What amount of heat would be involved in condensing 14.7 g of CH₃OH? (∆Hvap = 38.0 kJ/mol)

Question

What amount of heat would be involved in condensing 14.7 g of CH₃OH? (∆Hvap = 38.0 kJ/mol). I need the answer in kj

Answer 1

The heat of condensation is the negative of the heat of vaporization and is given by the negative of the product of the substance's latent heat of vaporization times the number of moles of the substance. So, we can write that the heat evolved by condensation is given by ΔHcond. = - ΔHvap. x moles CH3OH.

Moles of CH3OH = Mass CH3OH / Molar Mass CH3OH = 14.7 g / 32.0 g/mole = 0.469 moles.

ΔHcond. = -38.0 kj / mole x 0.469 moles = - 17.8 kj.

Note that the latent heat of vaporization is the heat of vaporization per mole of substance ( in this case 38.0 kj/mole).

Answer 2

First, we need to figure out how many moles of CH₃OH we have - this can be done by finding the molecular weight.

4 H = 4*(1 g/mol)

1 C = 1*(12 g/mol)

1 O = 1*(16 g/mol)

CH₃OH = 32 g/mol

The next step is to convert from grams (g) to moles (mol)

(14.7 g) / (32 g/mol) = 0.459 moles CH₃OH

To make sure your math is correct, double-check that your units cancel out to leave your final answer in moles.

Next, we need to determine if heat is being added or taken away. Because we are condensing, the CH₃OH is going from a gas to a liquid. This means that it is cooling, and thus heat is being removed from the system. Therefore, our ∆H_vap is going to be negative.

To solve for how much heat is needed, we can follow stoichiometry to multiply ∆H_vap by the number of moles to leave us with our final answer in kJ.

(-38.0 kJ/mol) * (0.459 mols) = -17.5 kJ

To double-check that the answer makes sense, we can see that the heat added is negative, which is the same as taking heat away. Because we are going from a gas to a liquid, it makes sense that heat needs to be removed from the system.