If a reaction is reversible, when can it be said to have reached equilibrium?

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Answer 1 · 2017-04-30T14:10:51

When the rate of the forward reaction is EQUIVALENT to the rate of the reverse reaction.............

For the reaction,

$A(g)+B(g)rightleftharpoonsC(g) + D(g)$

There is certainly a $"forward rate"=k_f[A][B]$

And a backwards rate, $"reverse rate"=k_r[C][D]$ .

And by definition, $"chemical equilibrium"$ specifies NOT the cessation of chemical change, BUT $"EQUALITY of FORWARD"$ and $"REVERSE RATES."$

And thus at equilibrium, $k_f[A][B]=k_r[C][D]$ , AND

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

And, clearly, if the forward rate is FASTER than the reverse rate, products are favoured at equilibrium..........and vice versa.

The quotient, $k_f/k_r$ is better known as $K_"eq"$ , the thermodynamic equilibrium constant, which is a constant according to temperature.

And so.........#k_f/k_r=K_"eq"=([C][D])/([A][B])#, and this is an equation with which you will get VERY familiar.........