Outage Notification

For questions where it is asking for the heat released or absorbed as water changes phases, we are not given the specific heat of ice, water and vapor, nor the values for heat of fusion and vaporization.  I used the values that were on the slides from the lectures for the calculations, but all the values are off just a little from the answer key, but they are more than rounding errors.  I am wondering if the values you are using to create the key are slightly different from the table from the lecture.  I did notice that the lecture slide has specific heat of ice as 2.06 J/gC while most internet references use 2.09 J/gC, but even with that, the values I calculate are slightly off from the keys.  #4. 250kJ vs 266 kJ, #5 154 kJ vs 152 kJ, #6 83166 J vs 82350 J.  # 10, #12 and #14 are also slightly different but less so. For #15 I cannot get the answer key values.  I calculate 68.8 kJ and the key says 40.5 kJ.  Also #18.  I calculate 15,041 J and the key is 20,000 J (is this due to rounding?) 

#21 is asking what is the equation for the standard enthalpy of formation of N2O3.  I would think the answer to this should be N2 (g)  + 3/2 O2 (g) –> N2O3 (g)  but the key says N2O (g) + O2 (g) –> N2O3  Shouldn’t the reactants be in elemental gas form? 

# 22 I think there is supposed to be an arrow before the 2SO3 (g) instead of a + sign.  Also I don’t know how we are supposed to solve this without some values.  Should we be given delta H of formation for the SO3 in order to determine the delta H naught? Same for #27

#24 I calculate methane’s delta H formation naught to be -75.8 kJ, using the information provided.  The answer key says -210.0 kJ

#2 c of Free response.  The first part of the question that is asking for the enthalpy of change is fine.  However, the second question might be missing information?  It is asking what is the final temperature it reaches if there is a change in energy of +52.088J?  What is “it”?  I assume it's a calorimeter, but do we know the amount of water?  Do don’t know the starting temperature either.  The answer key does not have an answer to this question.  

#3 of free response.  I have an issue with question b.  It is asking if the substance has greater heat of vaporization or heat of fusion.  I agree that based on data provided, it would appear that the heat of fusion is greater, but I think in the lecture, the lecturer made a point of saying that all substances have greater heat of vaporization than heat of fusion because something along the lines of it takes more energy to completely separate molecules from each other (vaporization) than to simply allow them to move more freely past each other (melting).
 For part c, it is asking to interpret whether the liquid or solid has lower specific heat…  We actually can’t tell because we were not told what temperature the solid was at start before it melted after three minutes.  I think you meant to say it started at zero degrees C since that is what the graph in the key shows, but that was not given.  If it started at -50 degrees or something, then solid will have lower specific heat.

Hi Yoko,
Thank you again for your comments! Detailed below are my responses.
* I added a table to the practice problems with ice, water and vapor specific heats and the heats of fusion and vaporization.
For #4, you are correct, the answer is 251 kJ. For #5, you are correct the answer is 154 kJ. For #6, the answer is 83,196 J.  For #15, the answer is 68.7 kJ. For #18 the answer is 15.02 kJ.
I revised the answer for #21 to N2(g) + 3/2 O2 --> N2O3.
For # 22, I replaced the + sign with an arrow. I added a table that includes heats of formation for S, O2 and SO3 for this question and #27.
For #24, you are correct, the answer is -75.8 kJ. 
For free response question 2, I removed the 2nd question that was part of part c.
For free response question 3, I changed the time to vaporize from 3 minutes to 6 minutes which makes the heat of vaporization greater than the heat of fusion which makes more sense.  For part c the steeper the slope of the phase change the lower the specific heat.  The steeper the curve the greater the temperature change with the addition of heat, the lower the specific heat.