**Since this post was originally written in January 2012, the AP exam has changed. One of the changes was to remove equation #2 below from the equations & constants sheet. As such, I think that knowledge of it, and the consequences associated with it, are unlikely to be tested quantitatively on the exam in the future, but nevertheless, I still feel that understanding that conditions other than standard ones will cause ∆G to take on new values is a useful reference point.
Since K is the equilibrium constant, we are at equilibrium, the amounts of products and reactants in the mixture are fixed, and the sign of ∆G° can be thought of as a guide to the ratio of the amount of products to the amount of reactants at equilibrium and therefore the thermodynamic favorability of the reaction.
If it so happens that products and reactants are equally favored at equilibrium, then ∆G° is zero, BUT ∆G° is not *necessarily* ZERO at equilibrium.
Since Q is NOT the K, and we are NOT necessarily at the equilibrium position, the sign of ∆G can be thought of as a predictor about which way the reaction (that has reactants and products defined by Q), will go.
If ∆G° is negative at equilibrium, then we will have lots of products at equilibrium, meaning Q needs to be bigger (greater than 1) to approach K. As Q gets larger (i.e., as we get more products), the term ‘RT ln Q’ gets increasingly positive, and eventually adding that term to a negative ∆G°, will make ∆G = 0, equilibrium will be established and no further change occurs.
It is possible that Q could already be too large and therefore ∆G is positive. IF so, then the reaction will need to from more reactants, reduce the value of Q, and allow ∆G to reach zero, i.e., allow equilibrium to be established.
If ∆G° is positive at equilibrium, then we will have lots of reactants at equilibrium, meaning Q needs to be smaller (less than 1) to approach K. As Q gets smaller (i.e., as we get more reactants), the term ‘RT ln Q’ gets increasingly negative, and eventually adding that term to a positive ∆G°, will make ∆G = 0, equilibrium will be established and no further change occurs.
It is possible that Q could already be too small and therefore ∆G is negative, IF so, then the reaction will need more products, increase the value of Q, and allow ∆G to reach zero, i.e., allow equilibrium to be established.
In short, it is ∆G (NOT ∆G°) that will be zero at equilibrium and the sign of it (generated by the combination of ∆G° and RT ln Q in Equation #2), will define which way the reaction proceeds.
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The prime usually denotes a standard free energy that corresponds to an apparent equilibrium constant where the concentration (or activity) of one or more constituents is held constant.
For example, for $ce{HA <=>A- + H+}$ the equilibrium constant is $K=frac{[ce{A-}][ce{H+}]}{[ce{HA}]}$ and the corresponding standard free energy change is $Delta G^circ=-RT ln(K)$. If you know the value of $K$ and the start concentration of $ce{HA}$ then you can compute the equilibrium concentrations of $ce{HA}$, $ce{A-}$, and $ce{H+}$.
However, if the pH is held constant then $[ce{H+}]$ is no longer a free variable and the apparent equilibrium constant is $K=frac{[ce{A-}]}{[ce{HA}]}$ and the corresponding standard free energy change is $Delta G^circ=-RT ln(K)$. So $Delta G^circ=-RT ln(K/[ce{H+}])=Delta G^circ+RTln[ce{H+}]$
In biochemistry there is often several important constituents in addition to $ce{H+}$ that are held constant such as $ce{Mg++}$, phosphate, etc.
One form of the fundamental equation of thermodynamics is:
$$dU = TdS – P dV + sum_{i}mu_i dN_i$$
In this equation, the total internal energy has canonical variables $V$, $S$, and $dN_i$, where $S$ is the total entropy (in units of $frac{mathrm J}{mathrm K}$), $V$ is the total volume, and $N_i$ is the number of moles of each molecular species present. $T$ is temperature; $P$ is pressure, and $mu_i$ is the chemical potential of species $i$. The equation implies that if we were to know an equation that gave $U$ as a function of $S$, $V$, and all the $N_i$ we would know everything about the system. However, this is inconvenient for two reasons. First, $S$ and $V$ are extensive variables. Make the system bigger without changing its composition, and $S$ and $V$ increase. Second and more importantly, it is often difficult to hold $S$ constant when doing experiments. The same is true of $V$. (We live in a constant pressure atmosphere.)
Taking the Legendre transform of $U$ with respect to variables $S$ and $V$ gives a new fundamental equation:
$$dG = -S dT + V dP + sum_{i}mu_i dN_i$$
This equation means that if we knew a function that gave the Gibbs free energy as a function of $T$, $P$, and all the $N_i$, we could easily compute all thermodynamic properties of the system.
Say were interested in the thermodynamics of ATP hydrolysis:
$ce{ATP + H2O <=> ADP + Pi}$
This equation is really better written as
$ce{A-P3O10H3 + H2O -> A-P2O7H2 + H3PO4}$
But of course in a buffer at pH 7, the conditions where many biochemical reactions occur, there really wont be $ce{H3PO4}$ etc., there will be dissociation of protons $ce{H+}$ and formation of anions like $ce{H2PO4-}$ etc. So now all those reactions will have to be tracked too. The number of protons released by ATP is not the same as released by inorganic phosphate, and this is generally true. During a reaction, it is difficult to hold the number $N_{ce{H+}}$ of protons constant, but through judicious choice of buffers etc. it is possible to hold the chemical potential of protons constant (i.e. do experiments at constant pH). Under such conditions, it makes sense to continue the Legendre transformations one step further:
$$dG^prime = -S dT + V dP – N_{ce{H+}} dmu_{ce{H+}} + sum_{i eq ce{H+}}mu_i dN_i$$
$Delta G^{circ prime}$ is a Legendre transform of $Delta G^{circ}$ with respect to the number of protons in the system.
Robert Albertys paper from 1994 is a good place for further reading.
The prime has nothing to do with whether a concentration is held constant. The prime is used to indicate that some species, typically protons or ions such as Mg, are being set to a value other than the standard 1M concentration for use in a reference free energy. Although the prime has been adopted by some in the biochemical community and was endorsed in 1994 by IUPAC, the prime is not used in other communities for good reason – it is much better to explicitly state what concentrations are being used for the reference free energy. When using a reference free energy other than the standard free energy, concentrations can vary or be held fixed. In most biochemical situations, it is useful to keep the pH fixed since buffers are used. But this is a separate issue.
The biochemistry convention does not assume all solutions are 1 M. If you did this then you would have [H+] = 1 M. This simply never happens in biochemical reactions. Instead, we assume pH = 7. We also assume that water has an activity of 1 even though its concentration is 55 M.
Delta G naught means that the reaction is under standard conditions (25 celsius, 1 M concentraion of all reactants, and 1 atm pressure). Delta G naught prime means that the pH is 7 (physiologic conditions) everything else is the same. The concentration of [H+] now isnt 1 molar because 1 molar concentration would be an extremely low pH (0). Delta G naught prime is just like Delta G naught but for biology.
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What does it mean if Delta G naught is positive?
Which means that the *reverse* reaction is favored, which means the forward reaction isn’t favored. Thus, when K1, the forward reaction isn’t spontaneous, producing a positive delta G nought.
If it so happens that products and reactants are equally favored at equilibrium, then ∆G° is zero, BUT ∆G° is not *necessarily* ZERO at equilibrium.
If ∆G° is negative at equilibrium, then we will have lots of products at equilibrium, meaning Q needs to be bigger (greater than 1) to approach K. As Q gets larger (i.e., as we get more products), the term ‘RT ln Q’ gets increasingly positive, and eventually adding that term to a negative ∆G°, will make ∆G = 0, equilibrium will be established and no further change occurs.
Since Q is NOT the K, and we are NOT necessarily at the equilibrium position, the sign of ∆G can be thought of as a predictor about which way the reaction (that has reactants and products defined by Q), will go.
Since K is the equilibrium constant, we are at equilibrium, the amounts of products and reactants in the mixture are fixed, and the sign of ∆G° can be thought of as a guide to the ratio of the amount of products to the amount of reactants at equilibrium and therefore the thermodynamic favorability of the reaction.
In short, it is ∆G (NOT ∆G°) that will be zero at equilibrium and the sign of it (generated by the combination of ∆G° and RT ln Q in Equation #2), will define which way the reaction proceeds.
FAQ
What is the difference between Delta G and Delta G naught prime?
What does positive delta G naught mean?
How do you calculate delta G from Delta G naught prime?
What is Delta G naught equation?
