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Heat capacity of NaCl

  1. Oct 28, 2015 #1
    1. The problem statement, all variables and given/known data

    The heat capacity of the solid NaCl from 500 K to 1074 K is given by [52.996 J*K-1*mol-1 – (7.86*10-3J*K-2*mol-1)*T + (1.97*10-5J*K-3*mol-1)*T2 ] and that of liquid NaCl from 1074 K to 1500 K is given by [125.637 J*K-1mol-1 – (8.187*10-2 J*K-2*mol-1)*T + (2.85*10-5 J*K-3*mol-1)*T2 ]. If Δfus HΘT, m = 28.158 kJ*mol-1 at 1074K, Please determine HΘ1500K - HΘ500K for NaCl.

    2. Relevant equations


    3. The attempt at a solution

    I tried to solve this task, but I don't understand what they exactly want me to do.

    Do I have to calculate the ΔH from 500K to 1074K with the given heat capacities? And what is ΔfusHΘ?
     
  2. jcsd
  3. Oct 28, 2015 #2
    You need to calculate two integrals. The integral of the solid heat capacity function with respect to temperature, from 500 K to 1074 K, and the integral of the liquid heat capacity function from 1074 K to 1500 K. ΔfusH° is the heat of fusion of NaCl. The first integral is the sensible heat required by solid NaCl to reach its melting point, then some amount of latent heat is required to melt NaCl at 1074 K, and the second integral is the sensible heat required by liquid NaCl to reach 1500 K. Add these three heats (don't neglect minus signs) and you'll have solved the problem.
     
  4. Oct 28, 2015 #3
    This is exactly the same kind of problem we did with the enthalpy chance of H2O in going from (373,l) to (393,v). The differences are that we are dealing with melting instead of vaporization, and the heat capacities depend on temperature. Since the heat capacities do depend on temperature, you have to integrate the heat capacity over the relevant temperature change to get the change in enthalpy.

    Chet
     
  5. Oct 29, 2015 #4
    I don't exactly understand what you mean with "integrate the heat capacity over the relevant temperature change", but maybe that's because I'm not a native English speaker.

    What is the change in the formula compared to the H2O problem?

    Does that mean, that H(T2)=H(T1)+Cp*(T2-T1) becomes H(T2)=H(T1)+(Cp(product)-Cp(starting material))*(T2-T1)?
     
  6. Oct 29, 2015 #5
    No.

    $$H(1074,s)=H(500,s)+\int_{500}^{1074}{C_p(T,s)dT}$$
    $$H(1074,l)=H(1074,s)+\Delta H_{fus}$$
    $$H(1500,l)=H(1074,l)+\int_{1074}^{1500}{C_p(T,l)dT}$$

    Do you see the similarity to our water problem now?

    Chet
     
  7. Oct 30, 2015 #6
    Yes I do, thank you!

    I try to calculate it when I'm home
     
  8. Oct 30, 2015 #7
    I got:

    34'186.93 J/mol for H(1074,s), while I set H(500,s) = 0

    315'766.93 J/mol for H(1074,l)

    and

    549'775.31 J/mol for H(1500,l)

    Are these values correct? They seem pretty big compared to those of the H20, but since it is a salt, I think it's possible?

    Now the task says: "Please determine HΘ1500K - HΘ500K for NaCl."

    Does this mean, that I have to add H(1074,s)+H(1074,l)+H(1500,l) to get de final ΔHΘ, which I think is asked?
     
  9. Oct 30, 2015 #8
    It takes a lot of energy to "loosen" (and break) ionic bonds.
    Yes.
     
  10. Oct 30, 2015 #9
    If H(1074,s)=34,186.93 J/mol, shouldn't H(1074,l) = 34186.93 + 28158 J/mol?
    Apparently not.

    The values for water were per kg, right?
    No. This difference is equal to H(1500,l)-H(500,s).

    Chet
     
  11. Oct 31, 2015 #10
    Right, I saw my mistake.

    Now I got:

    34'186.93 J/mol for H(1074,s), while I set H(500,s) = 0

    34186.93 + 28158 J/mol = 62'344.93 J/mol for H(1074,l)

    and

    296353.31 J/mol for H(1500,l)

    This means, that ΔHΘ = H(1500,l)-H(500,s) = 296353.31 J/mol - 0 J/mol = 296353.31 J/mol?
     
  12. Oct 31, 2015 #11
    Yes, that's better. I didn't check your arithmetic on the integrations, so your answer is correct if the integrations were done correctly. A way to check is to calculate the heat capacity at the average temperature for each temperature interval, and multiply each by the corresponding temperature difference. The results should be reasonable approximations to the values of the integrals.

    Chet
     
  13. Oct 31, 2015 #12
    Ok, thanks for the help!
     
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