Final answer:
The quantity of heat needed to convert 25.0 g of ethanol from 23.0°C to vapor at 78.3°C is calculated by combining the heat required to raise the temperature to its boiling point and the heat required for vaporization, resulting in a total of 24.7235 kJ. This value does not exactly match the provided options, indicating a possible discrepancy.
Step-by-step explanation:
To calculate the quantity of heat required to convert 25.0 g of ethanol from a liquid at 23.0°C to a vapor at its boiling point, 78.3°C, we need to perform two calculations: one for the heat required to raise the temperature of the ethanol from 23.0°C to 78.3°C (using its heat capacity), and one for the heat required to vaporize the ethanol at its boiling point (using its enthalpy of vaporization).
First, we find the heat required to raise the temperature:
q = m × c × ΔT
q = (25.0 g) × (2.46 J/g°C) × (78.3°C - 23.0°C)
q = 3395.5 J or 3.3955 kJ
Next, we convert grams of ethanol to moles to use the enthalpy of vaporization:
Molar mass of ethanol (C2H5OH) = 46.07 g/mol
m = 25.0 g / 46.07 g/mol = 0.5427 mol
The heat required for vaporization is:
q₂ = n × ΔHαααp
q₂ = 0.5427 mol × 39.3 kJ/mol
q₂ = 21.328 kJ
Finally, we sum the two quantities of heat:
Total heat q = q₁ + q₂
Total heat q = 3.3955 kJ + 21.328 kJ
Total heat q = 24.7235 kJ
This value is not exactly matching any of the options provided, suggesting there may have been a calculation error or a discrepancy in the given answer choices.