Is there a difference between equilibrium and steady state?
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The term equilibrium is used in the context of reversible reactions that reach a point where concentrations no longer change. The term steady-state is used in enzyme kinetics when the concentration of the enzyme-substrate complex no longer changes (or hardly changes, in case of a quasi steady state). It is also used to describe multi-step biochemical pathways. Is there a difference between the two, given that both concern a situation where concentrations don't change over time?
equilibrium kinetics
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The term equilibrium is used in the context of reversible reactions that reach a point where concentrations no longer change. The term steady-state is used in enzyme kinetics when the concentration of the enzyme-substrate complex no longer changes (or hardly changes, in case of a quasi steady state). It is also used to describe multi-step biochemical pathways. Is there a difference between the two, given that both concern a situation where concentrations don't change over time?
equilibrium kinetics
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add a comment |
$begingroup$
The term equilibrium is used in the context of reversible reactions that reach a point where concentrations no longer change. The term steady-state is used in enzyme kinetics when the concentration of the enzyme-substrate complex no longer changes (or hardly changes, in case of a quasi steady state). It is also used to describe multi-step biochemical pathways. Is there a difference between the two, given that both concern a situation where concentrations don't change over time?
equilibrium kinetics
$endgroup$
The term equilibrium is used in the context of reversible reactions that reach a point where concentrations no longer change. The term steady-state is used in enzyme kinetics when the concentration of the enzyme-substrate complex no longer changes (or hardly changes, in case of a quasi steady state). It is also used to describe multi-step biochemical pathways. Is there a difference between the two, given that both concern a situation where concentrations don't change over time?
equilibrium kinetics
equilibrium kinetics
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Karsten TheisKarsten Theis
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Yes, equilibrium and steady-state are distinct concepts.
A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant.
Steady-state implies a system that is not at equilibrium (entropy increases). A species is said to be at steady state when the rate of reactions (or more general, processes) that form the species is equal to the rate of reactions (or processes) that remove the species.
In both cases, there are rates ($mathrm{rate}_1$ and $mathrm{rate}_2$) that are equal. For an equilibrium, the forward and reverse rate of the same reaction are equal to each other. For a steady state, the rates of processes leading to increase of the concentration of a species are equal to the rates of processes leading to decrease of the concentration of the same species.
$$ce{A <=>[rate_1][rate_2] B} vs ce{source->[rate_1]C->[rate_2]sink} $$
For an equilibrium, all concentrations are constant over time. For a steady-state, there is a net reaction, so some amounts change (the amount of source and sink), while at least one species - the one at steady state - has a constant concentration as long as the conditions of steady state prevail.
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1 Answer
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$begingroup$
Yes, equilibrium and steady-state are distinct concepts.
A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant.
Steady-state implies a system that is not at equilibrium (entropy increases). A species is said to be at steady state when the rate of reactions (or more general, processes) that form the species is equal to the rate of reactions (or processes) that remove the species.
In both cases, there are rates ($mathrm{rate}_1$ and $mathrm{rate}_2$) that are equal. For an equilibrium, the forward and reverse rate of the same reaction are equal to each other. For a steady state, the rates of processes leading to increase of the concentration of a species are equal to the rates of processes leading to decrease of the concentration of the same species.
$$ce{A <=>[rate_1][rate_2] B} vs ce{source->[rate_1]C->[rate_2]sink} $$
For an equilibrium, all concentrations are constant over time. For a steady-state, there is a net reaction, so some amounts change (the amount of source and sink), while at least one species - the one at steady state - has a constant concentration as long as the conditions of steady state prevail.
$endgroup$
add a comment |
$begingroup$
Yes, equilibrium and steady-state are distinct concepts.
A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant.
Steady-state implies a system that is not at equilibrium (entropy increases). A species is said to be at steady state when the rate of reactions (or more general, processes) that form the species is equal to the rate of reactions (or processes) that remove the species.
In both cases, there are rates ($mathrm{rate}_1$ and $mathrm{rate}_2$) that are equal. For an equilibrium, the forward and reverse rate of the same reaction are equal to each other. For a steady state, the rates of processes leading to increase of the concentration of a species are equal to the rates of processes leading to decrease of the concentration of the same species.
$$ce{A <=>[rate_1][rate_2] B} vs ce{source->[rate_1]C->[rate_2]sink} $$
For an equilibrium, all concentrations are constant over time. For a steady-state, there is a net reaction, so some amounts change (the amount of source and sink), while at least one species - the one at steady state - has a constant concentration as long as the conditions of steady state prevail.
$endgroup$
add a comment |
$begingroup$
Yes, equilibrium and steady-state are distinct concepts.
A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant.
Steady-state implies a system that is not at equilibrium (entropy increases). A species is said to be at steady state when the rate of reactions (or more general, processes) that form the species is equal to the rate of reactions (or processes) that remove the species.
In both cases, there are rates ($mathrm{rate}_1$ and $mathrm{rate}_2$) that are equal. For an equilibrium, the forward and reverse rate of the same reaction are equal to each other. For a steady state, the rates of processes leading to increase of the concentration of a species are equal to the rates of processes leading to decrease of the concentration of the same species.
$$ce{A <=>[rate_1][rate_2] B} vs ce{source->[rate_1]C->[rate_2]sink} $$
For an equilibrium, all concentrations are constant over time. For a steady-state, there is a net reaction, so some amounts change (the amount of source and sink), while at least one species - the one at steady state - has a constant concentration as long as the conditions of steady state prevail.
$endgroup$
Yes, equilibrium and steady-state are distinct concepts.
A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant.
Steady-state implies a system that is not at equilibrium (entropy increases). A species is said to be at steady state when the rate of reactions (or more general, processes) that form the species is equal to the rate of reactions (or processes) that remove the species.
In both cases, there are rates ($mathrm{rate}_1$ and $mathrm{rate}_2$) that are equal. For an equilibrium, the forward and reverse rate of the same reaction are equal to each other. For a steady state, the rates of processes leading to increase of the concentration of a species are equal to the rates of processes leading to decrease of the concentration of the same species.
$$ce{A <=>[rate_1][rate_2] B} vs ce{source->[rate_1]C->[rate_2]sink} $$
For an equilibrium, all concentrations are constant over time. For a steady-state, there is a net reaction, so some amounts change (the amount of source and sink), while at least one species - the one at steady state - has a constant concentration as long as the conditions of steady state prevail.
edited 1 hour ago
answered 2 hours ago
Karsten TheisKarsten Theis
2,394328
2,394328
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