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 5 added 8 characters in body edited Jul 14 '18 at 8:38 Loong♦ 36.8k99 gold badges9292 silver badges196196 bronze badges Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction: $$\ce{A + B <=> C + D}$$ We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$ Using Nernst we can now establish a relation between $$K$$ and the left-to-right change in Gibbs Free Energy, $$\Delta G^0$$: $$\Delta G^0=-RT\ln K$$ (Of course, as you know $$\Delta G^0=\Delta H-T\Delta S$$) Evaluate this Nernst function to conclude that the more negative $$\Delta G^0$$ is, the higher the value of $$K$$: $$K=e^\frac{-\Delta G^0}{RT}$$$$K=\mathrm e^\frac{-\Delta G^0}{RT}$$ For very negative values of $$\Delta G^0$$, $$K\to+\infty$$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $$0$$. Then we can write: $$\ce{A + B -> C + D}$$ Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $$K$$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'. Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction: $$\ce{A + B <=> C + D}$$ We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$ Using Nernst we can now establish a relation between $$K$$ and the left-to-right change in Gibbs Free Energy, $$\Delta G^0$$: $$\Delta G^0=-RT\ln K$$ (Of course, as you know $$\Delta G^0=\Delta H-T\Delta S$$) Evaluate this Nernst function to conclude that the more negative $$\Delta G^0$$ is, the higher the value of $$K$$: $$K=e^\frac{-\Delta G^0}{RT}$$ For very negative values of $$\Delta G^0$$, $$K\to+\infty$$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $$0$$. Then we can write: $$\ce{A + B -> C + D}$$ Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $$K$$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'. Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction: $$\ce{A + B <=> C + D}$$ We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$ Using Nernst we can now establish a relation between $$K$$ and the left-to-right change in Gibbs Free Energy, $$\Delta G^0$$: $$\Delta G^0=-RT\ln K$$ (Of course, as you know $$\Delta G^0=\Delta H-T\Delta S$$) Evaluate this Nernst function to conclude that the more negative $$\Delta G^0$$ is, the higher the value of $$K$$: $$K=\mathrm e^\frac{-\Delta G^0}{RT}$$ For very negative values of $$\Delta G^0$$, $$K\to+\infty$$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $$0$$. Then we can write: $$\ce{A + B -> C + D}$$ Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $$K$$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'. 4 mathjax edited Mar 10 '18 at 15:10 Gaurang Tandon 5,43288 gold badges3030 silver badges7070 bronze badges Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction: $$A+B\ce{<=>}C+D$$$$\ce{A + B <=> C + D}$$ We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$ Using Nernst we can now establish a relation between $$K$$ and the left-to-right change in Gibbs Free Energy, $$\Delta G^0$$: $$\Delta G^0=-RT\ln K$$ (Of course, as you know $$\Delta G^0=\Delta H-T\Delta S$$.) Evaluate this Nernst function to conclude that the more negative $$\Delta G^0$$ is, the higher the value of $$K$$: $$K=e^\frac{-\Delta G^0}{RT}$$ For very negative values of $$\Delta G^0$$, $$K\to+\infty$$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $$0$$. Then we can write: $$A+B \rightarrow C+D$$$$\ce{A + B -> C + D}$$ Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $$K$$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'. Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction: $$A+B\ce{<=>}C+D$$ We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$ Using Nernst we can now establish a relation between $$K$$ and the left-to-right change in Gibbs Free Energy, $$\Delta G^0$$: $$\Delta G^0=-RT\ln K$$ (Of course, as you know $$\Delta G^0=\Delta H-T\Delta S$$.) Evaluate this Nernst function to conclude that the more negative $$\Delta G^0$$ is, the higher the value of $$K$$: $$K=e^\frac{-\Delta G^0}{RT}$$ For very negative values of $$\Delta G^0$$, $$K\to+\infty$$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $$0$$. Then we can write: $$A+B \rightarrow C+D$$ Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $$K$$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'. Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction: $$\ce{A + B <=> C + D}$$ We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$ Using Nernst we can now establish a relation between $$K$$ and the left-to-right change in Gibbs Free Energy, $$\Delta G^0$$: $$\Delta G^0=-RT\ln K$$ (Of course, as you know $$\Delta G^0=\Delta H-T\Delta S$$) Evaluate this Nernst function to conclude that the more negative $$\Delta G^0$$ is, the higher the value of $$K$$: $$K=e^\frac{-\Delta G^0}{RT}$$ For very negative values of $$\Delta G^0$$, $$K\to+\infty$$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $$0$$. Then we can write: $$\ce{A + B -> C + D}$$ Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $$K$$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'. 3 added 12 characters in body edited Nov 19 '17 at 15:00 Gert 1,65122 silver badges1111 bronze badges Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction: $$A+B\ce{<=>}C+D$$ We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$ Using Nernst we can now establish a relation between $$K$$ and the left-to-right change in Gibbs Free Energy, $$\Delta G$$$$\Delta G^0$$: $$\Delta G=-RT\ln K$$$$\Delta G^0=-RT\ln K$$ (Of course, as you know $$\Delta G=\Delta H-T\Delta S$$$$\Delta G^0=\Delta H-T\Delta S$$.) Evaluate this Nernst function to conclude that the more negative $$\Delta G$$$$\Delta G^0$$ is, the higher the value of $$K$$: $$K=e^\frac{-\Delta G}{RT}$$$$K=e^\frac{-\Delta G^0}{RT}$$ For very negative values of $$\Delta G$$$$\Delta G^0$$, $$K\to+\infty$$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $$0$$. Then we can write: $$A+B \rightarrow C+D$$ Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $$K$$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'. Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction: $$A+B\ce{<=>}C+D$$ We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$ Using Nernst we can now establish a relation between $$K$$ and the left-to-right change in Gibbs Free Energy, $$\Delta G$$: $$\Delta G=-RT\ln K$$ (Of course, as you know $$\Delta G=\Delta H-T\Delta S$$.) Evaluate this Nernst function to conclude that the more negative $$\Delta G$$ is, the higher the value of $$K$$: $$K=e^\frac{-\Delta G}{RT}$$ For very negative values of $$\Delta G$$, $$K\to+\infty$$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $$0$$. Then we can write: $$A+B \rightarrow C+D$$ Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $$K$$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'. Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction: $$A+B\ce{<=>}C+D$$ We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$ Using Nernst we can now establish a relation between $$K$$ and the left-to-right change in Gibbs Free Energy, $$\Delta G^0$$: $$\Delta G^0=-RT\ln K$$ (Of course, as you know $$\Delta G^0=\Delta H-T\Delta S$$.) Evaluate this Nernst function to conclude that the more negative $$\Delta G^0$$ is, the higher the value of $$K$$: $$K=e^\frac{-\Delta G^0}{RT}$$ For very negative values of $$\Delta G^0$$, $$K\to+\infty$$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $$0$$. Then we can write: $$A+B \rightarrow C+D$$ Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $$K$$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'. 2 added 47 characters in body edited Nov 19 '17 at 14:51 Gert 1,65122 silver badges1111 bronze badges 1 answered Nov 19 '17 at 14:44 Gert 1,65122 silver badges1111 bronze badges