What is the change in entropy for the vaporization of 18 g of H₂O at 100°C? At this temperature, the molar enthalpy of vaporization for water is 44.0 kJ/mol. A) 44 kJ/mol B) 440 J/K C) 117.9 J/K D) -41 kJ/K E) -440 kJ/K

Question

Quizplus · Accepted Answer

To calculate the change in entropy for the vaporization of water at 100°C, we can use the following formula:

ΔS = q/T

where ΔS is the change in entropy, q is the heat absorbed during vaporization, and T is the temperature at which vaporization occurs.

First, we need to calculate the amount of heat absorbed during the vaporization of 18 g of water:

q = (18 g / 18.02 g/mol) x (44.0 kJ/mol) = 43.2 kJ

Next, we can substitute the values into the formula:

ΔS = (43.2 kJ) / (373 K) = 0.117 J/K

When rounded to the nearest tenth, the answer is 0.1 J/K, which is equivalent to 117.9 J/K. Therefore, the correct answer is C) 117.9 J/K.

What Is the Change in Entropy for the Vaporization of 18