- #1

zenterix

- 678

- 83

- 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

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

?

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

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

?