Carbochlorination Reactor Mass Balance

Hello! Welcome back to our study of the Kroll process.

In the last lesson, we took a qualitative look at the first major step in titanium production: the conversion of titanium dioxide (TiO₂) ore into crude titanium tetrachloride (TiCl₄) via carbochlorination, and the subsequent purification steps. We learned what happens in the chlorinator and why purification is so critical.

Today, we move from the "what" to the "how much." We will apply the mass balance principles you encountered in Module 2 to the carbochlorination reactor. Our goal is to quantify the process, enabling you to calculate the precise amount of raw materials—ore, carbon, and chlorine—needed to produce a target quantity of TiCl₄. This is the foundational calculation for building a techno-economic model of a Kroll plant, as it directly determines the primary material costs.

1. The Principle of Mass Balance

Before we dive into the numbers, let's briefly recall the core principle. The law of conservation of mass states that mass cannot be created or destroyed. For a chemical process running at a steady state, this means:

Total Mass In = Total Mass Out

When chemical reactions occur, we track individual chemical species (atoms or molecules). For any given element (like Titanium, Carbon, Oxygen, or Chlorine), the total mass of that element entering the reactor must equal the total mass of that element leaving it. We use chemical reaction stoichiometry—the "recipe" for the reaction—to track how elements are rearranged from input compounds (reactants) to output compounds (products).

Your background in econometrics and modeling will be useful here. Think of a mass balance as a deterministic model of a physical system governed by strict conservation laws, much like balancing financial accounts or modeling flows in an economic system.

2. The Chemistry of Carbochlorination

To build our model, we need the specific chemical reactions that occur in the chlorinator. The process is designed to convert TiO₂, but the impurities in the ore also react, consuming additional reagents and creating waste products.

The following paper provides an excellent summary of the key reactions.

From the reading, we can extract the most important reactions. There is a subtle point regarding the carbon oxides produced. In our previous lesson, we used the reaction producing carbon monoxide (CO). The paper shows two possibilities:

1. Reaction producing CO₂:
   ${\text{TiO}}_2(s)+\text{C}(s)+2{\text{Cl}}_2(g)\to {\text{TiCl}}_4(g)+{\text{CO}}_2(g)$
2. Boudouard Reaction:
   ${\text{CO}}_2(g)+\text{C}(s)\rightleftharpoons 2\text{CO}(g)$

At the high temperatures of the chlorinator (~1000°C), the Boudouard reaction equilibrium favors the formation of CO. For practical purposes, these two reactions can be combined into a single overall reaction, which we will use for our calculations, consistent with the last lesson:

${\text{TiO}}_2(s)+2\text{C}(s)+2{\text{Cl}}_2(g)\to {\text{TiCl}}_4(g)+2\text{CO}(g)$

Impurities in the ore, such as iron oxide (Fe₂O₃) or magnesium oxide (MgO), also undergo carbochlorination. The general reaction from the paper (Eq. [3]) is:

$2{\text{MO}}_x(s)+x\text{C}(s)+2x{\text{Cl}}_2(g)\to 2{\text{MCl}}_{2x}(l/g)+x{\text{CO}}_2(g)$

For an impurity like iron(III) oxide (Fe₂O₃), where M=Fe and x=1.5, this becomes:

${\text{Fe}}_2{\text{O}}_3(s)+3\text{C}(s)+3{\text{Cl}}_2(g)\to 2{\text{FeCl}}_3(g)+3\text{CO}(g)$
(Note: We've again assumed the final carbon product is CO for consistency at high temperature).

These impurity reactions are critical because they consume extra carbon and chlorine, which is a direct cost to the operation.

3. Worked Example: Mass Balance on a Chlorinator

Let's put this into practice by calculating the feedstock requirements for a hypothetical plant.

Problem Statement:
Calculate the required feed rates (in kg/hr) of rutile ore, coke, and chlorine gas to produce 1,000 kg/hr of crude TiCl₄.

Process Assumptions:

• Basis of Calculation: 1,000 kg/hr of TiCl₄ produced.
• Feedstock (Rutile Ore): 95% TiO₂, 1.5% Fe₂O₃, 3.5% inert minerals (e.g., SiO₂).
• Reductant (Coke): 90% Carbon (C), 10% inert ash.
• Conversion Efficiency: 98% of the TiO₂ fed to the reactor is converted into TiCl₄.
• Excess Reactant: Chlorine gas is fed at 10% in excess of the total theoretical requirement to ensure complete reaction of the valuable ore.

Molar Masses (in kg/kmol):

• TiCl₄: 189.7
• TiO₂: 79.9
• Fe₂O₃: 159.7
• Cl₂: 70.9
• C: 12.0
• FeCl₃: 162.2

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Step 1: Calculate Moles of TiCl₄ Product

This is our starting point, or "basis."

$${\text{Moles of TiCl}}_4=\frac{1000\text{ kg/hr}}{189.7\text{ kg/kmol}}=5.27\text{ kmol/hr}$$

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Step 2: Calculate Reactants for TiCl₄ Production

Using the main reaction stoichiometry TiO₂ + 2C + 2Cl₂ → TiCl₄ + 2CO, we find the moles of TiO₂ that must react:

$${\text{Moles of TiO}}_2\text{ reacted}=5.27{\text{ kmol/hr TiCl}}_4\times \frac{1{\text{ kmol TiO}}_2}{1{\text{ kmol TiCl}}_4}=5.27\text{ kmol/hr}$$

Now, we account for the 98% conversion efficiency. To get 5.27 kmol/hr to react, we must feed more into the reactor.

$${\text{Total TiO}}_2\text{ fed}=\frac{5.27\text{ kmol/hr}}{0.98}=5.38\text{ kmol/hr}$$

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Step 3: Calculate Ore and Impurity Feed Rates

The total mass of TiO₂ we need is:

$${\text{Mass of TiO}}_2\text{ fed}=5.38\text{ kmol/hr}\times 79.9\text{ kg/kmol}=430.0\text{ kg/hr}$$

Since the ore is only 95% TiO₂, the total ore feed rate is:

$$\text{Total Ore Feed Rate}=\frac{430.0\text{ kg/hr}}{0.95}=452.6\text{ kg/hr}$$

This ore feed also contains Fe₂O₃. Let's find out how much:

$${\text{Mass of Fe}}_2{\text{O}}_3\text{ fed}=452.6\text{ kg/hr}\times 0.015=6.79\text{ kg/hr}$$ $${\text{Moles of Fe}}_2{\text{O}}_3\text{ fed}=\frac{6.79\text{ kg/hr}}{159.7\text{ kg/kmol}}=0.0425\text{ kmol/hr}$$

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Step 4: Calculate Total Reagent Consumption

Now we calculate the total carbon and chlorine needed for both the main reaction and the impurity reaction.

Carbon (C):

• For TiO₂: $$5.27{\text{ kmol/hr TiCl}}_4\times \frac{2\text{ kmol C}}{1{\text{ kmol TiCl}}_4}=10.54\text{ kmol/hr}$$
• For Fe₂O₃ (using Fe₂O₃ + 3C + 3Cl₂ → 2FeCl₃ + 3CO): $$0.0425{\text{ kmol/hr Fe}}_2{\text{O}}_3\times \frac{3\text{ kmol C}}{1{\text{ kmol Fe}}_2{\text{O}}_3}=0.13\text{ kmol/hr}$$
• Total Carbon Required: $$10.54+0.13=10.67\text{ kmol/hr}$$

Chlorine (Cl₂):

• For TiO₂: $$5.27{\text{ kmol/hr TiCl}}_4\times \frac{2{\text{ kmol Cl}}_2}{1{\text{ kmol TiCl}}_4}=10.54\text{ kmol/hr}$$
• For Fe₂O₃: $$0.0425{\text{ kmol/hr Fe}}_2{\text{O}}_3\times \frac{3{\text{ kmol Cl}}_2}{1{\text{ kmol Fe}}_2{\text{O}}_3}=0.13\text{ kmol/hr}$$
• Total Theoretical Chlorine Required: $$10.54+0.13=10.67\text{ kmol/hr}$$

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Step 5: Calculate Final Feedstock Rates

Finally, we convert the molar requirements back to mass, accounting for coke purity and the excess chlorine.

Total Coke Feed Rate:
The coke provides the carbon. Since coke is 90% carbon:

$$\text{Mass of Carbon}=10.67\text{ kmol/hr}\times 12.0\text{ kg/kmol}=128.0\text{ kg/hr}$$ $$\text{Total Coke Feed Rate}=\frac{128.0\text{ kg/hr}}{0.90}=142.2\text{ kg/hr}$$

Total Chlorine Feed Rate:
We need 10% excess chlorine over the theoretical requirement:

$$\text{Total Chlorine Fed (moles)}=10.67\text{ kmol/hr}\times 1.10=11.74\text{ kmol/hr}$$ $$\text{Total Chlorine Feed Rate}=11.74\text{ kmol/hr}\times 70.9\text{ kg/kmol}=832.4\text{ kg/hr}$$

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Summary of Results:
To produce 1,000 kg/hr of TiCl₄, our plant requires:

• Rutile Ore: 452.6 kg/hr
• Coke: 142.2 kg/hr
• Chlorine Gas: 832.4 kg/hr

Notice that for every kilogram of product, we need about 1.4 kg of raw materials (excluding inerts that pass through). This ratio is a key performance indicator for the process.

4. Feedstock Quality and Economic Value

This calculation clearly demonstrates why feedstock quality is so important. The 1.5% Fe₂O₃ impurity, though small, created an additional demand for expensive coke and chlorine. Now, imagine using a lower-grade ore with 10% or 20% impurities. The reagent consumption would increase significantly, as would the amount of waste chloride (FeCl₃) that needs to be handled and disposed of.

This is the central idea behind a "value-in-use" model, which evaluates a feedstock not just on its purchase price, but on the total operational cost it incurs.

As you can see from the paper, switching to a higher-purity feedstock reduces the consumption of both coke and chlorine and lowers waste disposal costs. This is the direct economic consequence of the mass balance we just calculated. An operator or investor would weigh the higher price of a premium ore against these savings in reagents and waste management to determine the most profitable strategy.

Conclusion

Today, we've translated the qualitative chemistry of carbochlorination into a quantitative engineering calculation. You now have the tools to determine the precise feedstock requirements for the first stage of the Kroll process.

Key Takeaways:

• Mass balance calculations are based on the conservation of mass and reaction stoichiometry.
• The process involves defining a basis, calculating reactant consumption for both the main product and side-reactions from impurities, and accounting for real-world factors like conversion efficiency and excess reagents.
• Feedstock impurities directly increase the consumption of expensive reagents (coke, chlorine) and the generation of waste, impacting overall profitability.
• These calculations form the bedrock of any serious techno-economic analysis of a chemical process.

In our next lesson, we will move to the heart of the Kroll process. We'll take the pure TiCl₄ we have conceptually produced and examine the thermodynamics of reducing it with magnesium to create the final titanium metal sponge.