What is delta H in kJ for the condensation of 25 g

Question

of CH3OH at 25 degrees C? (Delta Hvap = 38 kJ/mol)

Please explain.. thanks!

Answer 1

First we need to convert grams of methanol to moles.

25g CH3OH (mol/32g) = .78125 moles of CH3OH

Next we need to convert delta H vaporization to delta H condenstation. This is kinda simple.

Vaporization is:

CH3OH(l) + heat --> CH3OH(g)

Condensation is simply the reverse reaction:

CH3OH(g) --> heat + CH3OH(l)

Remember this rule of thumb the enthalpy change for the reverse reaction is the negative of the forward reaction, so:

delta Hvap = -(delta Hcon)  delta H con = -38kJ/mol

Since we know the number of moles we can find the energy released.

.78125moles of CH3OH *(-38kJ/mol) = -29.7kJ

Notice how the answer is negative, that's because energy is exiting the methanol.

Answer 2

Atomic weights: C=12  H=1  O=16  CH3OH=32

25gCH3OH x 1molCH3OH/32gCH3OH x (-38kJ/1molCH3OH) = -30 kJ to two significant figures

Boiling methanol is endothermic by +38 kJ/mole, so the condensation is exothermic by the same amount. Methanol boils at 64C, so this is only condensation at 25C. Nevertheless the same amounts of energy are involved in each process.