Question

The heat of vaporization of carbon disulfide is 27.4 kJ/mol at its boiling point of 46.3 °C. What is the change in enthalpy when 12.5 g CS2 condenses at 46.3 °C?

Answer 1

delta H for condensation = -27.4 KJ/mol

Molar mass of CS2,

MM = 1*MM(C) + 2*MM(S)

= 1*12.01 + 2*32.07

= 76.15 g/mol

mass(CS2)= 12.5 g

use:

number of mol of CS2,

n = mass of CS2/molar mass of CS2

=(12.5 g)/(76.15 g/mol)

= 0.1641 mol

Now use:

Q = number of mol * delta H

= 0.1641 mol * (-27.4 KJ/mol)

= -4.50 KJ

Answer: -4.50 KJ