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

To evaporate 100 g of ammonia at -33.3 °C, 136.76 kJ of energy is required.

Step-by-step explanation:

To calculate the amount of energy required to evaporate 100 g of ammonia at -33.3 °C, we need to use the molar enthalpy of vaporization (∆Hvap). The ∆Hvap for ammonia is 23.3 kJ/mol. First, we need to convert the mass of ammonia to moles using its molar mass. The molar mass of ammonia is 17.03 g/mol. So, 100 g of ammonia is equal to 100/17.03 = 5.87 mol. Now, we can calculate the energy required:

Energy required = ∆Hvap × number of moles = 23.3 kJ/mol × 5.87 mol = 136.76 kJ

Answer 2

Answer: 137 kJ

Step-by-step 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}}=(100g)/(17g/mol)=5.9moles`

1 mole of ammonia requires heat = 23.3 kJ

Thus 5.9 moles of ammonia require heat =
`(23.3)/(1)* 5.9=137kJ`

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