Final answer:
The heat required to evaporate one mole of water from liquid to gas is the enthalpy of vaporization, which is approximately 2260 kJ per mole. Therefore, the correct answer is 2260 kJ.
Step-by-step explanation:
The question is asking us to determine how much heat is required to evaporate one mole of water from liquid to gas form using the values from an appendix, which provides standard molar enthalpies of formation.
The key concept to understand here is that the heat required for the phase change from liquid to gas is known as the enthalpy of vaporization (∆Hvap), which is different from the enthalpy of formation (∆Hf).
A cited excerpt from the provided reference materials indicates that when 1 mole of gaseous water (H2O(g)) forms from its elements, 242 kJ of heat are released, which implies that the reverse process would require 242 kJ to be absorbed to convert 1 mole of water from gas back to its elements.
However, since we are dealing with the evaporation of liquid water to gaseous water, we need to find the heat of vaporization which is not provided explicitly in the question but is commonly known to be approximately 2260 kJ for one mole of water at its boiling point.
Given this widely accepted value, the answer to the question of how much heat is required to evaporate one mole of water from liquid to gas phase would be 2260 kJ, corresponding to option c).