The Enthalpy Games: Sunrise on the Calculations

DOFPro Team

Introduction

The Enthalpy Games: The Ballad of Integrals and Means discussed how internal energy and enthalpy vary as functions of \(T\) and \(P\).

The Enthalpy Games: Sunrise on the Calculations
does a worked example using nitrogen and data from NIST.

Example: Calculate \(\Delta H\) for 10 moles of nitrogen going from \(500\ \mathrm{K}\) to \(1000\ \mathrm{K}\). Compare the results for an approximate heat capacity, the mean heat capacity formula, integrating the heat capacity, and using the data table from the NIST Chemistry WebBook.

Approximate Heat Capacity

\[\begin{align} C_p &= \dfrac{7}{2} R = 3.5 \cdot 8.314\ (\mathrm{J/mol\ K}) \\&= 29.1\ (\mathrm{J/mol\ K}) = 0.0291\ (\mathrm{kJ/mol\ K}) \end{align}\]

\[\begin{align} \Delta \hat{H} &= C_p \Delta T = 0.0291\ (\mathrm{kJ/mol}) \cdot (1000 - 500) \\&= 14.55\ (\mathrm{kJ/mol}) \end{align}\]

\(\Delta H = n \Delta \hat{H} = 10\ \mathrm{mol} \cdot 14.55\ (\mathrm{kJ/mol}) = 145.5\ (\mathrm{kJ})\)

Shomate Equation, Average Heat Capacity

N.I.S.T. equations use \(T/1000\) instead of \(T\) to keep the constants scaled between \(50\) and \(0.05\)

For the derived formulas, plug in

\(\hspace{120px}T_\text{NIST} = T\ (\mathrm{K})/1000\)

\(\hspace{120px}T_{1_\text{NIST}} = 500/1000 = 0.5\)

\(\hspace{120px}T_{2_\text{NIST}} = 1000/1000 = 1.0\)

\[\begin{align} \bar{C}_p = a &+ \frac{b}{2} (T_1+T_2) + \frac{c}{3} (T_1^2 +T_1T_2 +T_2^2) \\&+ \frac{d}{4} [(T_1+T_2)(T_1^2+T_2^2)] + \dfrac{e}{T_1T_2} \end{align}\]

For \(\mathrm{N_2}\)

\(\hspace{80px}a = 19.50583,\)

\(\hspace{80px}b = 19.88705,\)

\(\hspace{80px}c = -8.598535,\)

\(\hspace{80px}d = 1.369784,\)

\(\hspace{80px}e = 0.527601\)

\(\hspace{80px}\bar{C}_p = 31.10\ (\mathrm{J/mol\ K}) = 0.03110\ (\mathrm{kJ/mol\ K})\)

\[\begin{align} \Delta \hat{H} &= C_p \Delta T = 0.03110\ (\mathrm{kJ/mol}) \cdot (1000 - 500) \\&= 15.55\ (\mathrm{kJ/mol}) \end{align}\]

\(\hspace{80px}\Delta H = n \Delta \hat{H} = 10\ \mathrm{mol} \cdot 15.55\ (\mathrm{kJ/mol}) = 155.5\ (\mathrm{kJ})\)

Shomate Equation, Integrate Heat Capacity

\[\Delta \hat{H} = \int\limits_{T_1}^{T_2} C_p (T) dT = \int\limits_{T_1}^{T_2} (a + b T + c T^2 + d T^3 + \dfrac{e}{T^2})dT\]

\[\begin{align} \Delta \hat{H}= a (T_2 - T_1) &+ \frac{b}{2} (T_2^2 - T_1^2) + \frac{c}{3} (T_2^3 - T_1^3) \\&+ \frac{d}{4} (T_2^4 - T_1^4) + e \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \end{align}\]

For \(\mathrm{N_2}\) \(a = 19.50583,\ b = 19.88705,\ c = -8.598535,\)

\(\hspace{325px}d = 1.369784,\ e = 0.527601\)

\(\Delta \hat{H} = 15.55\ (\mathrm{kJ/mol})\)

\(\Delta H = n \Delta \hat{H} = 10\ \mathrm{mol} \cdot 15.55\ (\mathrm{kJ/mol}) = 155.5\ (\mathrm{kJ})\)

\(\hat{H} = 0\) @ \(298.15\ \mathrm{K}\ (0\ ^\circ\mathrm{C})\)

\(\hat{H}_{500} = 5.91\ \mathrm{kJ/mol}\)

\(\hat{H}_{1000} = 21.46\ \mathrm{kJ/mol}\)

\(\Delta \hat{H} = \hat{H}_{1000} - \hat{H}_{500}\)

\(\ \ \ \ \ \ \ =21.46 - 5.91\)

\(\ \ \ \ \ \ \ = 15.55\ \mathrm{kJ/mol}\)

\(\Delta H = n \Delta \hat{H}\)

\(\ \ \ \ \ \ \ = 10\ \mathrm{mol} \cdot 15.55\ (\mathrm{kJ/mol})\)

\(\ \ \ \ \ \ \ = 155.5\ \mathrm{kJ}\)

\(\mathrm{N}_2\) data from NIST Chem WebBook

Table is equally spaced. Can average from \(500\ \mathrm{K}\) to \(1000\ \mathrm{K}\)

\(\hspace{200px}\bar{C}_p = 31.11\ (\mathrm{J/mol\ K}) = 0.003111\ (\mathrm{J/mol\ K})\)

\[\begin{align} \Delta \hat{H} &= C_p \Delta T = 0.03111\ (\mathrm{kJ/mol}) \cdot (1000 - 500) \\&= 15.56\ (\mathrm{kJ/mol}) \end{align}\]

Spreadsheet

Valuable Numbers to Know

Approximate Heat Capacities

Gas Type	\(C_v\)	\(C_p\)
Monatomic	\(\frac{3}{2}R\)	\(\frac{5}{2}R\)
Diatomic	\(\frac{5}{2}R\)	\(\frac{7}{2}R\)

Water values

\(C_p\text{ in } \left[\mathrm{\dfrac{Btu}{lb_m\ ^{\circ} F}}\right] \text{ or } \left[\mathrm{\dfrac{cal}{g\ ^{\circ} C}}\right]\)

 Steam \(\approx 0.5\)

 Water \(\approx 1.0\)

      Ice \(\approx 0.5\)

\(\Delta \hat{H}_v \approx 1000\ \left[\mathrm{\dfrac{Btu}{lb_m}}\right]\)

\(\ \Delta \hat{H}_m \approx 150\ \left[\mathrm{\dfrac{Btu}{lb_m}}\right]\)

The Takeaways

1. For moderate accuracy, the approximate heat capacities for monatomic and diatomic gases are adequate.
2. It is more efficient to calculate \(\Delta \hat{H}\) directly from the integral of \(C_p\) than to calculate \(\bar{C}_p\) and then use \(\Delta \hat{H} = \bar{C}_p \Delta T\).
3. There are multiple ways to calculate \(\Delta H\). The one you choose depends on what data you can find.

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The DOFPro Team