K·mol)).
energy = moles × molar heat capacity × temperature change
energy = 0.542 mol × 112.3 J/(K·mol) × (78.3 - (-10)) °C
energy = 0.542 mol × 112.3 J/(K·mol) × 88.3 °C
energy ≈ 54.87 kJ
Finally, we can calculate the total energy released:
total energy = energy from cooling vapor + energy from phase change + energy from cooling liquid
total energy = (-548 J) + 20.93 kJ + 54.87 kJ
total energy ≈ 75.25 kJ
Thus, the correct answer would be approximately 75.25 kJ, not 27.030 kJ.