Answer :
Final answer:
The change in energy (∆E) when 2.00 moles of a liquid is vaporized at its boiling point (85.0 °C) and 1.00 atm pressure is 78.2 kJ.
Explanation:
In chemistry, ∆E usually refers to change in energy. ∆Hvap represents the heat (enthalpy) of vaporization. It's the energy required to convert one mole of a substance from a liquid state into a gas state at a constant temperature and pressure. Here, the ∆Hvap is given as 39.1 kJ/mol at 85.0 °C and at 1.00 atm pressure.
To calculate the change in energy (∆E), you should use the formula:
∆E = n * ∆Hvap.
Where n is the number of moles. In this case, you have 2.00 moles
∆Hvap = 39.1 kJ/mol.
Thus, ∆E = 2.00 mol * 39.1 kJ/mol = 78.2 kJ.
So, the energy required to vaporize 2.00 moles of the substance at its boiling point and under 1.00 atm pressure is 78.2 kJ.
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