When atmospheric pressure is 700. mmHg, water boils at 97.714 degrees C. Calculate entropy change of water if 50.0 grams is vaporized under these conditions. The heat of vaporization of water can be found in almost any reference text, and Cp H2O= 33.6 J/molK.
I found the heat of vaporization to be 9.72 kcal/mol at 100. degrees C in my text book, and I know that I can find change in entropy with the equation:
delta-G= delta-H - T(delta-S)
I'm just having trouble putting all the pieces together.
I was thinking something like this:
50.0 g H20 x 1 mol/18.02 g= 2.774 mol H2O
I was going to then multiply this value by 9.72 kcal/mol, but then I remembered that this value was for 100. degrees C, not 97.714 degrees C like stated in the problem. Is my thought process on the right track here? How do I manipulate the value that I have for this problem?
Your thought process is on the right track. To calculate the entropy change of water when 50.0 grams is vaporized, you can follow these steps:
1. Convert the mass of water into moles. You correctly calculated that 50.0 grams of H2O is equal to 2.774 moles of H2O using the molar mass of water (18.02 g/mol).
2. In order to use the heat of vaporization value, you need to convert it to joules. Since 1 kcal = 4184 J, you can convert the heat of vaporization from 9.72 kcal/mol to 9.72 kcal/mol x 4184 J/kcal = 40,673 J/mol.
3. Now, you need to account for the temperature difference. The heat of vaporization value you found in your textbook is for water boiling at 100 degrees C, but the problem states that water is boiling at 97.714 degrees C. To adjust for this, you need to use the heat capacity formula: delta-H = Cp * (T2 - T1). Since Cp H2O is given as 33.6 J/molK, you can calculate the difference in temperature as follows: delta-T = 97.714 degrees C - 100 degrees C = -2.286 degrees C. Note that negative sign denotes a decrease in temperature.
4. Calculate the entropy change using the formula: delta-S = (delta-H - delta-T * Cp) / T. Plug in the values:
delta-S = (40,673 J/mol - (-2.286 degrees C * 33.6 J/molK)) / (97.714 + 273.15 K).
5. Convert the delta-S value from J/(molK) to J/K by multiplying it by the number of moles of water (2.774 mol). This gives you the entropy change in joules per Kelvin.
By following these steps, you can calculate the entropy change of water when 50.0 grams is vaporized under the given conditions.