Question about the equation ΔG = ΔG° + RT ln(Q)

[galpinj]

Hey everyone,

 

Why do we include ΔG° in this equation? Why does the Gibbs Free energy at standard conditions matter when finding the Gibbs Free energy of a certain reaction? I know that it is important for comparison and can help to determine which way a reaction will go, but I do not see why it is a fundamental part of the equation. Why does the free energy of a reaction have to rely on the free energy from the standard form? That seems very strange to me, and you don't see that kind of dependence in ΔG = ΔH - TΔS. Is the equation ΔG = ΔG° + RT ln(Q) somehow derived from ΔG = ΔH - TΔS?

 

Thank you for any help!

Edited August 3, 2016 by galpinj

[studiot]

This is an important question so I am going to work through it at some length.

It is important to distinguish between the various Gs and G⁰ s and their deltas.

 

G on its own refers to the Gibbs Free Energy function, which is a function of state.

 

G⁰ is just the constant of integration, and is made specific to each reactant and reaction by choosing a stated condition.

 

The integration is not of ΔG = ΔH - TΔS as you thought but of either Maxwell's relation

 

[math]{\left( {\frac{{\partial G}}{{\partial P}}} \right)_T}[/math]

Which brings out the fact that this is developed for gaseous reactants at constant temperature

or of the equation

[math]dG = VdP - SdT[/math]

Since T is constant, SdT = 0 and we are left with the following

[math]{G_T} = \int {Vdp = nRT\int {\frac{{dP}}{P}} } = nRT{\log _e}\left( P \right) + {\rm{constant}}[/math]

The state function of temperature has a value at any T given by integrating the above

Working per mole n = 1 and we set the constant equal to G⁰

^Thus

[math]{\rm{Constant}} = {G^0}[/math]

^and

[math]G = RTLo{g_e}\left( P \right) + {G^0}[/math]

This is the general thermodynamic equation.

To apply it to a particular chemical reaction say

[math]A + B \to C + D[/math]

we calculate the change in G as ΔG by summing the values for the reactants products and subtracting the values for the products reactants

Edit oops !

[math]\Delta G = {G_C} + {G_D} - {G_A} - {G_B}[/math]

Each of these four components individually follows the general law so substituting this in

[math]\Delta G = \left( {RTLo{g_e}\left( {{P_C}} \right) + G_C^0} \right) + \left( {RTLo{g_e}\left( {{P_D}} \right) + G_D^0} \right) - \left( {RTLo{g_e}\left( {{P_A}} \right) + G_A^0} \right) - \left( {RTLo{g_e}\left( {{P_B}} \right) + G_B^0} \right)[/math]

Collecting terms and rearranging

[math]\Delta G = \left( {G_C^0 + G_D^0 - G_A^0 - G_B^0} \right) + \left( {RTLo{g_e}\left( {{P_C}} \right) + RTLo{g_e}\left( {{P_D}} \right) - RTLo{g_e}\left( {{P_A}} \right) - RTLo{g_e}\left( {{P_B}} \right)} \right)[/math]

This shows the various reactant and product contributions to [math]\Delta {G^0}[/math] in the first bracket

Doing some algebra on the logs yields

[math]\Delta G = \Delta {G^0} + \left( {RTLo{g_e}\left( {\frac{{{P_C}{P_D}}}{{{P_A}{P_B}}}} \right)} \right)[/math]

Defining the reaction coefficient, Q as

[math]Q = \left( {\frac{{{P_C}{P_D}}}{{{P_A}{P_B}}}} \right)[/math]

and substitution yields

[math]\Delta G = \Delta {G^0} + RTLo{g_e}\left( Q \right)[/math]

Which is your equation for this particular reaction.

Remember always that G on its own is general, but introducing G⁰ makes it specific to a particular reaction.

Edited August 3, 2016 by studiot

[studiot]

I see that you were online after I posted yesterday so presumably you saw my reply.

I note also that you are a biologist, although you have posted this in the Chemistry section so I if you have any questions arising, don't hesitate to ask.

 

Some further notes.

 

If the temperature is not constant then you also have to integrate the SdT term. This can be performed at constant pressure.

 

My analysis was for gases and the pressures are therefore partial pressures if there is more than one.

 

Pressures are a measure of (molar) concentration and the reaction quotient or coefficient, q, can also be described in these terms, which is good for non gaseous components.

 

You have used Q for the quotient, which is becoming common, when K used to be used.

There is a danger of confusing the q used in the First Law with Q which is (partly) why K was originally chosen.