Question

The molar enthalpy of vaporization (∆Hvap) for ammonia (NH3) is 23.3 kJ/mol (at −33.3 °C). How much energy is required to evaporate 100. g of ammonia at this temperature?

Answer 1

Answer: 137 kJ

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{100g}{17g/mol}=5.9moles$

1 mole of ammonia requires heat = 23.3 kJ

Thus 5.9 moles of ammonia require heat =$\frac{23.3}{1}\times 5.9=137kJ$

Thus the energy is required to evaporate 100 g of ammonia at this temperature is 137 kJ