Question

Calculate the enthalpy change that occurs when 1.00 kg of acetone condenses at its boiling point (329.4 K). The standard enthalpy of vaporization of acetone is 29.1 kJ · mol-1 .
1.) −501 kJ
2.) −29.1 kJ
3.) +501 kJ
4.) +29.1 kJ
5.) −2.91 × 104 kJ

Answer 1

Answer:

1.) -501 kJ.

Explanation:

The enthalpy of vaporization is defined as the amount of energy that must be added to a liquid substance, to transform a quantity of that substance into a gas.

ΔHvap = 29.1 kJ/mol

Δq = n * ΔHvap

Where,

n = number of moles

ΔHvap = standard enthalpy of vaporization

Δq = enthalpy change of vaporization

Molar mass of acetone, C3H6O = (3*12) + (6*1) + (1*16)

= 58 g/mol

Number of moles = mass/molar mass

= 1000/58

= 17.2412 mol.

Δq = 17.2412 * 29.1

= +501.7 kJ.

Δq = 501 kJ is required to be added to evaporate it so Δq = - 501 kJ is required to condense it

Answer 2

Answer:

Option 1 is correct.

ΔH for the condensation of 1 kg of Acetone is - 501 KJ.

Explanation:

Molar mass of acetone = 58.08 g/mol

Number of moles of acetone that condenses = mass/molar mass = 1000/58.08 = 17.22 moles

1 mole of Acetone vaporizer with an enthalpy change of 29.1 KJ.

1 mole of Acetone will condense with an enthalpy change of - 29.1 KJ (since condensation is the exact reverse of vaporization)

That is, vaporization is given by,

Acetone(l) ----> Acetone(g) ΔH = 29.1 KJ

Then condensation will be given by,

Acetone(g) ----> Acetone(l) ΔH = -29.1 KJ

So,

1 mole of Acetone will condense with an enthalpy change of - 29.1 KJ

17.22 moles of Acetone will condense with an enthalpy change of 17.22 × - 29.1 KJ = - 501 KJ.

The first option is correct.