How do we define the thermodynamic equilibrium for a chemical reaction?

1 Answer
Oct 26, 2016

Answer:

Good question, I will make an attempt to explain.

Explanation:

We write the general equilibrium as:

#A+BrightleftharpoonsC+D#

As with any reaction there is a rate forward, #"rate forward"=k_f[A][B]#, and of course a rate backward, #"rate backward"=k_r[C][D]#.

Now, by definition, the chemical condition of equilibrium is defined when the forward and reverse rates are equal:

#"i.e." " rate forward "=" rate backward"#,

#"i.e."# #k_f[A][B]=k_r[C][D]#

And upon rearrangement,

#k_f/k_r=([C][D])/([A][B])#

And the quotient #k_f/k_r=K_c#, otherwise known as the equilibrium constant for the reaction. Generally #K_c# is a constant for a given temperature. #K_c# can be large (i.e. the products are favoured at equilibrium) or small (the reactants are favoured). #K_c# may be formally related to the thermodynamic properties of the reaction, though I am not going to do it here.

Now #K_c# must be measured, and as a constant it cannot be altered. However, a chemist or engineer can certainly manipulate the equilibrium. For instance, if we remove (somehow) the products of the reaction, #C# and #D#, the equilibrium will have to re-establish itself, and it does this by moving to the right as written to satisfy the equilibrium equation, and to re-establish equilibrium concentrations of #C# and #D#. On the other hand, if we pump more reactant into the equilibrium, the equilibrium will move in a forward direction to cope with increased #[A]# and #[B]#.

There should be many answers here that deal with equilibria. If there is a specific problem or query, ask, and someone will help you.