find the molar enthalpy of vaporization for a substance, given that 3.21 mol of the substance absolve 28.4 kj of energy as heat when the substance changes from liquid to gas? show the work
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To find the molar enthalpy of vaporization (ΔHvap), we need to use the equation:
ΔHvap = q / n
Where:
- ΔHvap is the molar enthalpy of vaporization (in kJ/mol)
- q is the energy absorbed as heat (in kJ)
- n is the number of moles of the substance
In this case, we are given that 3.21 mol of the substance absorbs 28.4 kJ of energy as heat when it changes from liquid to gas.
Substituting the given values into the equation, we have:
ΔHvap = 28.4 kJ / 3.21 mol
Simplifying this equation, we get:
ΔHvap ≈ 8.85 kJ/mol
So, the molar enthalpy of vaporization for the substance is approximately 8.85 kJ/mol.