Join thousands of students who trust us to help them ace their exams!
Multiple Choice
The normal boiling point of water is 100.0 °C, and its molar enthalpy of vaporization is 40.67 kJ/mol. What is the change in entropy in the system in J/K when 39.3 grams of steam at 1 atm condenses to a liquid at the normal boiling point?
A
75.6 J/K
B
98.5 J/K
C
150.3 J/K
D
117.2 J/K
0 Comments
Verified step by step guidance
1
First, identify the key information given in the problem: the molar enthalpy of vaporization (ΔH_vap) is 40.67 kJ/mol, and the normal boiling point of water is 100.0 °C. The mass of steam is 39.3 grams.
Convert the mass of steam to moles using the molar mass of water (approximately 18.02 g/mol). Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).
Calculate the change in entropy (ΔS) using the formula: \( \Delta S = \frac{\Delta H_{\text{vap}}}{T} \), where \( T \) is the temperature in Kelvin. Convert the boiling point from Celsius to Kelvin by adding 273.15.
Substitute the values into the entropy change formula. Remember to convert the enthalpy from kJ/mol to J/mol by multiplying by 1000.
Multiply the entropy change per mole by the number of moles calculated in step 2 to find the total change in entropy for the given mass of steam.