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

• krootox217
In summary: I'll check the integrations when I have time.So the final answer is:In summary, the change in enthalpy, ΔHΘ, for NaCl from 500 K to 1500 K is 296353.31 J/mol.
krootox217

## 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.

## 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?

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.

krootox217
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

krootox217
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?

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:

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.

krootox217
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?
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

krootox217
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?

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

krootox217
Ok, thanks for the help!

## What is the heat capacity of NaCl at different temperatures?

The heat capacity of NaCl (sodium chloride) varies depending on the temperature range. At temperatures ranging from 500K to 1074K, the heat capacity is 81.5 J/mol*K. At higher temperatures, ranging from 1074K to 1500K, the heat capacity decreases to 74.5 J/mol*K.

## How does the heat capacity of NaCl change with increasing temperature?

The heat capacity of NaCl decreases with increasing temperature. This is because at higher temperatures, the molecules of NaCl are already in a more excited state and have less capacity to absorb additional heat.

## Why does the heat capacity of NaCl decrease at higher temperatures?

The decrease in heat capacity at higher temperatures is due to the increase in molecular vibrations and rotations within the NaCl structure. These vibrations and rotations contribute to the overall internal energy of the substance, making it less responsive to additional heat energy.

## How does the heat capacity of NaCl compare to other substances?

The heat capacity of NaCl is relatively low compared to other substances. This is due to its simple crystal lattice structure which limits the number of degrees of freedom for molecular motion. Other substances with more complex structures tend to have higher heat capacities.

## What factors can affect the heat capacity of NaCl?

The heat capacity of NaCl can be affected by factors such as impurities, crystal defects, and changes in pressure or volume. Impurities and crystal defects can disrupt the regular lattice structure of NaCl and alter its heat capacity. Changes in pressure or volume can also affect the amount of energy required to raise the temperature of NaCl.

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