Rate law expression with solids and liquids

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SUMMARY

The discussion focuses on the incorporation of solids and liquids in rate law expressions, particularly in the context of chemical kinetics and equilibrium. It is established that solids and liquids are not included in the equilibrium constant (K) because their concentrations are considered constant. However, for kinetics, the reaction rate is influenced by the surface area of solids, complicating their inclusion in rate law expressions. The conversation highlights the need for further exploration of these concepts, especially in relation to reaction mechanisms and surface chemistry.

PREREQUISITES
  • Understanding of chemical kinetics and equilibrium concepts
  • Familiarity with rate law expressions and their components
  • Knowledge of surface area effects in heterogeneous reactions
  • Basic principles of surface chemistry and heterogeneous catalysis
NEXT STEPS
  • Research "Surface Area Effects in Heterogeneous Reactions" for insights on solid reactants
  • Study "Kinetics of Solid-State Reactions" to understand the role of solids in rate laws
  • Explore "Principles of Surface Chemistry" for foundational knowledge on solid-liquid interactions
  • Examine "Reaction Mechanisms in Kinetics" to grasp how mechanisms influence rate laws
USEFUL FOR

Chemistry students, educators, and researchers interested in the complexities of reaction kinetics and equilibrium involving solid and liquid reactants.

yolo123
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Hello Forum!

I was wondering:

In the rate law expressions, I could never find an example with a solid or a liquid in the reactants in my textbook (Chemistry by Zumdahl). I searched Chemistry (Raymond Chang) and Principles of Molecular Chemistry, without any success.

What happens to solids and liquids who act as reactants in reactions? How do I incorporate them into the rate law? Are they ALWAYS already included in the k constant?
 
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Solids and liquids are not incorporated into the K constant because their concentrations are assumed to be constant (which is a pretty good approximation; as you consume either a solid or a liquid, you decrease the total amount but do not generally affect their concentrations).
 
k (kinetics) or K (equilibrium)?

For equilibrium presence of the solid is enough - no matter how much of the solid there is, the result (in terms of the final equilibrium position) is always the same.

For kinetics it is much more complicated, as reaction rate depends on the surface area of the solid. But I have never seen it addressed in the introductory courses.
 
Oops, true, I should have specified that I meant equilibrium Ks only. As Borek says, Ks for kinetics differ.
 
Not to mention, the rate law also depends on the reaction mechanism. So, it is impossible to guess on your own unless provided with some incentive.
 
I meant k as in the rate law: for example, rate=k[NO2]^2.
 
"For kinetics it is much more complicated, as reaction rate depends on the surface area of the solid. But I have never seen it addressed in the introductory courses."
Hi, it's three years too late but I'm hoping someone could elaborate on the complicated concept here. How do you include the order of the reaction of solid in the reaction rate expression? This is the only post on the internet where someone has hinted that there is an answer to this question. If you can't answer, it will still be helpful if you can point me to some resources as I have no idea what the answer to this is and I am tasked to find one. Any help will be appreciated, thank you.

(This is for both interest and homework.)​
 
In general, the reaction rate will depend on the surface area of the solid exposed to reactants. How the surface area changes over time will depend on the exact geometry of the material. Likely you would be able to find more information by searching for textbooks on surface chemistry or heterogeneous catalysis.
 
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