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The standard free energy change for equilibrium problems is relative to both reactants AND products starting at 1 molar.

Hess’s law takes the sum of the values for products minus the sum of the values for reactants. My interpretation is that the result of such a calculation is for changing from a state of all reactants to all products. So even if I use standard values the initial state does not involve 1 molar for BOTH products and reactants but instead involves an initial state of only reactants.therefore I can not equate $\Delta G = \Delta H-T \Delta S$ to standard $\Delta G= RT \ln K$ if the enthalpy I calculated came from using Hess' law. It would be like comparing apples to oranges. Is my logic correct?

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The molar Gibbs free energy change of reaction refers to $\Delta G$ for conversion of a stoichiometric amount of reagents to products for one mole of reaction as written. For a reaction $\ce{aA + bB +...->pP +qQ +...}$ this means $a$ moles of $\ce{A}$ react with $b$ moles of $\ce{B}$ and so on to form $p$ moles of $\ce{P}$, etc, with reagents and products at the same T and p. An underscore "m", as in $\Delta_m G$, is used to indicate a molar change. An "r" subscript, as in $\Delta_r G$, is also commonly used. When the reaction is carried out under standard conditions, a circle or Plimsoll symbol, as in $\Delta_r G^\circ$, indicates standard states for reactants and products*.

It is correct to write

$$\Delta_r G^\circ = \Delta_r H^\circ -T \Delta_r S^\circ \\ = -RT \ln K \tag1$$

which follows from

$$\Delta G = \Delta H -T \Delta S \tag2$$

and

$$\Delta_r G =\Delta_r G^\circ + RT \ln Q \tag3$$

and $Q=K$ at equilibrium, when $\Delta_r G=0$.

In going from Eqs 2 and 3 to Eq 1 you are in fact also applying Hess' law by equating the result of two particular cycles between the reactants and products, as illustrated in the following figure: enter image description here One converts reactants to products in their standard states directly (Eq 2), while the other (derived from Eq 3) takes the reactants in their standard states to the equilibrium state, at which conversion to products occurs with $\Delta_r G = 0$, and then equilibrium products are converted to their standard states (hat tip to jimchmst).


*The definition of $\Delta_r G^\circ$ assumes all products and all reactants are ideal, which implies that these are isolated (not mixed). The 1M standard state is typically hypothetical (an abstraction with thermodynamic properties determined through extrapolation from measurable states). The behavior of a real 1M solution is often non-ideal in the thermodynamic sense.

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  • $\begingroup$ The standard free energy change is the change from the condition where all involved substances, reactants AND products, are in their standard states to equilibrium activities. If reversed it is the free energy from equilibrium to the standard states. Nonstandard conditions are evaluated by calculating the appropriate reaction quotient. The last paragraph makes no sense, calculation of state functions is independent of the path. Reactions run to equilibrium not completion. $\endgroup$
    – jimchmst
    Commented Jan 1 at 0:24
  • $\begingroup$ ΔG=ΔG∘+RTlnQ ΔG∘=−nRTlnK when Q is 1 which occurs when both reactants and products are 1moles/L. Hess's law says ΔH∘f=ΔH∘f,products−ΔH∘f,reactant. and you can do that with gibbs free energy as well: ΔG∘f=ΔG∘f,products−ΔG∘f,reactant. However my interpretation of the math involved in Hess's law is that the reaction goes from a state of ONLY reactants(zero products) to ONLY products(zero reactants). So if I found ΔG∘f using Hess's law and then I wanted to find the equilibrium constant then I am saying it wouldn't make sense to equate it to −RTlnK which had products AND reactants start at 1M $\endgroup$
    – Alfred
    Commented Jan 1 at 1:25
  • $\begingroup$ @Alfred Standard values are useful because they deal with pure substances or dilute solutions where conditions are well defined. In a reaction flask, say at equilibrium, conditions (concentrations in particular) can differ significantly from std states. $\endgroup$
    – Buck Thorn
    Commented Jan 1 at 10:20
  • $\begingroup$ That is not what I said! Standard states are defined activities are 1.. The standard free energy change for a reaction is the free energy change from the condition of all involved components both products and reactants are in standard states to the condition of the equilibrium activities. Write out the equation and do the math: delta G -G[0] = RTlnKeq -RTlnQ. At equilbrium deltaG= O This is true when Q=1 and that is when both products and reactants have activity =1 $\endgroup$
    – jimchmst
    Commented Jan 2 at 19:11
  • $\begingroup$ @jimchmst Ok, I think I see what you were driving at. I included something along the lines of your comment into my answer. Hope it's clearer now. $\endgroup$
    – Buck Thorn
    Commented Jan 5 at 19:15

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