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Calcination energy requirement for CO₂ capture

by Steven Baltakatei Sandoval

Created on 2023-11-13T15:50-08 under a CC BY-SA 4.0 License. Updated on 2023-11-13T17:29-08.

Abstract

The minimum energy required to capture $1 ton$ of carbon dioxide (CO₂) via the calcination reaction ${CaCO}_{3} (s) \to CaO (s) + {CO}_{2} (g)$ is calculated by a heat of formation analysis.

1Background

The New York Times reported on 2023-11-09 of a direct air capture (DAC) plant in the United States operated by Heirloom Carbon Technologies (Plumer, 2023). Heirloom's technology involves capturing carbon dioxide (CO₂) via adsorption to plate holding calcium oxide (CaO) powder which is converted to calcium carbonate ( ${CaCO}_{3}$ ) over time via the calcination reaction (Equation 1).

{CaCO}_{3} (s) \to CaO (s) + {CO}_{2} (g)

(1)

2Analysis

The amount of energy required to drive a reaction or that is liberated by a reaction can be calculated by summing the difference between the standard heat of formation of its products with its reactants (Equation 2).

Δ H^{\circ} = \sum Δ H_{f}^{⦵} (products) - \sum Δ H_{f}^{⦵} (reactants)

(2)

Each reactant in Equation 1 has an associated standard heat of formation that may be found in the literature (Coker, 2001). The heat of formation of a substance is the enthalpy change associated with a chemical reaction for forming the substance from its constituent elements. For example, $C (s) + O_{2} (g) \to {CO}_{2} (g)$ is the reaction for forming carbon dioxide gas. Each formation reaction has an associated enthalpy change associated with (e.g. $Δ H_{f}^{⦵} = - 393, 509 J$ for carbon dioxide gas). The calcination reaction in equation 1 has three associated formation reactions:

\begin{array}{ccccc} Ca (s) + C (s) + \frac{3}{2} O_{2} (g) & \to & {CaCO}_{3} (s) & Δ H_{f}^{⦵} = - 289.5 \\ Ca (s) + \frac{1}{2} O_{2} (g) & \to & CaO (s) & Δ H_{f}^{⦵} = - 151.7 \\ C (s) + O_{2} (g) & \to & {CO}_{2} (g) & Δ H_{f}^{⦵} = - 289.5 \end{array}

These equations and their associated standard heat of formations may be summed to produce the desired calcination reaction and its standard heat of reaction.

\begin{array}{ccccc} {CaCO}_{3} & \leftarrow & Ca (s) + C (s) + \frac{3}{2} O_{2} (g) & Δ H_{f}^{⦵} = 289.5 \\ Ca (s) + \frac{1}{2} O_{2} (g) & \to & CaO (s) & Δ H_{f}^{⦵} = - 151.7 \\ C (s) + O_{2} (g) & \to & {CO}_{2} (g) & Δ H_{f}^{⦵} = - 289.5 \\ {CaCO}_{3} & \to & CaO (s) + {CO}_{2} (g) & Δ H^{⦵} = 43.7 \end{array}

In other notation:

$Δ H^{⦵}$	$=$	$[Δ H_{f}^{⦵} (CaO) + Δ H_{f}^{⦵} ({CO}_{2})] - [Δ H_{f}^{⦵} ({CaCO}_{3})]$	(3)
	$=$	$[(- 151.7 \frac{kcal}{mol}) + (- 94.052 \frac{kcal}{mol})] - [- 289.5 \frac{kcal}{mol}]$	(4)
	$=$	$[- 151.7 - 94.052] + [289.5] \frac{kcal}{mol}$	(5)
$Δ H^{⦵}$	$=$	$43.748 \frac{kcal}{mol}$	(6)
$Δ H^{⦵}$	$=$	$43.7 \frac{kcal}{mol}$	(7)

This figure may be expressed in standard units of $\frac{kJ}{mol}$ .

$Δ H^{⦵}$	$=$	$43.7 \frac{kcal}{mol}$
	$=$	$(43.7 \frac{kcal}{mol}) \cdot (\frac{4.184 kJ}{kcal})$	(8)
	$=$	$182.8408 \frac{kJ}{mol}$	(9)
$Δ H^{⦵}$	$=$	$183 . \frac{kJ}{mol}$	(10)

\begin{array}{lll} {CaCO}_{3} (s) \to CaO (s) + {CO}_{2} (g) & Δ H^{\circ} = 183 . \frac{kJ}{mol} \end{array}

(11)

To calculate the amount of energy required to capture 1 ton of CO₂, use the molar mass of CO₂ ( $44.01 \frac{g}{mol}$ ) to calculate the number of mols in 1 ton, then apply equation 11. Test: ton CO₂ .

$1 ton {CO}_{2}$	$=$	$(1 ton {CO}_{2}) \cdot (\frac{1000000 g {CO}_{2}}{1 ton {CO}_{2}}) \cdot (\frac{1 mol {CO}_{2}}{44.01 g {CO}_{2}})$	(12)
	$=$	$\overline{2272} 0 mol {CO}_{2}$	(13)

Now to calculate the heat released by carrying out the reaction described by equation 11:

$Δ H^{⦵}$	$=$	$(\overline{2272} 0 mol {CO}_{2}) \cdot (1) \cdot (\frac{183 . kJ}{mol})$	(14)
$Δ H^{⦵}$	$=$	$\overline{415} 7760 kJ$	(15)
$Δ H^{⦵}$	$≅$	$4160000 kJ$	(16)
$Δ H^{⦵}$	$≅$	$4.16 GJ$	(17)

In other words, the energy required to drive the calcination reaction described in equation 11 is:

Δ H^{⦵} = 4.16 \frac{GJ}{ton {CO}_{2}}

(18)

To calculate how much energy is required by a person as a function of their per capita CO₂ emissions (e.g. $\dot{m} = 14.44 \frac{ton {CO}_{2}}{year}$ per person in the United States (European, 2023)), the following equation may be used:

E = (4.16 \frac{GJ}{ton {CO}_{2}}) \cdot \dot{m}

(19)

Where:

\begin{array}{rcl} \dot{E} & : & energy flow required \\ \dot{m_{}} & : & mass flow of CO 2 . \end{array}

So, for the case of an average US resident:

$\dot{E}$	$=$	$(4.16 \frac{GJ}{ton {CO}_{2}}) \cdot \dot{m}$	(20)
	$=$	$(4.16 \frac{GJ}{ton {CO}_{2}}) \cdot (14.44 \frac{ton {CO}_{2}}{year})$	(21)
	$=$	$(4.16 \frac{GJ}{ton {CO}_{2}}) \cdot (14.44 \frac{ton {CO}_{2}}{year}) \cdot (\frac{year}{365.25 \times 24 \times 3600 s}) \cdot (\frac{10^{9} J}{GJ}) \cdot (\frac{W}{(\frac{J}{s})})$	(22)
$\dot{E}$	$=$	$\overline{190} 35 W$	(23)
$\dot{E}$	$=$	$19.0 kW$	(24)

For comparison, this figure is approximately equivalent to 10 typical hair dryers running continuously. The figure does not take into account inefficiencies introduced by waste heat from the $CaO (s)$ regeneration process or the energy cost of hauling the saturated ${CaCO}_{3} (s)$ solids to and from the high-temperature regeneration sites.

3Conclusion

Capturing CO₂ by the calcination reaction requires at least $4.16 \frac{GJ}{ton {CO}_{2}}$ .

4Works Cited

Coker, A. Kayode. (2001). “Modeling of Chemical Kinetics and Reactor Design”, “Appendix: Heats and Free Energies of Formation”. Elsevier. ISBN: 978-0-08-049190-5. OCLC: 476059966 .
European Commission and Joint Research Centre; Crippa, M; et. al.(2023). “GHG emissions of all world countries – 2023”. Publications Office of the European Union. https://doi.org/10.2760/953322 .
Plumer, Brad. (2023). “In a U.S. First, a Commercial Plant Starts Pulling Carbon From the Air”. New York Times. Accessed 2023-11-13. https://www.nytimes.com/2023/11/09/climate/direct-air-capture-carbon.html .

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