How much energy (in kilojoules) is released when 29.0 g of ethanol vapor at 88.0 ∘C is cooled to -10.0 ∘C? Express your answer to two significant figures.
q = ______________ kJ
Ethanol has mp = -114.5 ∘C, bp = 78.4 ∘C, ΔHvap = 38.56 kJ/mol and ΔHfusion = 4.60 kJ/mol. The molar heat capacity is 113 J/(K⋅mol) for the liquid and 65.7 J/(K⋅mol) for the vapor.
To calculate the energy released (q) when cooling ethanol vapor from 88.0 °C to -10.0 °C, we need to consider the heating and cooling steps involved.
First, we need to calculate the energy required to cool the ethanol vapor from its initial temperature (88.0 °C) to its condensation temperature (bp = 78.4 °C). We can use the formula:
q1 = n * ΔHvap
where q1 is the energy required, n is the number of moles of ethanol, and ΔHvap is the enthalpy of vaporization of ethanol.
To calculate the number of moles (n) of ethanol, we can use the formula:
n = mass / molar mass
where mass is the given mass of ethanol (29.0 g) and molar mass is the molar mass of ethanol (46.07 g/mol).
Next, we need to calculate the energy released during the condensation of ethanol at its boiling point (78.4 °C) to its freezing point (mp = -114.5 °C). We can use the formula:
q2 = n * ΔHfusion
where q2 is the energy released, n is the number of moles of ethanol, and ΔHfusion is the enthalpy of fusion of ethanol.
Finally, we need to calculate the energy released to cool the liquid ethanol from its freezing point (mp = -114.5 °C) to the final temperature (-10.0 °C). We can use the formula:
q3 = n * Cp * ΔT
where q3 is the energy released, n is the number of moles of ethanol, Cp is the molar heat capacity of the liquid ethanol, and ΔT is the change in temperature.
Once we have calculated q1, q2, and q3, we can add them together to get the total energy released (q).
Remember to convert the molar heat capacities from J/(K⋅mol) to kJ/(K⋅mol) for consistent units.
Let's calculate the values:
1. Calculate the number of moles (n) of ethanol:
n = mass / molar mass
n = 29.0 g / 46.07 g/mol
n ≈ 0.630 mol
2. Calculate q1, the energy required to cool the vapor from 88.0 °C to 78.4 °C:
q1 = n * ΔHvap
q1 = 0.630 mol * 38.56 kJ/mol
q1 ≈ 24.336 kJ
3. Calculate q2, the energy released during condensation at boiling point:
q2 = n * ΔHfusion
q2 = 0.630 mol * 4.60 kJ/mol
q2 ≈ 2.898 kJ
4. Calculate q3, the energy released to cool the liquid from freezing point to final temperature:
q3 = n * Cp * ΔT
Cp = 113 J/(K⋅mol) * (1 kJ/1000 J) = 0.113 kJ/(K⋅mol)
ΔT = -10.0 °C - (-114.5 °C) = 104.5 °C
q3 = 0.630 mol * 0.113 kJ/(K⋅mol) * 104.5 °C
q3 ≈ 7.309 kJ
5. Calculate the total energy released (q):
q = q1 + q2 + q3
q ≈ 24.336 kJ + 2.898 kJ + 7.309 kJ
q ≈ 34.543 kJ
Therefore, the energy released when 29.0 g of ethanol vapor is cooled from 88.0 °C to -10.0 °C is approximately 34.5 kJ (rounded to two significant figures).
To calculate the amount of energy released when cooling ethanol vapor, we need to consider the following steps:
1. Calculate the energy required to cool ethanol vapor from 88.0 ∘C to its boiling point at 78.4 ∘C.
2. Calculate the energy released when the ethanol vapor condenses into a liquid.
3. Calculate the energy released when the liquid ethanol cools from its boiling point to -10.0 ∘C.
Step 1: Calculate the energy required to cool ethanol vapor from 88.0 ∘C to 78.4 ∘C.
First, we need to calculate the molar heat capacity for the vapor. As given, the molar heat capacity of ethanol vapor is 65.7 J/(K⋅mol).
ΔT = final temperature - initial temperature = 78.4 ∘C - 88.0 ∘C = -9.6 K
Heat = molar heat capacity × moles × ΔT
We need to calculate the number of moles of ethanol vapor. To do this, we can use the molar mass of ethanol, which is 46.07 g/mol.
moles = mass / molar mass = 29.0 g / 46.07 g/mol = 0.6291 mol
Heat = 65.7 J/(K⋅mol) × 0.6291 mol × -9.6 K = -380.2 J
Step 2: Calculate the energy released when the ethanol vapor condenses into a liquid.
The enthalpy of vaporization, ΔHvap, is given as 38.56 kJ/mol. We need to convert it to J/mol.
ΔHvap = 38.56 kJ/mol × 1000 J/1 kJ = 38560 J/mol
The number of moles of ethanol can be calculated as before: moles = 0.6291 mol.
Heat = -38560 J/mol × 0.6291 mol = -24224.6 J
Step 3: Calculate the energy released when the liquid ethanol cools from its boiling point to -10.0 ∘C.
Next, we need to calculate the molar heat capacity for the liquid ethanol, which is 113 J/(K⋅mol).
ΔT = final temperature - initial temperature = -10.0 ∘C - 78.4 ∘C = -88.4 K
Heat = molar heat capacity × moles × ΔT
Using the same number of moles as before: 0.6291 mol
Heat = 113 J/(K⋅mol) × 0.6291 mol × -88.4 K = -4977.17 J
Finally, we add up the heats from each step:
Total Heat = Heat from Step 1 + Heat from Step 2 + Heat from Step 3
Total Heat = -380.2 J + (-24224.6 J) + (-4977.17 J) = -29081 J
To express the result to two significant figures, we convert the joules to kilojoules by dividing by 1000:
q = -29081 J / 1000 = -29.1 kJ
Therefore, the energy released when 29.0 g of ethanol vapor cools from 88.0 ∘C to -10.0 ∘C is -29.1 kJ.