# How do we use enthalpy of formation in this calculation?

• Chemistry
• zenterix
zenterix
Homework Statement
Calculate
Relevant Equations
$$\overline{H}^{\circ}(2000\text{K})-\overline{H}^\circ (0\text{K})$$

for ##\mathrm{H(g)}##.
This is a problem from the book "Physical Chemistry" by Silbey, Alberty, and Bawendi.

The end of the book says the answer is ##41.572\text{kJ/mol}##.

Here is how to obtain this answer

There is a table a the end of the book with the following entry

The fourth column is ##\overline{H}_T^\circ-\overline{H}_{298}^\circ##.

As far as I understand, what this tells us is the following.

$$dH_P=\left (\frac{\partial H}{\partial T}\right )_P dT=C_PdT$$

$$\Delta H_P=\int_{T_1}^{T_2}C_PdT$$

where the subscript ##P## denotes constant pressure.

The notation "##^\circ##" in the formulas indicates we are at specific (standard) conditions. For a pure gaseous substance, this means constant pressure of ##1## bar.

If we have just one mole of the gas, we use the overline notation to indicate molar quantities.

$$\Delta \overline{H}^\circ_P=\int_{T_1}^{T_2}\overline{C}^{\circ}_PdT$$

If ##T_1=298\text{K}## and ##T_2=T## then

$$\Delta \overline{H}^\circ_P=\overline{H}_T^\circ-\overline{H}_{298}^\circ=\int_{298}^{T}\overline{C}^{\circ}_PdT$$

Therefore, to solve the problem we can use the calculation

$$(\overline{H}_{2000}^\circ-\overline{H}_{298}^\circ)-(\overline{H}_0^\circ-\overline{H}_{298}^\circ)$$

$$=35.375-(-6.197)$$

$$=41.572\text{kJ}$$

Okay, this is fine, but my question is if it is possible to arrive at this result using enthalpies of formation and if it is possible then how to do it?

What I thought about so far is the following, which seems incorrect

It seems we have the following reaction

$$\mathrm{H(g,0K,1\ bar)\rightarrow H(g,2000K,1\ bar)}\tag{1}$$

This is just the heating of ##\text{H}## in gas state.

We have

$$\Delta H^\circ_f(H,g,2000\text{K})=226.898\text{kJ}$$

$$\Delta H^\circ_f(H,g,0\text{K})=216.035\text{kJ}$$

I'm not sure what the equation is for the chemical reactions that have these enthalpies, but my guess is

$$\frac{1}{2}H_2(g)\rightarrow H(g)$$

Based on (1), why doesn't the following calculation

$$\Delta H^\circ_f(H,g,2000\text{K})-\Delta H^\circ_f(H,g,0\text{k})=12.74\text{kJ}$$

give us the same thing as

$$\overline{H}^{\circ}(2000\text{K})-\overline{H}^\circ (0\text{K})$$

?

Draw a Hess's Law cycle for (1) involving the formation of the species from the elements. What additional information do you need?

zenterix
Here is the cycle. We are interested in the enthalpy of reaction 1.

$$\Delta H^\circ_{rx}-\Delta H^\circ_f(H,g,2000\text{K})+\Delta H_3+\Delta H_f^\circ(H,g,0\text{K})=0$$

$$\Delta H^\circ_{rx}=\Delta H^\circ_f(H,g,2000\text{K})-\Delta H_f^\circ(H,g,0\text{K})-\Delta H_3$$

$$=12.74\text{kJ}-\int_{2000}^0\frac{1}{2}\text{mol}\cdot 28.82\mathrm{\frac{J}{K\cdot mol}}dT$$

$$=12.74\text{kJ}-(-28.82\text{kJ})$$

$$=41.56\text{kJ}$$

Which seems to be correct.

What I had wrong in my original calculation, ie just the difference in enthalpies of formation is shown in the cycle: we can only simply take this difference when the reactants and products are at the same temperature.

We break the bonds of the reactant (##H(g)## at ##0\text{K}##) and we get ##\frac{1}{2}H_2(g)## at ##0\text{K}## plus a change in energy (##-216.07\text{kJ}##).

The enthalpy of formation of the product (##H(g)## at ##2000\text{K}##) seems to be based on ##\frac{1}{2}H_2(g)## at ##2000\text{K}##. This enthalpy of formation is ##+228.82\text{kJ}##.

Thus, to use the ##\frac{1}{2}H_2(g)## at ##0\text{K}## we have to heat it up which requires ##28.82\text{kJ}##.

Therefore, to get from the reactant to ##\frac{1}{2}H_2(g)## at ##2000\text{K}## we use ##-216.07+28.82=-187.25\text{kJ}##.

We now add in the energy ##+228.82## required to form ##H(g)## at ##2000\text{K}## and the total energy is ##+41.56\text{kJ}##.

Last edited:
Why do you write 228.82 kJ for the heat of formation of H at 2000K? This is not the value in your table.
Is Cp,m of H2 equal to 5R/2 over the whole range 0-2000K?
Also, what is the standard state of H2 at 0K? What else do you have to consider?

I actually have no idea anymore where I got the ##228.82\text{kJ}## figure from. You are correct. It is not in the table in the OP.

I am looking at other books, etc, so I suspect I saw it in some other table. It is surprising that the result turned out to be correct numerically.

Using the values in the table and the same calculation from my previous post, the result comes out to ##39.68\text{kJ}## which does not match the correct result.

You are right, there is a huge temperature change and so the assumption of constant pressure heat capacity
is perhaps this is why the result came out to only ##39.68\text{kJ}##.

As for the standard state of ##H_2## at ##0K##, I am not sure. I would guess it is just ##H_2## gas at ##1\text{bar}##.

Using the values of heat capacity at each different temperature for ##H_2(g)## as shown in the table, I get the following result

$$\Delta H_{rx}^\circ=\Delta H_f^\circ (H,g,2000\text{K})-\Delta H_f^\circ(H,g,0\text{K})-\Delta H_3$$

$$=226.898\text{kJ}-216.035\text{kJ}-(-31.94\text{kJ})$$

$$=42.8\text{kJ}$$

where

$$\Delta H_3=\int_{2000}^{1000} \frac{1}{2}\cdot 34.28dT+\int_{1000}^{500} \frac{1}{2}\cdot 30.205dT+\int_{500}^{298} \frac{1}{2}\cdot 29.26dT+\int_{298}^{0} \frac{1}{2}\cdot 28.836dT$$

$$=-31.94\text{kJ}$$

mjc123 said:
What else do you have to consider?

As far as I can tell, the only extra to consider is the changing heat capacities as I did above. Am I still missing something?

The result of ##42.8\text{kJ}## does not match the desired result of ##41.57\text{kJ}##.

Hydrogen is a solid at 0K. You have to think about the heat capacities of the solid and liquid, and the heats of fusion and evaporation.

You see how much more complicated it is using heats of formation! If you have the tabulated H° values, use them!

Where can I find data about heat capacities of hydrogen gas from near 0K up to 2000K?

It seems that to compute ##\Delta H_3## in my previous post we need to do

$$\mathrm{H_2(s,0K)\rightarrow H_2(s,14K)\rightarrow H_2(l,14K)\rightarrow H_2(l,20K)\rightarrow H_2(g,20K)\rightarrow H_2(g,2000K)}\tag{1}$$

The table entry for ##H_2## is

Which doesn't seem too useful for computing the sum in (1).

Oh well, if you've got it tabulated, you can get it from the values in column 4 (assuming the correct standard state at 0K). At any rate, this gives the book answer.

mjc123 said:
Oh well, if you've got it tabulated, you can get it from the values in column 4 (assuming the correct standard state at 0K). At any rate, this gives the book answer.
Don't we need to consider phase transitions? These are not part of the table though right?

Depends. The standard state of H2 at 0K is solid, so if (despite the heading "H2(g)") they are using this standard state at 0K, the phase transitions will be included in the value of H°(0)-H°(298). If on the other hand they do mean H2(g) at 0K, then they must have in mind the cycle you drew in post 3. Either way, the book values give the book answer.

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