Re: chemical equilibrium

b201b402cd21c3@xxxxxxxxx wrote:
Consider the reaction:

A -> B + C

If the reactor already has substantial B and C in it, the reaction
will not proceed as far to the right as it would without B and C (Le
Chatelier).

What if the reactor has substantial C, but no B in it? Will the
reaction still proceed as though it had been devoid if both B and C?

It depends on "how substantial". Both activities of C and B will be in the numerator of the expression for the dG (they are multiplied). So either of them can drive dG to become positive and reaction will not occure:


For forward reaction
dG = R*T*ln(K)
where in your case

K = [B]*[C] / [A]

So if [B]*[C] gets larger than [A],
the K value under ln() will become larger than 1,
and so ln(K) will become positive. But for reaction
to proceed dG needs to be negative.

Of cause this is an oversimplification for constant pressure, temperature etc, and it does not consider the TdS factor.


See details here:
http://en.wikipedia.org/wiki/Chemical_equilibrium



B201B402CD21C3
.

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