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Consider the reaction

$$\ce{A -> B}$$

The reaction Gibbs free energy, $\Delta_\mathrm{r} G$ is given by the following equation $$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT \ln Q$$

Now what is the difference between $\Delta G$ and $\Delta_\mathrm{r} G$ ?

Which one of the above represents $G_{\text{products}} - G_{\text{reactants}}$ ?

I read in my book $\Delta_\mathrm{r} G$ is the slope of $G$ plotted against the extent of reaction at different instants. And at equilibrium, it is zero. So can we say that it has value at a particular instant while $\Delta G$ is for a process ($\ce{A}$ to $\ce{B}$ in my question) i.e. is for a time interval?

Does $\Delta G$ for the above reaction represent change in Gibbs Energy when 1 mole of $\ce{A}$ reacts completely to form $1$ mole of $\ce{B}$ ?

orthocresol
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CLAP or SLAP II
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3 Answers3

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Not for the faint-hearted: There is an excellent, but very mathsy, article here: J. Chem. Educ. 2014, 91, 386 describing the difference.

The Gibbs free energy change, $\Delta G$

You are quite right in saying that $\Delta G$ represents a change in a system over a time interval. The notation $\Delta G$ itself implies that it is a difference in $G$ between two things: an initial state and a final state. Let's use a real example in order to make things clearer. Consider the thermal decomposition of ammonium nitrate:

$$\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$$

If someone were to ask you "what is $\Delta G$ for this reaction?", you should really, technically, be telling them: it is not well-defined. That is because of two things. Firstly, the reaction conditions, e.g. temperature and pressure, are not specified.

But more importantly, a balanced equation does not tell you exactly how much ammonium nitrate is reacting. In your beaker, you could have $1~\mathrm{mol}$ of $\ce{NH4NO3}$ reacting according to the above equation, but you could equally well have $2~\mathrm{mol}$, or you could have $5000~\mathrm{mol}$ (I must say that's one huge beaker though), or you could have $0.001~\mathrm{mmol}$.

The stoichiometric coefficients in the equation do not indicate the amounts of compounds reacting.

So, a more proper question would read like this:

$\pu{2 mol}$ of ammonium nitrate fully decompose according to the balanced equation $\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. Calculate $\Delta G$ for this process.

Great. Because the decomposition is complete, we now know that our final state is $2~\mathrm{mol}~\ce{N2O(g)} + 4~\mathrm{mol}~\ce{H2O(g)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$, and our initial state is $2~\mathrm{mol}~\ce{NH4NO3(s)}$ at $298~\mathrm{K}$ and $1~\mathrm{bar}$. So,

$$\begin{align} \Delta G &= [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot G_\mathrm{m}(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot G_\mathrm{m}(\ce{NH4NO3(s)})] \end{align}$$

We can't quite find the absolute molar Gibbs free energies, so the best we can do is to use Gibbs free energies of formation.

$$\begin{array}{c|c} \text{Compound} & \Delta_\mathrm{f}G^\circ / \mathrm{kJ~mol^{-1}}\text{ (at }298~\mathrm{K}\text{)} \\ \hline \ce{NH4NO3(s)} & -183.87 \\ \ce{N2O(g)} & +104.20 \\ \ce{H2O(g)} & -228.57 \end{array}$$ $$\scriptsize \text{(data from Atkins & de Paula, }\textit{Physical Chemistry}\text{ 10th ed., pp 975-7)}$$

(Note that the use of standard formation Gibbs free energies is only because of the conditions specified in the question, which conveniently corresponds to the standard state. If we specify different conditions, we can still find $\Delta G$, but we would have to use different data.) So:

$$\begin{align} \Delta G &= [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{N2O(g)})] + [(4~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{H2O(g)})] - [(2~\mathrm{mol})\cdot \Delta_\mathrm{f} G^\circ(\ce{NH4NO3(s)})] \\ &= [(2~\mathrm{mol})(+104.20~\mathrm{kJ~mol^{-1}})] + [(4~\mathrm{mol})(-228.57~\mathrm{kJ~mol^{-1}})] - [(2~\mathrm{mol})(-183.87~\mathrm{kJ~mol^{-1}})] \\ &= -1073.62~\mathrm{kJ} \end{align}$$

Note that we have units of kJ. Since $\Delta G$ is the difference between the Gibbs free energy of one state and another, $\Delta G$ has to have the same units as $G$, which is units of energy.

Now, this does not necessarily mean that $\Delta G = G_\text{products} - G_\text{reactants}$. For example, if I changed my question to be:

In the Haber process, 100 moles of $\ce{N2}$ and 300 moles of $\ce{H2}$ are reacted at $800~\mathrm{K}$ and $200~\mathrm{bar}$ according to the equation $\ce{N2 + 3H2 -> 2NH3}$. Only 10% of the starting materials are converted under these conditions. Calculate $\Delta G$ for the process. (These numbers are made up.)

then, your final state would not be the pure products. Your final state is not 200 moles of $\ce{NH3}$. Your final state is $90~\mathrm{mol}~\ce{N2}$, $270~\mathrm{mol}~\ce{H2}$, and $20~\mathrm{mol}~\ce{NH3}$.

In general, one could write, for a chemical reaction,

$$\Delta G = \sum_i (\Delta n_i) G_{\mathrm{m},i}$$

where $\Delta n_i$ is the change in the amount of compound $i$ (in moles), and $G_{\mathrm{m},i}$ is the molar Gibbs free energy of the pure compound $i$, under the $T$ and $p$ conditions specified. Going back to our ammonium nitrate example, we would have $\Delta n_{\ce{NH4NO3}} = -2~\mathrm{mol}$, $\Delta n_{\ce{H2O}} = +2~\mathrm{mol}$, and $\Delta n_{\ce{H2O}} = +4~\mathrm{mol}$.

The Gibbs free energy change of reaction, $\Delta_\mathrm{r} G$

As you have correctly stated, $\Delta_\mathrm{r}G$ is the slope of a graph of $G_\mathrm{syst}$ against the extent of reaction, commonly denoted $\xi$. This is the easiest way of interpreting $\Delta_\mathrm{r}G$. This question contains a slightly fuller derivation and explanation of what $\Delta_\mathrm{r}G$ means.

However, $\Delta_\mathrm{r}G$ does indeed, somewhat, refer to the instantaneous difference between the "molar Gibbs free energies" of the products and reactants. This is different from $\Delta G$ in three main ways.

First, $\Delta G$ is the difference between the Gibbs energy of the entire system at two points in time. Here, we are interpreting $\Delta_\mathrm{r}G$ as the difference between the "molar Gibbs free energies" of two components of the system: reactants and products. In other words, the system generally contains both reactants and products, and $\Delta_\mathrm{r}G$ may be thought of as the difference between the Gibbs energies of the product part of the system, and the reactant part of the system, even though you cannot separate them in the laboratory.

Secondly, it is the difference between the chemical potentials, not the molar Gibbs free energies. The molar Gibbs free energy is simply defined by $G_i/n_i$; the chemical potential is a partial derivative and is defined by $\mu_i = (\partial G/\partial n_i)_{T,p,n_j}$. In the case where there are no other species present (i.e. species $i$ is pure), then the chemical potential is identical to the molar Gibbs free energy.

Lastly, it is weighted by the stoichiometric coefficients $\nu_i$ instead of the change in the amount $\Delta n_i$. The stoichiometric coefficient is a dimensionless quantity, which is negative for reactants and positive for products. So, in the ammonium nitrate decomposition as written at the very top, we have

$$\nu_{\ce{NH4NO3}} = -1; \qquad \nu_{\ce{N2O}} = +1; \qquad \nu_{\ce{H2O}} = +2$$

and our expression for $\Delta_\mathrm{r} G$ is

$$\begin{align} \Delta_\mathrm{r} G &= \sum_i \nu_i \mu_i \\ &= \nu_{\ce{N2O}}\mu_{\ce{N2O}} + \nu_{\ce{H2O}}\mu_{\ce{H2O}} - \nu_{\ce{NH4NO3}}\mu_{\ce{NH4NO3}} \\ &= \mu_{\ce{N2O}} + 2\mu_{\ce{H2O}} - \mu_{\ce{NH4NO3}} \end{align}$$

Now, note the units again. The chemical potential is a partial derivative of the Gibbs free energy (units $\mathrm{kJ}$) with respect to the amount of $i$ (units $\mathrm{mol}$), and so it must have units $\mathrm{kJ~mol^{-1}}$. And from our above expression, $\Delta_\mathrm{r}G$ must also have units of $\mathbf{kJ~mol^{-1}}$, since the stoichiometric coefficients are dimensionless.

Does it matter if you start with $1~\mathrm{mol}$ or $2~\mathrm{mol}$ of ammonium nitrate? The answer is now, no. The amount of starting material does not affect $\mu_i$, nor does it affect $\nu_i$. Therefore, $\Delta_\mathrm{r}G$ is independent of the amount of starting material.

Why is $\Delta_\mathrm{r} G$ an instantaneous difference between the chemical potentials? Well, that is because the chemical potentials of the reactants and products, $\mu_i$, are changing continuously as the reaction occurs. Therefore, if we want to calculate $\Delta_\mathrm{r} G$, we have to take a "snapshot" of the reaction vessel: otherwise it makes absolutely no sense to speak of $\mu_i$ because we wouldn't know which value of $\mu_i$ to use.

Compare this with $\Delta G$ above: the quantities of $G_{\mathrm{m},i}$ are constants that do not vary depending on the extent of reaction. Therefore, we do not need to specify a particular extent of reaction to calculate $\Delta G$.

A diagram to sum up

Diagram

  • Note that, to adequately define what $\Delta G$ is, you need a starting point and an ending point. I showed two possibilities for $\Delta G$; there are infinitely many more.
  • Let's say the reaction goes to completion. If you double your starting material and double your product, the difference between $G_\text{products}$ and $G_\text{reactants}$ will also be doubled. So, it's important to specify!
  • Likewise, to adequately define $\Delta_\mathrm{r} G$, you have to define the single specific point at which you intend to calculate $\Delta_\mathrm{r} G$.
  • If you double your starting material and double your product, the curve is stretched by a factor of 2 along the y-axis, but it is also stretched by a factor of 2 along the x-axis because $\xi$ is also doubled. (For a mathematical explanation see the definition of $\xi$ given in the linked question earlier.) So, while the difference $\Delta G$ is doubled, the gradient $\Delta_\mathrm{r} G$ remains unchanged.
  • The units should also be clear from this. Since $\Delta G$ is a difference between two values of $G$, it has to have units of $\mathrm{kJ}$. On the other hand, $\Delta_\mathrm{r}G$ is a gradient and therefore has units of $\mathrm{kJ/mol}$.

A caveat

Unfortunately, the notation $\Delta G$ is often loosely used and treated as being synonymous with $\Delta_\mathrm{r}G$. You will therefore see people give $\Delta G$ units of $\mathrm{kJ~mol^{-1}}$. For more information refer to Levine, Physical Chemistry 6th ed., p 343. I would personally recommend making a distinction between the two.

orthocresol
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    "To adequately define $\Delta_\mathrm{r} G$, you have to define the single specific point at which you intend to calculate $\Delta_\mathrm{r} G$".

    Consider following definition.

    "The standard Gibbs energy (change) $\Delta_\mathrm{r} G^\circ$ for a chemical reaction is the change in G for converting stoichiometric numbers of moles of the separated pure reactants, each in its standard state at T, into the separated pure products in their standard states at T."

    As $\Delta_\mathrm{r} G^\circ$ is special case of $\Delta_\mathrm{r} G^\circ$ when $Q=1$, then why it is defined as above?

     – CLAP or SLAP II May 03 '16 at 12:42
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    Please don't edit the question. I have rolled back the changes. Anyway, let me use an analogy. If you have a function $y = x^2$, $\mathrm{d}y/\mathrm{d}x = 2x$. The value of the derivative $\mathrm{d}y/\mathrm{d}x$ varies with $x$ and therefore we cannot assign it a specific numerical value, unless we specify one particular value of $x$ that we want to evaluate the derivative at. Likewise, we cannot find a numerical value for $\Delta_\mathrm{r}G$ unless we specify the point at which we want it to be evaluated. – orthocresol May 03 '16 at 12:50
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    For $\Delta_\mathrm{r}G^\circ$, the additional condition that the $\circ$ imposes is that the particular point we pick must correspond to $Q = 1$. For $\Delta_\mathrm{r} G$, we have to pick a point, but it does not have to be the point corresponding to $Q = 1$. – orthocresol May 03 '16 at 12:52
  • Brilliant Answer by the way. Exactly what i was looking for. Thank you very much, Sir. – CLAP or SLAP II May 03 '16 at 12:52
  • I read in my book that

    The first term on the right in the equality is the standard reaction Gibbs energy, $\Delta_\mathrm{r} G^\circ$ : $\Delta_\mathrm{r} G^\circ$= {cμC + dμD} − {aμA + bμ B } Because the standard states refer to the pure materials, the standard chemical potentials in this expression are the standard molar Gibbs energies of the (pure) species. Therefore, eqn is the same as $\Delta_\mathrm{r} G^\circ$ = {cGm(C) + dGm(D)} − {aGm(A) + bGm(B)}

    Is this why we can define $\Delta_\mathrm{r} G^\circ$ like difference in Standard Molar Gibbs Energies?

     – CLAP or SLAP II May 03 '16 at 12:58
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    Yes, that is correct. Because $\mu_i^\circ = G_{\mathrm{m},i}$ you can replace the chemical potentials with molar Gibbs energies in the expression for $\Delta_\mathrm{r} G^\circ$. – orthocresol May 03 '16 at 12:58
  • Last question So We can define $\Delta_\mathrm{r} G^\circ$ for a process ( converting stoichiometric numbers of moles of the separated pure reactants, each in its standard state at T, into the separated pure products in their standard states at T) and it made sense But we cannot do so for $\Delta_\mathrm{r} G$ and it has a physical meaning only at a instant? – CLAP or SLAP II May 03 '16 at 13:13
  • I would hesitate to say that that is a "process", because you are not converting 1 mol of something to another, you are converting 1 of something to another ($\nu_i$ being dimensionless). But I get what you mean. If you want to look at it that way, then $\Delta_\mathrm{r} G$ at a particular instant can also be thought of as a process of converting the reactants, at the particular state at that instant, into the products, at the particular state at that instant. This state needs to be defined by multiple parameters, including $T$, $p$, and all $n_i$'s of species present. – orthocresol May 03 '16 at 13:19
  • (...) So, when we add the standard state symbol in $\Delta_\mathrm{r} G^\circ$, all that changes is that "that instant" corresponds to the time where $Q = 1$, and the "particular states" correspond to standard states. – orthocresol May 03 '16 at 13:20
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    "Then ΔrG at a particular instant can also be thought of as a process of converting the reactants, at the particular state at that instant, into the products, at the particular state at that instant" Can you give an example? – CLAP or SLAP II May 03 '16 at 13:26
  • Using the Haber process example above, if the reaction proceeds to equilibrium, there will be 90 mol N2, 270 mol H2, and 20 mol NH3. The "state" of N2 is therefore completely specified by: temperature 800 K, pressure 200 bar, amount of N2 present 90 mol, amount of H2 present 270 mol, and amount of NH3 present 20 mol. Changing any one of these variables will affect the value of the chemical potential; the state is specified by the variables $T$, $p$, $n_i$, $n_j$, $n_k$, $\cdots$ – orthocresol May 03 '16 at 13:41
  • Compare this to our "standard state" at 800 K (standard state does not specify temperature, so it must be additionally specified): temperature 800 K, pressure 1 bar, amount of N2 present does not matter, amount of H2 present 0 mol, and amount of NH3 present 0 mol (since N2 must be pure). The amount of N2 present does not matter because you divide by it anyway (whereas in the earlier case it matters because it is not a division, but a partial derivative). – orthocresol May 03 '16 at 13:43
  • Sir, I have understood everything. But this thing that "ΔrG at a particular instant can also be thought of as a process of converting the reactants, at the particular state at that instant, into the products, at the particular state at that instant" is not getting into my mind. I tried reading books but it was not explained. I have read your other answer couple of time but i am still having this as doubt. Can you please add a detailed explanation to this in this answer? It would be really helpful. Why it can be thought so? – CLAP or SLAP II May 03 '16 at 16:30
  • In my books they have written $\Delta_\mathrm{r} G^\circ$ as the energy which can be extracted from system but I think that should be $\Delta G$ and not $\Delta_\mathrm{r} G^\circ$. Because as you wrote $\Delta G$ represents change in Gibbs Free Energy for a process and not Reaction Gibbs Energy. For example see this http://i.imgur.com/UEqrLYg.png Why $\Delta_\mathrm{r} G^\circ$ is for a reaction and not for an instant? – CLAP or SLAP II May 03 '16 at 16:41
  • @CLAPorSLAPII Come over here: http://chat.stackexchange.com/rooms/39235/thermodynamics – orthocresol May 03 '16 at 16:43
  • @orthocresol In the formula for calculating equilibrium constant $$K=e^{\frac{-\Delta G}{RT}}$$ what $\Delta G$ represents? – ado sar Mar 24 '20 at 19:00
  • @adosar that is not correct: it should be $\Delta_\mathrm r G^\circ$. Standard Gibbs energy of reaction. see https://chemistry.stackexchange.com/q/41862/16683 – orthocresol Mar 24 '20 at 19:09
  • @But if it is $\Delta_\mathrm r G^\circ$ then what it represents? A derivative with respect to the extent of reaction? I have always related the above equation to stability. So it must turn out that this stability factor is hidden in $\Delta_\mathrm r G^\circ$. I am completely lost. Does there exist a free book for chemical thermodynamics? – ado sar Mar 24 '20 at 19:19
  • @adosar I don't have much left to say. What I wanted to say has already been said in this post and the one in my last comment. Yes it is a derivative with respect to $\xi$. When you want to get a derivative, you must also specify a point at which you want to evaluate it. The $^\circ$ means that you evaluate this at the point where each species is in their standard state. If this point is towards the left of the minimum (see graph above), then the slope $\Delta_\mathrm r G^\circ < 0$, $K > 0$, products are favoured, which makes sense because the minimum (equilibrium) lies to the right. – orthocresol Mar 24 '20 at 19:41
  • @adosar Of course it is about stability; equilibrium is all about stability, but simply saying that doesn't give us any insight as to the problem. It is a big word that has no meaning on its own. The crux of the matter is that there are two points on the graph that are of interest. (1) the standard-state where activities are equal to unity. (2) the equilibrium, which is the minimum, i.e. lowest $G$. If 1 lies to the left of 2, then $K > 0$ as discussed previously. If 1 lies to the right of 2, then $K < 0$. In both cases, 2 is more stable than 1! – orthocresol Mar 24 '20 at 19:49
  • @orthocresol Appreciated the answer. Thanks. – ado sar Mar 25 '20 at 13:05
  • @orthocresol But how we can calculate ΔG for a sytem if it is not in equilibrium? All the thermodynamic variables must be measured at equilibrium. How can we measure arbitrarily two G energies at non-equlibrium states? – ado sar Oct 17 '20 at 09:56
  • @CLAPorSLAPII "The standard Gibbs energy (change) ΔrG∘ for a chemical reaction is the change in G for converting stoichiometric numbers of moles of the separated pure reactants, each in its standard state at T, into the separated pure products in their standard states at T." How we can convert all reactants into products if the reaction will reach equilibrium? – ado sar Oct 17 '20 at 09:59
  • @orthocresol What an excellent answer. As a new thermodynamics student, I'm having trouble due to lousy definitions. I really appreciate that you go into minute detail and explain things from the ground up. Thanks! – Boson Aug 05 '22 at 20:24
  • @orthocresol Just one doubt-in the expression for $ \Delta_r G$-can I take the chemical potential=$G_m$ of the reactant/product at that particular time if the reaction is at constant temperature and/or constant pressure? – Boson Aug 05 '22 at 20:31
  • @Boson I'm not sure. I'd probably be inclined to say no, because $\mu$ is like $dy/dx$ and $G_m$ is like $y/x$ -- if $y = f(x)$ then the value of $f(x)/x$ is not necessarily equal to the value of $f'(x)$. But I don't know, I'm not really very good at thermodynamics. – orthocresol Aug 05 '22 at 20:38
  • @orthocresol In an earlier comment to this you said "Yes, that is correct. Because $∘=m,i$, you can replace the chemical potentials with molar Gibbs energies in the expression for $Δr∘$ Could this be extended to non-standard state conditions but with constant T and P so it can be used for $Δr$? – Boson Aug 05 '22 at 20:42
  • @orthocresol Btw, I think you are actually very good at thermodynamics, at least for undergrads. I've read several of your other answers on thermodynamics and they always strike me as both elaborate and understandable, which is really helpful in thermodynamics which is quite confusing to me due to lousy definitions. On introspection, my question is actually for a rather niche case and not really important. Thanks again! – Boson Aug 05 '22 at 20:49
  • I think the fact that I contradicted myself suggests that I don't really understand things ;) Well, yes, chemical potentials aren't associated with standard states, they are defined for any system so that shouldn't be an issue. – orthocresol Aug 05 '22 at 21:21
  • @orthocresol 1 more question-if the reaction is at constant T and P, and we take $=$, then in the expression $Δ_r=∑_G{m,}$ would $Δr$ be constant throughout the reaction because $$ is obviously constant and $G_{m,}$ is also constant since T and P is constant(all above for ideal gases)? – Boson Aug 06 '22 at 08:17
  • @orthocresol Also, is $ \Delta_r G$ analogous to $ \Delta_r H$? If yes, then why is there no mention of extent of reaction here ? – Boson Aug 06 '22 at 11:16
  • @Boson $G$ is constant at constant $T$ and $p$ if there is no reaction occurring. You are probably thinking of the equation $\mathrm{d}G = -S\mathrm{d}T + V\mathrm{d}p$ but this is only true for a system which is not undergoing chemical change. In general one has $\mathrm{d}G = \mu \mathrm{d}n - S\mathrm{d}T + V\mathrm{d}p$ (in which case, you can see that the no-reaction case simply has constant $n$, so reduces to the original). I'm not entirely sure about the link to enthalpy. Sorry, but I'm going to stop here; if you have further questions I gently urge you to post a new one. – orthocresol Aug 06 '22 at 11:21
  • @orthocresol Could we continue in chat? – Boson Aug 06 '22 at 11:26
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Let's see how this plays out for the case of a batch reactor in which a reversible reaction of an ideal gas is occurring at constant temperature and total pressure: aA + bB --> cC + dD. Let the initial concentrations of the four species in the reactor be $n_A,\ n_B,\ n_C,\ and\ n_D$. At time t, the concentrations of these 4 species are: $$n_A=n_{A0}-a\xi $$ $$n_A=n_{B0}-b\xi $$ $$n_A=n_{C0}+c\xi $$ $$n_A=n_{A0}+d\xi $$where $\xi$ is the "extent of reaction" at time t. If we are willing to make the assumption that the free energy of the system can be precisely determined at any time during the process (even though the process is irreversible/spontaneous), we can express the free energy at time t during the reaction by:$$G=n_A\mu_A+n_B\mu_B+n_C\mu_C+n_D\mu_D$$where $\mu$'s are the chemical potentials of the four species in the reactor mixture. For an ideal gas mixture, the chemical potential of the 4 species in the mixture are given by: $$\mu_A=\mu_A^0(T)+RT\ln{p_A}=\mu_A^0(T)+RT\ln{(Px_A)}$$ $$\mu_B=\mu_B^0(T)+RT\ln{p_B}=\mu_B^0(T)+RT\ln{(Px_B)}$$ $$\mu_C=\mu_C^0(T)+RT\ln{p_C}=\mu_C^0(T)+RT\ln{(Px_C)}$$ $$\mu_D=\mu_D^0(T)+RT\ln{p_D}=\mu_D^0(T)+RT\ln{(Px_D)}$$where the $\mu^0$'s are the molar free energies of the pure components at temperature T and a pressure of 1 Bar, the p's are the partial pressures of the various species in the reaction mixture at time t, P is the total pressure in the reactor (constant), and the x's are the mole fractions at time t. For example: $$x_A=\frac{n_A}{n_A+n_B+n_C+n_D}$$ If we focus on the change in free energy G between time t and time t + dt, we have:$$dG=(dn_A\mu_A+dn_B\mu_B+dn_C\mu_C+dn_D\mu_D)+(n_Ad\mu_A+n_Bd\mu_B+n_Cd\mu_C+n_Dd\mu_D)$$ If we make use of the various relationships presented above to evaluate the two terms in parenthesis in the above equation, we obtain the following results:$$(n_Ad\mu_A+n_Bd\mu_B+n_Cd\mu_C+n_Dd\mu_D)=0$$ $$(dn_A\mu_A+dn_B\mu_B+dn_C\mu_C+dn_D\mu_D)=(c\mu_C^0+d\mu_D^0-a\mu_A^0-b\mu_B^0+RT\ln{Q})d\xi$$where $$Q=\frac{(p_C)^c(p_D)^d}{(p_A)^a(p_B)^b}$$ So, we are left with :$$dG=(c\mu_C^0+d\mu_D^0-a\mu_A^0-b\mu_B^0+RT\ln{}Q)d\xi$$ The first four terms on the right hand side are usually exprfessed as $$c\mu_C^0+d\mu_D^0-a\mu_A^0-b\mu_B^0=\Delta G^0$$where $\Delta G^0$ can be seen to represent physically the change in free energy in going between the following two thermodynamic equilibrium states;

State 1: a moles of pure A and b moles of pure B, each in its own separate container at temperature T and pressure 1 bar

State 2: c moles of pure C and d moles of pure D, each in its own separate container at temperature T and pressure 1 bar

$\Delta G^0$ is typically referred to as the "standard free energy change for the reaction." So, in terms of $\Delta G^0$, we end up with:$$\frac{dG}{d\xi}=\Delta G^0+RT\ln{Q}$$ This equation shows that, on a graph of G versus $\xi$, the right hand side represents the slope. The slope becomes equal to zero, and equilibrium prevails, when the reaction has proceeded to the extent that Q = K, where K is the equilibrium constant.

Chet Miller
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In this context $\Delta G$ and $\Delta_\mathrm{r} G$ are different notations for two similar things, both are changes in Gibbs Free Energy. The "r" means, that you can only interpret it for a very specific equation.

szentsas
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