Consider the reaction $$\ce{N2O4 <=> 2 NO2}.$$ The forward direction of this reaction is non-spontaneous and under standard conditions $\Delta G^\circ = \pu{4.76 kJ/mol}$. Suppose we begin with $\pu{1.00 bar}$ of $\ce{N2O4}$ and no products. Then, we would expect the reaction to proceed to equilibrium in which some small concentration of products is formed. I am struggling to understand why it is not a contradiction that the reaction is non-spontaneous, but still proceeds in the forward direction (which indicates that the forward direction is spontaneous).
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Because the criterion of a negative value of standard Gibbs free energy change to signify a spontaneous reaction is only a rule of thumb. – Chet Miller Jun 01 '23 at 18:10
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@Johnny Smith, the spontaneity criterium is $\Delta G > 0$. Just recall that $ \Delta G = \Delta G^0 + \ln Q$$ – PAEP Jun 01 '23 at 18:11
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1@PAEP do you mean $\Delta G <0$? Also, doesn't assuming that the reactant begins with a partial pressure of 1 bar mean that $\Delta G = \Delta G^\circ$ or am I missing something. I appolgize in advance if this is a very stupid question. – Johnny Smith Jun 01 '23 at 18:18
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1"standard conditions" mean that there is 1 bar of products as well as reactants. $\Delta G^\circ >0$ simply means that the net reaction is from products to reactants when all start at 1 bar. – Andrew Jun 01 '23 at 18:21
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Yes, I meant $\Delta G^0 <0$. As @Andrew pointed, your initial conditions do not correspond to standard conditions. – PAEP Jun 01 '23 at 18:27
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2Chem+Math Expression formatting reference: MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. – Poutnik Jun 02 '23 at 05:14
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@PAEP You likely meant $\Delta G^\circ < 0$ because $\Delta G^0 \equiv 1.$ – andselisk Jun 02 '23 at 05:52
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See this answer also https://chemistry.stackexchange.com/questions/173788/can-free-energy-of-a-spontaneous-reaction-be-positive/173805#173805 – porphyrin Jun 02 '23 at 07:46
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A reaction will proceed to achieve an equilibrium state, this only happens when some amount of both reactants and products are present.
Consider a general reaction:
$$ \sum_ir_i\ce{R}_i \ce{->} \sum_jp_j\ce{P}_j $$
The equilibrium constant is given by:
$$ K = \dfrac{\prod_{j}[\ce{P}_j]_\text{eq}^{p_j}}{\prod_i[{\ce{R}_{i}}]_\text{eq}^{r_i}} $$
If no product is formed then:
$$ \prod_{j}[\ce{P}_j]_\text{eq}^{p_j} = 0 \implies K = 0 $$
Calculating the free energy for such a reaction:
$$ \Delta G^\circ = -\mathrm{R}T\ln{K} \implies \lim_{K \rightarrow 0} \Delta G^\circ = \infty $$
Thus, for a process in which no products are formed, the standard free energy is infinity, which isn't logical, and some products are always formed, regardless of whether the reaction is spontaneous, $\Delta G^\circ <0$, or non-spontaneous, $\Delta G^\circ >0$.
ananta
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Is this the thinking that caused the alchemists to fail in turning Pb into Au? There never was any product! Now the reason is clear! DeltaG = +infinity – jimchmst Jun 06 '23 at 08:29