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Standard state conditions for standard Gibbs free energy change stipulate (among other conditions) that the partial pressure of each gas is $\pu{1 bar},$ e.g. for the reaction

$$\ce{2 A(g) + 3 B(g) <=> 4 C(g) + 5 D(g)}$$

for standard Gibbs free energy change calculations the partial pressures are equal:

$$p(\ce{A}) = p(\ce{B}) = p(\ce{C}) = p(\ce{D}) = \pu{1 bar}.$$

If we are calculating the standard enthalpy change for this reaction, are the partial pressures still $\pu{1 bar}$ for every gas? Or is the external pressure is constant and is $\pu{1 bar}?$

Because if the external pressure is $\pu{1 bar},$ then the partial pressure for every gas can't be $\pu{1 bar}$: if $\pu{2 mol}$ of $\ce{A}$ are at $\pu{1 bar},$ then $\pu{3 mol}$ of $\ce{B}$ will have a higher pressure if they are all present in the same closed container.

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    $\begingroup$ It isn't the partial pressure. In the standard state, the species are taken to be pure, each at a total pressure of 1 atm. They are not mixed in the standard state. $\endgroup$ Sep 26, 2019 at 14:21
  • $\begingroup$ Or you refer to a mixture in standard state, in principle one could do that, and choose 1 bar as pressure, for instance. But mixture composition would be then a parameter, too. $\endgroup$
    – Alchimista
    Sep 26, 2019 at 14:40
  • $\begingroup$ See chemistry.stackexchange.com/questions/114673/… $\endgroup$
    – Karsten
    Sep 26, 2019 at 15:46
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    $\begingroup$ @Alchemista The term “standard state” for a gas implies pure species. $\endgroup$ Sep 26, 2019 at 16:05
  • $\begingroup$ External pressure is not relevant in a closed container. $\endgroup$
    – jimchmst
    Aug 26, 2023 at 22:42

2 Answers 2

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For this reaction, the standard state is a hypothetical one: All (gas) components are present as pure at standard pressure. Of course, you can create that state, but it would not react because each species is in its separate container.

You can still determine the standard Gibbs energy by extrapolation. The easiest way is to let a gas mixture come to equilibrium (so the Gibbs energy is zero), measure the concentrations, assume an ideal gas, and plug them into the equation below:

$$\Delta_\mathrm rG^⦵ = -RT\ln K \tag{1}$$

This may be derived from

$$\Delta_\mathrm rG = \Delta_\mathrm rG^⦵ + RT\ln Q \tag{2}$$

which is the "recipe" to calculate the Gibbs energy of reaction at any state from that of the standard state (assuming ideal gas or solution), combined with the fact that at equilibrium, $\Delta_\mathrm rG$ is equal to zero and $Q = K$.

[comment in the linked question] how can gases start out as pure component, they have to mix for the reaction to occur? Also, how can each of the products and reactants have 1 bar pressure, the pressure would depend on the no. Of moles of the different reactants and products(which are not the same for all reactions)?

The Gibbs energy of reaction is a state function. You can start with the standard state, mix the gases, let the reaction proceed for a while, and then separate the gases again. For each step, you have to measure the change in Gibbs energy, or calculate it. Then you add them all up to get the somewhat surprising net reaction of pure, separated substances reacting to make a bit more of the product, yielding pure, separated substances again. Curiously, you never actually have to do this experiment. The Gibbs energy of mixing is baked into the formulas above.

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The standard Free energy is the free energy change from reactants and product each in their standard state and the activities at equilibrium. Mix them all up introduce a catalyst or initiator and measure the energy change. Alternatively establish equilibrium, measure the activities of products and reactants and determine Keq. Delta G[0] = -RTlnKeq. Again the energy difference between the states of the reaction quotient equal to the equilibrium concentrations and the state where all activities are 1 [ln 1= 0]. For gases the pressure[fugacity] is equal to 1 for each gas. the total pressure [fugacity] is the sum of the pressures of each gas [volume is constant].

There is a subtle problem with gases that I do not understand. Upon mixing there is an entropy [and possibly small energy] of mixing and the mole fractions are no longer unity. This can be removed by establishing an electrochemical cell such as in the electrolysis of water. The O2 and H2 can be separated physically, connected electrically, each at activities of 1 and mole fractions of 1 in contact with products, activity =1. Now the standard potential can be measured and the equilibrium constant calculated. [taking into account the differences in the anode and cathode chambers]

Standard conditions may or may not be physically attainable. The energies can be calculated from equilibrium data, electrode potentials, and various Born-Haber cycles using cascaded reactions.

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