Is there a difference between equilibrium and steady state?Diffusion across a thin filmKinetics of...

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

Outlet with 3 sets of wires

How can I manipulate the output of Information?

Do items de-spawn?

Do I really need to have a scientific explanation for my premise?

What would be the most expensive material to an intergalactic society?

Trig Subsitution When There's No Square Root

Rationale to prefer local variables over instance variables?

how to modify custom status text color in UI component grid magento 2?

Does an unused member variable take up memory?

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

Which classes are needed to have access to every spell in the PHB?

Why couldn't the separatists legally leave the Republic?

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?

Specifying a starting column with colortbl package and xcolor

Drawing close together horizontal lines in Latex

Whose blood did Carol Danver's receive, Mar-vell's or Yon-Rogg's in the movie?

Crossing a border with an infant of a different citizenship

What do *foreign films* mean for an American?

Windows Server Datacenter Edition - Unlimited Virtual Machines

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

Was it really inappropriate to write a pull request for the company I interviewed with?

Why is a very small peak with larger m/z not considered to be the molecular ion?

Street obstacles in New Zealand



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












$begingroup$


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?










share|improve this question









$endgroup$

















    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?










    share|improve this question









    $endgroup$















      1












      1








      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?










      share|improve this question









      $endgroup$




      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






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 1 hour ago









      Karsten TheisKarsten Theis

      2,379328




      2,379328






















          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











          $endgroup$













            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


















            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















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            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











            $endgroup$


















              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











              $endgroup$
















                3












                3








                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











                $endgroup$



                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.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 1 hour ago

























                answered 1 hour ago









                Karsten TheisKarsten Theis

                2,379328




                2,379328






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Chemistry Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    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















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Why do type traits not work with types in namespace scope?What are POD types in C++?Why can templates only be...

                    Will tsunami waves travel forever if there was no land?Why do tsunami waves begin with the water flowing away...

                    Should I use Docker or LXD?How to cache (more) data on SSD/RAM to avoid spin up?Unable to get Windows File...