Can someone help me with the experimental tabulation or spread sheet data on enthalpy, temperature, and pressure for aluminum and oxygen combustion [closed]








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I need an experimental tabulation or spread sheet data of enthalpy, temperature, and pressure tabulated in columns for aluminum and oxygen combustion under non standard conditions.
Any references to books might help. Please I have tried checking it up I can't find any performed experiment whose data have been tabulated for further research purpose. I found something similar on NEST but it was for one bar pressure and enthalpy was a function of temperature. But I need for all temperature and pressure so I can interpolate the data for my research purpose.




The chemical reaction is as follows
$$ 4\ce{Al (s)} + 3\ce{O2 (g)} \rightarrow 2\ce{Al2O3 (s)} $$
We will abuse the fact that enthalpy is a **state function**. The process is illustrated in the following figure. The change we want is in red, but we will do it in blue:


[![enter image description here](https://i.stack.imgur.com/38dzT.png)](https://i.stack.imgur.com/38dzT.png)


**First Step-Reactants** The enthalpy differential for a pure fluid of constant composition is
$$ dH = c\_P dT + \left[V - T\left(\frac{\partial V}{\partial T}\right)\_P\right]dP
 \tag{1} $$
Given that we have two separated non-interacting phases, we can attack both species separately. Respectively:


1. **Solid $\ce{Al}$**: we will assume it is an incompressible fluid. If you find data of the volume expansivity $\beta$ in order to evaluate the partial derivative, then use it. If not, just get the density at room temperature and consider it constant. Integrating Eq. (1) yields
\begin{align}
 \Delta H\_{1} = \int\_{T\_1}^{T\_0} c\_\ce{P,Al(s)}(T) \; dT + V\_\ce{Al(s)}(P\_0 - P\_1) \tag{2}
\end{align}
2. **Gaseous $\ce{O2}$**: this is a non-polar, light gas, so the virial equation of state up to the second coefficient will suffice
\begin{align}
 Z &= 1 + \frac{BP}{RT} \\
 \frac{PV}{RT} &= 1 + \frac{BP}{RT} \rightarrow V = \frac{RT}{P} + B \tag{3} \\
\end{align}
Remember the virial coefficients are only function of temperature.


We will divide the enthalpy change of oxygen in two substeps. The first substep starts on $P\_1$ and ends on $P\_0$ at constant $T\_1$. Eq. (1) yields
\begin{align}
 \Delta H\_2 &= \int\_{P\_1}^{P\_0} \left[\frac{RT\_1}{P} + B(T\_1) - 
 T\_1\left(\frac{R}{P} - \frac{dB}{dT}(T\_1)\right)\right] \; dP \\
 \Delta H\_2 &= \int\_{P\_1}^{P\_0} \left(\frac{RT\_1}{P} + B(T\_1) - 
 \frac{RT\_1}{P} + T\_1\frac{dB}{dT}(T\_1)\right) \; dP \\
 \Delta H\_2 &= \left[B(T\_1) + T\_1\frac{dB}{dT}(T\_1)\right] 
 \int\_{P\_1}^{P\_0} \; dP \\
 \Delta H\_2 &= \left[B(T\_1) + T\_1\frac{dB}{dT}(T\_1)\right] 
 (P\_0 - P\_1) \tag{4} \\
\end{align}
The second substep starts on $T\_1$ and ends on $T\_0$ for constant $P\_0$. Eq. (1) yields
$$ \Delta H\_3 = \int\_{T\_1}^{T\_0} c\_\ce{P,O\_2(g)}(T) \; dT \tag{5} $$


Finally, the enthalpy change for the "reactant" step which I'll call $ \Delta H\_R $, is the sum of Eqs. (2), (4) and (5)
\begin{align}
 \Delta H\_R =& \int\_{T\_1}^{T\_0} c\_\ce{P,O\_2(g)}(T) \; dT +
 \left[B(T\_1) + T\_1\frac{dB}{dT}(T\_1)\right] (P\_0 - P\_1) + \\
 &\int\_{T\_1}^{T\_0} c\_\ce{P,Al(s)}(T) \; dT + V\_\ce{Al(s)}(P\_0 - P\_1) \tag{6}
\end{align}


**Second Step-Reaction** We on purpose made the reaction to happen at the conditions where the standard enthalpy of formation of the species are available. This one is easy, let me call this step $\Delta H\_\pu{RXN} $
\begin{align}
 \Delta H\_\pu{RXN} &= \sum\_j \nu\_j \Delta\_\pu{f} H^\Theta(j,\pu{state}) \\
 \Delta H\_\pu{RXN} &= 2\Delta\_\pu{f} H^\Theta(\ce{Al2O3},s) \tag{7} \\
\end{align}
Since oxygen in the gas state and aluminum in the solid state are in their most stable form.


**Third Step-Products** This is exactly the same as we did for aluminum, i.e., Eq. (2) but changing names. I will call this step $\Delta H\_P$


\begin{align}
 \Delta H\_P = \int\_{T\_0}^{T\_2} c\_\ce{P,Al2O3(s)}(T) \; dT + V\_\ce{Al2O3(s)}(P\_2 - P\_0) \tag{8}
\end{align}


**Conclusion** The enthalpy change of all the process is just the sum of Eqs. (6), (7), and (8), which I will resume because the equation is too large
$$ \boxed{\Delta H = \Delta H\_R + \Delta H\_\pu{RXN} + \Delta H\_P} \tag{9} $$
This gives you the enthalpy change as a function of pressure and temperature for this particular chemical reaction.


**References**


* [Density of aluminum](https://en.wikipedia.org/wiki/Aluminium).
* [Density and standard enthalpy of formation of aluminum oxide](https://en.wikipedia.org/wiki/Aluminium_oxide).
* Specific heat capacities for the three components, in their respective state, can be found on [NIST](https://www.nist.gov/).
* The second virial coefficient for oxygen can be obtained by statistical mechanics, given that its intermolecular potential is well established. I will give you the correlation needed. The Lennard-Jones potential is
\begin{equation}
 \Gamma(r) = 4\varepsilon\bigg[\left(\frac{\sigma}{r}\right)^{12} -
 \left(\frac{\sigma}{r}\right)^6\bigg] \tag{10}
\end{equation}
and for $O\_2$ [the parameters are](https://www.semanticscholar.org/paper/Molecular-Theory-Of-Gases-And-Liquids-Hirschfelder-Curtiss/c74bb1379abeec1f42b3cad633b95d19cfea69f4) $\sigma = \pu{3.433E-10 m}$ and $\varepsilon = \pu{1.560E-21 J}$. The second virial coefficient is given exactly by the following equation
\begin{equation}
 B(T) = 2\pi N\_A \int\_{0}^\infty \bigg[1 - \exp\bigg(-\frac{\Gamma(r)}{k\_B T}\bigg)\bigg]r^2 \; dr \tag{11}
\end{equation}
Coupling Eqs (10) and (11), and solved for many temperatures $T$, gives the value of $B$ at these temperatures. However, the function has no primitive, so the calculation has to be done numerically. The results and the correlation that I've just did are shown below, along with the derivative that is also needed:


[![enter image description here](https://i.stack.imgur.com/YnZYi.jpg)](https://i.stack.imgur.com/YnZYi.jpg)


As you see, there is virtually no difference. The function is the following:
\begin{equation}
 B(T) = 26.168 + 7.0889\cdot 10^{-3}T - \frac{8728}{T} - \frac{1.009 \cdot 10^6}{T^2} \hspace{1 cm} (\pu{50 K} \leq T \leq \pu{1000 K})
\end{equation}
with $T$ in $K$ and $B$ in $\pu{cm3/mol}$.


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