Chemistry Education: What is Thermodynamics of Chemical equilibrium?

Thermodynamics of Chemical Equilibrium:

The relation between the Gibbs energy and the equilibrium constant can be found by considering chemical potentials.

At constant temperature and pressure, the Gibbs free energy, G, for the reaction depends only on the extent of reaction: ξ (Greek letter xi), and can only decrease according to the second law of thermodynamics. It means that the derivative of G with ξ must be negative if the reaction happens; at the equilibrium the derivative being equal to zero.

$\left(\frac {dG}{d\xi}\right)_{T,p} = 0~$ : equilibrium

At constant temperature and volume, one must consider the Helmholtz free energy for the reaction: A; and at constant internal energy and volume, one must consider the entropy for the reaction: S.
In this article only the constant pressure case is considered. The constant volume case is important in geochemistry and atmospheric chemistry where pressure variations are significant. Note that, if reactants and products were in standard state (completely pure), then there would be no reversibility and no equilibrium. The mixing of the products and reactants contributes a large entropy (known as entropy of mixing) to states containing equal mixture of products and reactants. The combination of the standard Gibbs energy change and the Gibbs energy of mixing determines the equilibrium state.

In general an equilibrium system is defined by writing an equilibrium equation for the reaction

$\alpha A + \beta B \rightleftharpoons \sigma S + \tau T$

In order to meet the thermodynamic condition for equilibrium, the Gibbs energy must be stationary, meaning that the derivative of G with respect to the extent of reaction: ξ, must be zero. It can be shown that in this case, the sum of chemical potentials of the products is equal to the sum of those corresponding to the reactants. Therefore, the sum of the Gibbs energies of the reactants must be the equal to the sum of the Gibbs energies of the products.

$\alpha \mu_A + \beta \mu_B = \sigma \mu_S + \tau \mu_T \,$

where μ is in this case a partial molar Gibbs energy, a chemical potential. The chemical potential of a reagent A is a function of the activity, {A} of that reagent.

$\mu_A = \mu_{A}^{\ominus} + RT \ln\{A\} \,$ , ( $\mu_{A}^{\ominus}~$ is the standard chemical potential ).

Substituting expressions like this into the Gibbs energy equation:

$dG = Vdp-SdT+\sum_{i=1}^k \mu_i dN_i$ in the case of a closed system.

Now

$dN_i = \nu_i d\xi \,$ ( $\nu_i~$ corresponds to the Stoichiometric coefficient and $d\xi~$ is the differential of the extent of reaction ).

At constant pressure and temperature we obtain:

$\left(\frac {dG}{d\xi}\right)_{T,p} = \sum_{i=1}^k \mu_i \nu_i = \Delta_rG_{T,p}$ which corresponds to the Gibbs free energy change for the reaction .

This results in:

$\Delta_rG_{T,p} = \sigma \mu_{S} + \tau \mu_{T} - \alpha \mu_{A} - \beta \mu_{B} \,$ .

By substituting the chemical potentials:

$\Delta_rG_{T,p} = ( \sigma \mu_{S}^{\ominus} + \tau \mu_{T}^{\ominus} ) - ( \alpha \mu_{A}^{\ominus} + \beta \mu_{B}^{\ominus} ) + ( \sigma RT \ln\{S\} + \tau RT \ln\{T\} ) - ( \alpha RT \ln\{A\} + \beta RT \ln \{B\} )$ ,

the relationship becomes:

$\Delta_rG_{T,p}=\sum_{i=1}^k \mu_i^\ominus \nu_i + RT \ln \frac{\{S\}^\sigma \{T\}^\tau} {\{A\}^\alpha \{B\}^\beta}$

$\sum_{i=1}^k \mu_i^\ominus \nu_i = \Delta_rG^{\ominus}$ : which is the standard Gibbs energy change for the reaction. It is a constant at a given temperature, which can be calculated, using thermodynamical tables.

$RT \ln \frac{\{S\}^\sigma \{T\}^\tau} {\{A\}^\alpha \{B\}^\beta} = RT \ln Q_r$

( $Q_r ~$ is the reaction quotient when the system is not at equilibrium ).

Therefore

$\left(\frac {dG}{d\xi}\right)_{T,p} = \Delta_rG_{T,p}= \Delta_rG^{\ominus} + RT \ln Q_r$

At equilibrium $\left(\frac {dG}{d\xi}\right)_{T,p} = \Delta_rG_{T,p} = 0$

$Q_r = K_{eq}~$ ; the reaction quotient becomes equal to the equilibrium constant.

leading to:

$0 = \Delta_rG^{\ominus} + RT \ln K_{eq}$

and

$\Delta_rG^{\ominus} = -RT \ln K_{eq}$

Obtaining the value of the standard Gibbs energy change, allows the calculation of the equilibrium constant

Addition of reactants or products

For a reactional system at equilibrium: $Q_r = K_{eq}~$ ; $\xi = \xi_{eq}~$ .

If are modified activities of constituents, the value of the reaction quotient changes and becomes different from the equilibrium constant: $Q_r \neq K_{eq}~$

$\left(\frac {dG}{d\xi}\right)_{T,p} = \Delta_rG^{\ominus} + RT \ln Q_r~$
and
$\Delta_rG^{\ominus} = - RT \ln K_{eq}~$
then
$\left(\frac {dG}{d\xi}\right)_{T,p} = RT \ln \left(\frac {Q_r}{K_{eq}}\right)~$

If activity of a reagent $i~$ increases

$Q_r = \frac{\prod (a_j)^{\nu_j}}{\prod(a_i)^{\nu_i}}~$ , the reaction quotient decreases.

then

$Q_r < K_{eq}~$ and $\left(\frac {dG}{d\xi}\right)_{T,p} <0~$ : The reaction will shift to the right (i.e. in the forward direction, and thus more products will form).

If activity of a product $j~$ increases

then

$Q_r > K_{eq}~$ and $\left(\frac {dG}{d\xi}\right)_{T,p} >0~$ : The reaction will shift to the left (i.e. in the reverse direction, and thus less products will form).
Note that activities and equilibrium constants are dimensionless numbers.

Chemistry Education

Wednesday 29 January 2014

What is Thermodynamics of Chemical equilibrium?

Thermodynamics of Chemical Equilibrium:

Addition of reactants or products

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