Heat Capacity of NaCl: 500K-1074K & 1074K-1500K

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Discussion Overview

The discussion revolves around calculating the change in enthalpy (ΔH) for solid and liquid NaCl between the temperatures of 500 K and 1500 K, specifically focusing on the heat capacities of NaCl in solid and liquid states across specified temperature ranges. The context includes mathematical reasoning and integration of heat capacities to determine enthalpy changes.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents the heat capacity equations for solid and liquid NaCl and seeks clarification on how to calculate ΔH from 500 K to 1500 K.
  • Another participant explains that two integrals need to be calculated: one for the solid heat capacity from 500 K to 1074 K and another for the liquid heat capacity from 1074 K to 1500 K, along with the latent heat of fusion.
  • A participant draws a parallel to a previous problem involving water, noting the differences in phase change and heat capacity dependence on temperature.
  • Some participants express confusion about the integration process and the formulas used, particularly in relation to the water problem.
  • One participant shares their calculated values for enthalpy at various stages and questions their correctness, noting that the values seem large compared to those for water.
  • Another participant suggests that the large values are plausible due to the nature of ionic bonds in salts.
  • There is a discussion about the relationship between the enthalpy values at different states and the correct approach to calculating ΔH, with some participants correcting earlier misunderstandings.
  • Final calculations are presented, with one participant confirming their results and seeking validation on the arithmetic and integration process.

Areas of Agreement / Disagreement

Participants generally agree on the need to integrate the heat capacities and the approach to calculating ΔH, but there are varying levels of understanding regarding the integration process and the relationships between enthalpy values. Some confusion remains about the specifics of the calculations.

Contextual Notes

Participants express uncertainty about the integration of temperature-dependent heat capacities and the implications of phase changes on enthalpy calculations. There are also references to potential errors in arithmetic and integration steps that have not been resolved.

krootox217
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Homework Statement



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.

Homework Equations

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?
 
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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.
 
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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
 
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)?
 
krootox217 said:
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)?
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
 
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Yes I do, thank you!

I try to calculate it when I'm home
 
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?
 
krootox217 said:
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?
It takes a lot of energy to "loosen" (and break) ionic bonds.
krootox217 said:
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?
Yes.
 
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krootox217 said:
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)
If H(1074,s)=34,186.93 J/mol, shouldn't H(1074,l) = 34186.93 + 28158 J/mol?
and

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

Are these values correct?
Apparently not.

They seem pretty big compared to those of the H20, but since it is a salt, I think it's possible?
The values for water were per kg, right?
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?
No. This difference is equal to H(1500,l)-H(500,s).

Chet
 
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  • #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?
 
  • #11
krootox217 said:
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?
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
 
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  • #12
Ok, thanks for the help!
 

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