How to determine whether a single reaction is better than multi-stage reaction?

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SUMMARY

The discussion focuses on evaluating the efficiency of a single endothermic reaction with an enthalpy change of +13.23 kJ compared to a multi-stage reaction process. Key considerations include energy dissipation in multi-stage reactions, the necessity of maintaining temperature at 500K, and the implications of reaction balance. The conversation also touches on industrial factors such as cost, equilibrium positions, and the use of catalysts, emphasizing the importance of understanding energy dynamics in chemical processes.

PREREQUISITES
  • Understanding of enthalpy changes in chemical reactions
  • Familiarity with multi-stage reaction processes
  • Knowledge of industrial chemical processes and their economic considerations
  • Basic principles of thermodynamics and reaction kinetics
NEXT STEPS
  • Research the thermodynamics of multi-stage reactions in industrial applications
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  • Explore the role of catalysts in optimizing reaction efficiency
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Chemical engineers, industrial chemists, and students of physical chemistry seeking to understand the comparative efficiency of single versus multi-stage reactions in industrial settings.

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Homework Statement
Consider the following reactions

$$\mathrm{2C(s)+2H_2)(g)=2CO(g)+2H_2(g)}$$
$$\mathrm{CO(g)+H_2)(g)=CO_2(g)+H_2(g)}$$
$$\mathrm{CO(g)+3H_2(g)=CH_4(g)+H_2)(g)}$$

and their sum

$$\mathrm{2C(s)+2H_2)(g)=CH_4(g)+CO_2(g)}$$
Relevant Equations
From the standpoint of heat balance, would it be better to develop a process to carry out the overall reactions in three separate reactors or in a single reactor?
The above question is the final part of a problem in Silbey, Alberty, and Bawendi's Physical Chemistry. The main part of the problem had to do with various calculations of changes in enthalpy at 500K, which I completed successfully. The single reaction is endothermic with reaction enthalpy +13.23 kJ.

I'm not sure how to think about the question of if a single reaction is better than the three separate reactions.

What I have thought about so far is the following.

Each separate stage will have unrecoverable energy dissipation.

The multi-stage reaction only has +13.23 kJ of enthalpy change if each reaction occurs and stays at 500K. Lost heat will decrease temperature and so extra energy will be needed to maintain temperature.
 
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Are the equations supposed to look like that?
They don't look balanced and I can't tell if that's intentional.

Is this methane synthesis/the Sabatier reaction?
 
I think all the H2 on the left should be H2O...
Which then means I can relate it to an industrial process and the target product is not methane. IMHO.

To attempt to answer the question asked.

In industry they always consider:
Cost, equilibrium positions, high pressures vs atmospheric, expensive catalysts, by-products (possibly sellable or not), high temps vs room temperatures, expensive separation procedures, corrosive reagents, toxic reagents..... even Ni(CO)4 is used in industry!!!

This applies to all commercial processes, and your task with the equations, I think, is to understand the energy used in the multi-step and consider whether a one vessel process exists, its yield, advantages and disadvantages. Even if a one vessel reaction was available, is it more or less energy intensive with a good yield.

But I am trying to interpret your question, and I often find that hard to do in your posts, due to how they are worded.
 
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