Calculate difference in molar entropy

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    1. The problem statement, all variables and given/known data

    From Atkins' Physical Chemistry, 8th Edition. Problem 3.1.

    Calculate the difference in molar entropy (a) between liquid water and ice at -5 degrees C, (b) between liquid water and its vapour at 95 degrees C and 1.00 atm. The differences in heat capacities on melting and on vaporization are 37.3 J/K*mol and -41.9 J/K*mol, respectively. Distinguish between the entropy changes of the sample, the surroundings, and the total system, and discuss the spontaneity of the transitions at the two temperatures.

    2. Relevant equations

    (1) [tex]\Delta S = \frac{\Delta_{fus}H}{T}[/tex]

    (2) [tex]\Delta S_{trs} = \frac{\Delta_{trs}H}{T_{trs}}[/tex]

    (3) [tex]S(T_f)=S(T_i) + C_p ln(\frac{T_f}{T_i})[/tex]

    where [tex]C_p[/tex] is the Heat Capacity at constant pressure.

    3. The attempt at a solution

    The problem I am having is that I don't understand what the question is asking. It gives me the differences in heat capacity, but the temperature doesn't seem to be changing. The given heat capacities seems to imply that I should use equation three, but if I use the constant temperature the [tex]\Delta S[/tex] goes to zero, which is wrong.

    Alternatively, I think I could use equation 2 by using the molar heat of fusion (and vaporization) and then just dividing by the temperature. But I don't understand how the phase transition can occur at the temperatures given.

    Can someone help me understand what this question is asking for?
     
    Last edited: Jan 26, 2009
  2. jcsd
  3. Mapes

    Mapes 2,532
    Science Advisor
    Homework Helper
    Gold Member

    Break (a) into two steps: find the difference in entropy between liquid water and ice at 0°C. This is the difference due to the phase change alone. Now find the entropy difference between ice at 0°C and 5°C. This is the difference due to a temperature change.
     
  4. Thanks for your reply. I'll try that and see if it matches the answer in the back of the book.
     
  5. Here's what I've got so far:

    First Part of (a)

    Calculate change due to temperature increase:

    [tex]\Delta S_{sys,temp} = C_p * ln(\frac{T_f}{T_i})[/tex]
    [tex]\Delta S_{sys,temp} = \frac{37.3 J}{K*mol}ln(\frac{273.15}{268.15}) = \frac{0.689 J}{K*mol}[/tex]

    Calculate change due to phase transition:

    [tex]\Delta S_{sys, trs} = \frac{\Delta_{fus}H H_2O}{T} = \frac{\frac{6008 J}{K*mol}}{273.15 K} = \frac{22.0 J}{K*mol}[/tex]

    Add them together:

    [tex]\Delta S_{sys} = \Delta S_{sys,temp} + \Delta S_{sys, trs}[/tex]

    [tex]\frac{0.689 J}{K*mol} + \frac{22.0 J}{K*mol} = \frac{22.69 J}{K*mol}[/tex]

    The answer in the book says [tex]\frac{-21.3 J}{K*mol}[/tex]. I would have the right result if my [tex]\Delta S_{sys, trs}[/tex] was negative, but the table in the book gives the heat of vaporization as a positive quantity. Also, since the transition is melting ice into water, the [tex]\Delta H[/tex] and [tex]\Delta S[/tex] should be positive, right? Did I do something wrong with my equations?

    Second Part of (a)

    I don't understand how to find the [tex]\Delta S_{surr}[/tex]. I have the following equations:

    [tex]\Delta S = \int_{T_i}^{T_f} \frac{dq}{T}[/tex]

    [tex]\Delta S_{trs} = \frac{\Delta_{trs}H}{T_{trs}}[/tex]

    Here's what I've tried:

    Temperature:
    [tex]\Delta S_{surr,temp} = -q*ln(\frac{273.15}{268.15}) = \frac{-.013 J}{K*mol}[/tex]

    Transition:
    [tex]\Delta S_{surr,trs} = -\Delta S_{sys, trs} = -\frac{22.0 J}{K*mol}[/tex]

    Adding them together:

    [tex]\Delta S_{surr} = \Delta S_{surr,temp} + \Delta S_{surr,trs}=\frac{-22.01 J}{K*mol}[/tex]

    But the answer in the book says it should be [tex]\frac{21.7 J}{K*mol}[/tex]...even changing the signs of my calculated answers will not result in this number. What have I done wrong?
     
  6. Mapes

    Mapes 2,532
    Science Advisor
    Homework Helper
    Gold Member

    Whoops, I was thinking liquid water at 0°C and ice at -5°C and missed a third step. You need to take the liquid water down to -5°C also. Then you should have the right answer.
     
  7. I have something similar for 3.1 a), and I'll just fill in on what I have for it:

    This problem asks us to calculate various entropies as a function of two things -- temperature and phase. Since entropy is a state function, we can break the change of state into a few steps by changing one component at a time (as you have already done).

    [tex]
    \Delta S_{sys} = \Delta S_{sys}(-5^{\circ}C\rightarrow0^{\circ}C, water) + \Delta S_{sys}(0^{\circ}C, water\rightarrow ice) + \Delta S_{sys}(0^{\circ}C\rightarrow-5^{\circ}C, ice)
    [/tex]
    [tex]
    = \Delta C_{p,m (water)}*ln(\frac{273.15}{268.15})
    + \frac{-\Delta H_{fus}}{273.15 K} + \Delta C_{p,m (ice)}*ln(\frac{268.15}{273.15})
    [/tex]
    [tex]
    = \Delta C_{p}*ln(\frac{273.15}{268.15}) + \frac{-\Delta H_{fus}}{273.15}
    [/tex]
    [tex]
    = \frac{37.3 J}{K * mol}*ln(\frac{273.15}{268.15}) - \frac{\frac{6008 J}{mol}}{273.15 K}
    [/tex]
    [tex]
    = -21.306\frac{J}{K * mol}
    [/tex]

    (Note that [tex]\Delta C_{p} = \Delta C_{p,m (ice)} - \Delta C_{p,m (water)}[/tex])


    [tex]
    \Delta S_{surr} = \frac{-\Delta H_{fus}(268.15 K)}{268.15 K}
    [/tex]
    [tex]
    = \frac{\Delta H_{sys}(-5^{\circ}C\rightarrow0^{\circ}C, water) + \Delta H_{sys}(0^{\circ}C, water\rightarrow ice) + \Delta H_{sys}(0^{\circ}C\rightarrow-5^{\circ}C, ice)}{268.15 K}
    [/tex]
    [tex]
    = \frac{\Delta C_{p,m (water)}(273.15 K - 268.15 K) + \frac{6008 J}{mol} - \Delta C_{p,m (ice)}(268.15 K - 273.15 K)}{268.15 K}
    [/tex]
    [tex]
    = \frac{\frac{6008 J}{mol} - \Delta C_{p}(273.15 K - 268.15 K)}{268.15 K}
    [/tex]
    [tex]
    = \frac{\frac{6008 J}{mol} - \frac{37.3 J}{K * mol}(5 K)}{268.15 K}
    [/tex]
    [tex]
    = 21.710\frac{J}{mol}
    [/tex]
    [tex]
    \Delta S_{tot} = \Delta S_{sys} + \Delta S_{surr}
    = 0.40372\frac{J}{mol}
    [/tex]

    Because [tex]\Delta S_{sys} > 0[/tex], the transition of water to solid at [tex]-5^{\circ}C[/tex] is a spontaneous process (the reverse process isn't).

    The steps for solving part b is very similar, except I added the heat of vaporization (instead of adding its negative) for [tex]\Delta S_{sys}[/tex] and used 368.15K instead of 373.15K as my denominator when solving for [tex]\Delta S_{surr}[/tex]. You should find that the transition of water to vapor at [tex]95^{\circ}C[/tex] is not spontaneous.

    I apologize in advance for my unfamiliarity with Latex codes.
     
  8. Oops, my bad... the third step for computing [tex]\Delta S_{surr}[/tex] should be:

    [tex]
    = \frac{-\Delta C_{p,m (water)}(273.15 K - 268.15 K) + \frac{6008 J}{mol} - \Delta C_{p,m (ice)}(268.15 K - 273.15 K)}{268.15 K}
    [/tex]

    Told you I was bad with the Latex code... heh
     
  9. 1) I agree that [tex]\Delta H_[/tex] is positive when melting ice into water -- so on the other hand, when water solidifies into ice, [tex]\Delta H_[/tex] is negative, but [tex]\Delta S_[/tex] may not be. For this problem, I think we can assume that we're calculating the enthalpy changes that occur when transitioning from water to ice.

    2) There shouldn't be any need for using ln operators for computing [tex]-\Delta H_{fus}[/tex]. The formula of integration on P. 56 (it's from Chapter 2, eq. 2.35) seems to suggest that.
     
  10. You know, I went to the recitation today and the grad student guy didn't know how to solve this problem so he just talked for a long time. It was really annoying. He basically told me that the book might be wrong and that I should just try to solve the logic of the problem and not pay much attention to the numerical result.

    Yet, here you have solved it logically and matched the answer in the book. I wish you were the grad student at the recitation. haha. I was definitely using the wrong idea for the molar heat of fusion, so thank you for correcting that. I had been puzzling over the [tex]\Delta S_{surr}[/tex] part for hours but it seems straight-forward now.

    By the way your LaTeX code looks great. It's at least as good as mine, probably better.

    I owe you one. And when the economy improves, I'll have one to give you. =D
     
    Last edited: Jan 27, 2009
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