Question

The enthalpy of vaporization of water is 40.79 kJ/mol. How much heat is required to vaporize 2 moles of water at its boiling point?

• A. 40.79 kJ
• B. 81.58 kJ
• C. 20.39 kJ
• D. 10.39 kJ

Hint:

The enthalpy of vaporization represents the heat required to convert 1 mole of a liquid into vapor without changing temperature. Be sure to multiply by the number of moles for the total heat required.

Answer 1

The Correct Option is B

The heat required to vaporize water can be calculated using the formula: \[ Q = n \Delta H_{\text{vap}} \] Where: - \( n = 2 \, \text{mol} \) is the number of moles, - \( \Delta H_{\text{vap}} = 40.79 \, \text{kJ/mol} \) is the enthalpy of vaporization. Now, calculate the heat: \[ Q = 2 \times 40.79 = 81.58 \, \text{kJ} \] Thus, the heat required to vaporize 2 moles of water is 81.58 kJ.