Tjyylli

Minimizing with differential evolution

Vocabulary for giving just numbers, not a full answer

Is divide-by-zero a security vulnerability?

After `ssh` without `-X` to a machine, is it possible to change `$DISPLAY` to make it work like `ssh -X`?

Haman going to the second feast dirty

How to draw dashed arc of a circle behind pyramid?

Why does cron require MTA for logging?

How does Ehrenfest's theorem apply to the quantum harmonic oscillator?

Do items de-spawn?

Having the player face themselves after the mid-game

Giving a career talk in my old university, how prominently should I tell students my salary?

What ability score modifier does a javelin's damage use?

Can one live in the U.S. and not use a credit card?

Specifying a starting column with colortbl package and xcolor

What do you call someone who likes to pick fights?

How many characters using PHB rules does it take to be able to have access to any PHB spell at the start of an adventuring day?

What will happen if my luggage gets delayed?

Doesn't allowing a user mode program to access kernel space memory and execute the IN and OUT instructions defeat the purpose of having CPU modes?

I can't die. Who am I?

Does "Until when" sound natural for native speakers?

What's the 'present simple' form of the word "нашла́" in 3rd person singular female?

Crossing a border with an infant of a different citizenship

Why aren't there more Gauls like Obelix?

Am I understanding this Storm King's Thunder map wrong?

Is there a difference between equilibrium and steady state?

Diffusion across a thin filmKinetics of interstellar chemistry: Applying steady state to formation of H₂Rate-determining step and steady state approximation failureHow can a protein folding transition state have zero lifetime?How is the “quasi-steady-state” assumption of Michaelis-Menten kinetics different from the steady state assumption of ordinary kinetics?Quasi-steady state hypothesis applicationMeaning of steady state (kinetics)Has IUPAC been inaccurate in their 1994 definition of kinetic stationary state?Validity of the equation K_eq = ratio of rate constants of forward and reverse reactionsApplication of Michaelis-Menten rate law for biological systemsDifference between Henry constant and equilibrium constant

1

1

$begingroup$

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

share|improve this question

asked 1 hour ago

Karsten TheisKarsten Theis

2,379327

$endgroup$

add a comment | 

1

1

$begingroup$

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

share|improve this question

asked 1 hour ago

Karsten TheisKarsten Theis

2,379327

$endgroup$

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

share|improve this question

asked 1 hour ago

Karsten TheisKarsten Theis

2,379327

$endgroup$

share|improve this question

asked 1 hour ago

Karsten TheisKarsten Theis

2,379327

share|improve this question

asked 1 hour ago

Karsten TheisKarsten Theis

2,379327

asked 1 hour ago

Karsten TheisKarsten Theis

2,379327

asked 1 hour ago

Karsten TheisKarsten Theis

2,379327

1 Answer
1

active

oldest

votes

3

$begingroup$

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.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Karsten TheisKarsten Theis

2,379327

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "431"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f110794%2fis-there-a-difference-between-equilibrium-and-steady-state%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

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.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Karsten TheisKarsten Theis

2,379327

$endgroup$

add a comment | 

3

$begingroup$

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.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Karsten TheisKarsten Theis

2,379327

$endgroup$

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.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Karsten TheisKarsten Theis

2,379327

$endgroup$

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Karsten TheisKarsten Theis

2,379327

answered 1 hour ago

Karsten TheisKarsten Theis

2,379327

answered 1 hour ago

Karsten TheisKarsten Theis

2,379327